Contents Chapter 0
Chapter 0 Review
1
Chapter 1
Functions and Their Graphs
Chapter 2
Classes of Functions
114
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Contents Chapter 0
Chapter 0 Review
1
Chapter 1
Functions and Their Graphs
Chapter 2
Classes of Functions
114
Chapter 3
The Limit of a Function
171
Chapter 4
The Derivative of a Function
206
Chapter 5
Applications: Graphing Functions; Optimization
285
Chapter 6
The Integral of a Function and Applications
385
Chapter 7
Other Applications and Extensions of the Integral
439
Chapter 8
Calculus of Functions of Two or More Variables
462
Appendix A Graphing Utilities
58
510
iii
Chapter 0 Review 0.1 Real Numbers 1. (a) Natural Numbers: 2, 5 (b) Integers: – 6, 2, 5 (c) Rational numbers: – 6,
1 , – 1.333…, 2, 5 2
(d) Irrational numbers: π 1 (e) Real numbers: – 6, , – 1.333…, π, 2, 5 2 3. (a) Natural Numbers: 1
(b) Integers: 0, 1 (c) Rational numbers: 0, 1,
1 1 1 , , 2 3 4
(d) Irrational numbers: none 1 1 1 (e) Real numbers: 0, 1, , , 2 3 4 5. (a) Natural Numbers: none (b) Integers: none (c) Rational numbers: none
(d) Irrational numbers: (e) Real numbers:
2 , π,
2 , π,
7. Number: 18.9526 Rounded: 18.953 Truncated: 18.952
2 + 1, π +
2 + 1, π +
1 2
1 2 9.
Number: 28.65319 Rounded: 28.653 Truncated: 28.653
11. Number: 0.06291 Rounded: 0.063 Truncated: 0.062
13.
Number: 9.9985 Rounded: 9.999 Truncated: 9.998
15.
17.
521 = 34.73333… 15 Rounded: 34.733 Truncated: 34.733
21.
x+2=3·4
3 = 0.428571… 7 Rounded: 0.429 Truncated: 0.428
Number:
19. 3 + 2 = 5
Number:
2
SECTION 0.1
23. 3y = 1 + 2
25.
x–2=6
27.
29.
9–4+2=5+2=7
31. – 6 + 4 · 3 = – 6 + 12 = 6
33.
4+5–8=9–8=1
35.
37.
6 – [3 · 5 + 2 · (3 – 2)] = 6 – [15 + 2 · 1] = 6 – [15 + 2] = 6 – 17 = – 11
39. 2 · (3 – 5) + 8 · 2 – 1 = 2 · (-2) + 16 – 1 = – 4 + 16 – 1 = 12 – 1 = 11
41.
10 – [6 – 2 · 2 + (8 – 3)] · 2 = 10 – [6 – 4 + 5] · 2 = 10 – [7] · 2 = 10 – 14 =–4
43.
45.
4 + 8 12 = =6 5−3 2
x =6 2
4+
1 12 1 13 + = = 3 3 3 3
(5 – 3)
1 1 = (2) = 1 2 2
47.
3 10 30 2 ⋅ = = 5 21 105 7
49.
6 10 2 ⋅ 3 ⋅ 2 ⋅ 5 4 ⋅ = = 25 27 5 ⋅ 5 ⋅ 3 ⋅ 9 45
51.
3 2 3 ⋅ 5 2 ⋅ 4 15 + 8 23 + = + = = 4 5 4⋅5 4⋅5 20 20
53.
5 9 5 ⋅ 5 9 ⋅ 6 25 + 54 79 + = + = = 6 5 6⋅5 5⋅6 30 30
55.
5 1 5⋅2 1 ⋅ 3 10 + 3 13 + = + = = 18 12 18 ⋅ 2 12 ⋅ 3 36 36
57.
1 7 1⋅ 3 7 ⋅5 − − = 30 18 30 ⋅ 3 18 ⋅ 5 3 − 35 − 32 16 = = =− 90 90 45
59.
3 2 3⋅3 2⋅4 9 −8 1 − = − = = 20 15 20 ⋅ 3 15 ⋅ 4 60 60
61.
5 18 = 5 ⋅ 27 = 5 ⋅ 9 ⋅ 3 = 15 11 18 11 9 ⋅ 2 ⋅ 11 22 27
63. 6(x + 4) = 6x + 24
65.
x(x – 4) = x 2 – 4x
67. (x + 2)(x + 4) = (x + 2)x + (x + 2)4 = x2 + 2x + 4x + 8 = x2 + 6x + 8
69.
(x – 2)(x + 1) = (x – 2)x + (x – 2)1 = x2 – 2x + x – 2 = x2 – x – 2
SECTION 0.2 71. (x – 8)(x – 2) = (x – 8)x + (x – 8)(-2) = x2 – 8x – 2x + 16 = x2 – 10x + 16
73.
(x + 2)(x – 2) = (x + 2)x + (x + 2)(-2) = x2 + 2x – 2x – 4 = x2 – 4
75. Answers will vary.
77.
Answers will vary.
79. Subtraction is not commutative. Examples will vary.
81.
Division is not commutative. Examples will vary.
83. Explanations will vary. 85. There are no real numbers that are both rational and irrational. There are no real numbers that are neither rational nor irrational. Explanations will vary. 87. 0.9999 … = 1 To show that 0.9999 … = 1, we let n = 0.9999 …, then 10n = 9.9999 … 10n = 9.9999 … (1) n = 0.9999 … (2) 9n = 9.0000 … Subtract (2) from (1). Divide both sides by 9. n=1
0.2 Algebra Review 1.
3.
1 >0 2
5.
–1>–2
7.
π > 3.14
9.
1 = 0.5 2
11.
2 < 0.67 3
13.
x>0
15.
x–1
23.
d(C, D) = |C – D| = |0 – 1| = | – 1| = 1
25.
d(D, E) = |D – E | = |1 – 3| = | – 2 | = 2
3
SECTION 0.2 71. (x – 8)(x – 2) = (x – 8)x + (x – 8)(-2) = x2 – 8x – 2x + 16 = x2 – 10x + 16
73.
(x + 2)(x – 2) = (x + 2)x + (x + 2)(-2) = x2 + 2x – 2x – 4 = x2 – 4
75. Answers will vary.
77.
Answers will vary.
79. Subtraction is not commutative. Examples will vary.
81.
Division is not commutative. Examples will vary.
83. Explanations will vary. 85. There are no real numbers that are both rational and irrational. There are no real numbers that are neither rational nor irrational. Explanations will vary. 87. 0.9999 … = 1 To show that 0.9999 … = 1, we let n = 0.9999 …, then 10n = 9.9999 … 10n = 9.9999 … (1) n = 0.9999 … (2) 9n = 9.0000 … Subtract (2) from (1). Divide both sides by 9. n=1
0.2 Algebra Review 1.
3.
1 >0 2
5.
–1>–2
7.
π > 3.14
9.
1 = 0.5 2
11.
2 < 0.67 3
13.
x>0
15.
x–1
23.
d(C, D) = |C – D| = |0 – 1| = | – 1| = 1
25.
d(D, E) = |D – E | = |1 – 3| = | – 2 | = 2
3
4
SECTION 0.2
27.
d(A, E ) = |A – E | = | – 3 – 3| = | – 6 | = 6
29.
If x = – 2 and y = 3, then x + 2y = (– 2) + 2(3) = – 2 + 6 = 4
31.
If x = – 2 and y = 3, then 5xy + 2 = 5(– 2)(3) + 2 = – 30 + 2 = – 28
33.
If x = – 2 and y = 3, then
2 ( −2 ) 2x −4 4 = = = x − y ( −2 ) − 3 −5 5
35.
If x = – 2 and y = 3, then
3x + 2 y 3 ( −2 ) + 2 ( 3) −6 + 6 = = =0 2+ y 2+3 5
37.
If x = 3 and y = – 2, then | x + y | = | 3 + (– 2)| = | 1 | = 1
39.
If x = 3 and y = – 2, then | x | + | y | = | 3 | + | – 2 | = 3 + 2 = 5
41.
If x = 3 and y = – 2, then
43.
If x = 3 and y = – 2, then | 4x – 5y | = | 4(3) – 5(– 2)| = | 12 – (– 10)| = | 22 | = 22
45.
If x = 3 and y = – 2, then 4 x − 5 y = 4 ( 3) − 5 ( −2 ) = 12 − −10 = 12 − 10 = 2 = 2
47.
We must exclude values of x that would cause the denominator to equal zero. x≠0 (c)
49.
We must exclude values of x that would cause the denominator to equal zero. x2 – 9 ≠ 0 (x – 3)(x + 3) ≠ 0 x ≠ 3; x ≠ – 3 (a)
51.
We must exclude values of x that would cause the denominator to equal zero, but x2 + 1 can never equal zero, so no values are excluded.
53.
We must exclude values of x that would cause the denominator to equal zero. x3 – x ≠ 0 x(x2 – 1) ≠ 0 x(x – 1)(x + 1) ≠ 0 x ≠ 0; x ≠ 1; x ≠ – 1 (b), (c), and (d)
55.
The domain of the variable x is {x | x ≠ 5}.
57.
The domain of the variable x is { x | x ≠ – 4}.
x x
=
3 3
=
3 =1 3
SECTION 0.2 59.
61.
C=
5 5 ( F − 32 ) If F = 32°, then C = ( 32 − 32 ) = 0°. 9 9
C=
5 5 5 ( F − 32 ) If F = 77°, then C = ( 77 − 32 ) = ( 45) = 25°. 9 9 9
63.
(– 4) 2 = 16
67.
3–6 · 34 = 3–6+4 = 3–2 =
71. 75.
65.
3
2
73.
( − 4)
77.
= 8 2 ⋅ x 3 i 2 = 64 x 6
79.
x 2y 3 x 2 y 3 x = ⋅ 4 = x 2 − 1 ⋅ y 3− 4 = x ⋅ y − 1 = 4 xy x y y
81.
( − 2)
1 1 = 2 4 16
(3 – 2) – 1 = 3 (
69.
1 1 = 32 9
25 = 5
(8 x )
4 −2 =
(x
2
y
2
−1
)
− 2 ) i ( − 1)
= 16 = 4 2
=x
2i2
⋅y
2
8 3 −1 8 x 3z 8x 3 z ⋅x ⋅y ⋅z=− ⋅ =− 9 9 y 9y
83.
⎛ 3x − 1 ⎞ ⎜ −1 ⎟ ⎝ 4y ⎠
85.
If x = 2 and y = – 1, then 2 xy − 1 = 2 ⋅ 2 ⋅ (− 1) − 1 = − 4 .
87.
If x = 2 and y = – 1, then x 2 + y 2 = 2 2 + ( − 1) = 4 + 1 = 5 .
89.
If x = 2 and y = – 1, then ( xy ) = ( 2 ⋅ ( −1) ) = ( − 2 ) = 4 .
91.
If x = 2 and y = – 1, then
x2 = 22 = 4 = 2.
If x = 2 and y = – 1, then
x 2 + y 2 = 2 2 + ( − 1) = 4 + 1 = 5 .
95.
( − 1) i 2
=x y
x 4 ( yz ) − 8x 4 y 2 z 2 8 x4 y2 z2 8 = =− ⋅ ⋅ 3 ⋅ = − ⋅ x 4 −1 ⋅ y 2 −3 ⋅ z 2 −1 2 3 3 3 xy z 9 xy z 9 x y 9 z 3
=−
93.
= 32 = 9
−2
⎛ 3y ⎞ =⎜ ⎟ ⎝ 4x ⎠
−2
2
⎛ 4x ⎞ 4 2 ⋅ x 2 16 x 2 =⎜ ⎟ = 2 = 3 ⋅ y 2 9y 2 ⎝ 3y ⎠
2
2
2
If x = 2 and y = – 1, then x y = 2 − 1 =
2
2
1 . 2
4
−2
x4 = 2 y
5
6 97.
SECTION 0.2
If x = 2, then 2 x 3 − 3x 2 + 5 x − 4 = 2 ( 2 ) − 3 ( 2 ) + 5 ( 2 ) − 4 = 10 . 3
2
If x = 1, then 2 x 3 − 3x 2 + 5 x − 4 = 2 (1) − 3 (1) + 5 (1) − 4 = 0 . 3
2
99.
( 666 ) 4 ( 222 )
⎛ 666 ⎞ 4 =⎜ ⎟ = 3 = 81 ⎝ 222 ⎠
4
101.
(8.2 )
103.
( 6.1)
= 0.0044057 = 0.004
105.
( − 2.8)
107.
( − 8.11)
4
−3
−4
= 0.00023116 = 0.000
6
= 304,006.6714 = 304,006.671
6
= 481.890304 = 481.890
109. 454.2 = 4.542 × 10 2
111. 0.013 = 1.3 × 10 − 2
113. 32,155 = 3.2155 × 10 4
115. 0.000423 = 4.23 × 10 − 4
117. 6.15 × 10 4 = 61,500
119. 1.214 × 10 − 3 = 0.001214
121. 1.1 × 10 8 = 110,000,000
123. 8.1 × 10 − 2 = 0.081 125.
A = lw;
127.
C=πd;
129.
131.
A =
V =
domain: A > 0, l > 0, w > 0
3 2 ⋅x ; 4 4 3 πr ; 3
domain: C > 0, d > 0 domain: A > 0, x > 0
domain: V > 0, r > 0
133.
V = x3 ;
135.
C = 4000 + 2x (a) If x = 1000 watches are produced, it will cost C = 4000 + 2(1000) = $6000.00
domain: V > 0, x > 0
(b) If x = 2000 watches are produced, it will cost C = 4000 + 2(2000) = $8000.00 137.
(a) If actual voltage is x = 113 then |113 – 115| = | - 2 | = 2 Since 2 < 5, an actual voltage of 113 is acceptable. (b) If actual voltage is x = 109 then |109 – 115| = | - 6 | = 6
SECTION 0.3
Since 6 > 5, an actual voltage of 109 is not acceptable. 139.
(a) If the radius is x = 2.999, then | x – 3 | = | 2.999 – 3 | = | - 0.001 | = 0.001 Since 0.001 < 0.010 the ball bearing is not acceptable. (b) If the radius is x = 2.89, then | x – 3 | = | 2.89 – 3 | = | - 0.11 | = 0.11 Since 0.11 > 0.01, the ball bearing is not acceptable.
141.
1 1 ≠ 0.333 ; > 0.333 3 3 1 = 0.333 … ; 0.333… – 0.333 = 0.000333… 3
143.
The answer is no. Student answers should justify and explain why not.
145.
Answers will vary.
0.3 Polynomials and Rational Expressions 1.
(10 x
3.
( x + a)
5.
( x + 8)( 2 x + 1) = 2 x 2 + x + 16 x + 8 = 2 x 2 + 17 x + 8
7.
(x
2
5
− 8 x 2 ) + ( 3x 3 − 2 x 2 + 6 ) = 10 x 5 + 3x 3 − 10 x 2 + 6 2
− x 2 = ⎡⎣ x 2 + 2ax + a 2 ⎤⎦ − x 2 = 2ax + a 2
+ x − 1)( x 2 − x + 1) = x 2 ( x 2 − x + 1) + x ( x 2 − x + 1) − 1( x 2 − x + 1)
= x 4 − x 3 + x 2 + x 3 − x 2 + x − x 2 + x −1 = x 4 − x 2 + 2x − 1 9.
( x + 1)
3
− ( x − 1) = ( x 3 + 3 x 2 + 3 x + 1) − ( x 3 − 3 x 2 + 3 x − 1) 3
= x 3 + 3x 2 + 3x + 1 − x 3 + 3x 2 − 3x + 1 = 6x 2 + 2 11.
This is the difference of 2 squares. x 2 − 36 = ( x − 6 )( x + 6 )
13.
This is the difference of 2 squares. 1 − 4 x 2 = (1 − 2 x )(1 + 2 x )
15.
x 2 + 7 x + 10 = ( x + 5 )( x + 2 )
The product of 5 and 2 is 10; the sum of 5 and 2 is 7.
7
SECTION 0.3
Since 6 > 5, an actual voltage of 109 is not acceptable. 139.
(a) If the radius is x = 2.999, then | x – 3 | = | 2.999 – 3 | = | - 0.001 | = 0.001 Since 0.001 < 0.010 the ball bearing is not acceptable. (b) If the radius is x = 2.89, then | x – 3 | = | 2.89 – 3 | = | - 0.11 | = 0.11 Since 0.11 > 0.01, the ball bearing is not acceptable.
141.
1 1 ≠ 0.333 ; > 0.333 3 3 1 = 0.333 … ; 0.333… – 0.333 = 0.000333… 3
143.
The answer is no. Student answers should justify and explain why not.
145.
Answers will vary.
0.3 Polynomials and Rational Expressions 1.
(10 x
3.
( x + a)
5.
( x + 8)( 2 x + 1) = 2 x 2 + x + 16 x + 8 = 2 x 2 + 17 x + 8
7.
(x
2
5
− 8 x 2 ) + ( 3x 3 − 2 x 2 + 6 ) = 10 x 5 + 3x 3 − 10 x 2 + 6 2
− x 2 = ⎡⎣ x 2 + 2ax + a 2 ⎤⎦ − x 2 = 2ax + a 2
+ x − 1)( x 2 − x + 1) = x 2 ( x 2 − x + 1) + x ( x 2 − x + 1) − 1( x 2 − x + 1)
= x 4 − x 3 + x 2 + x 3 − x 2 + x − x 2 + x −1 = x 4 − x 2 + 2x − 1 9.
( x + 1)
3
− ( x − 1) = ( x 3 + 3 x 2 + 3 x + 1) − ( x 3 − 3 x 2 + 3 x − 1) 3
= x 3 + 3x 2 + 3x + 1 − x 3 + 3x 2 − 3x + 1 = 6x 2 + 2 11.
This is the difference of 2 squares. x 2 − 36 = ( x − 6 )( x + 6 )
13.
This is the difference of 2 squares. 1 − 4 x 2 = (1 − 2 x )(1 + 2 x )
15.
x 2 + 7 x + 10 = ( x + 5 )( x + 2 )
The product of 5 and 2 is 10; the sum of 5 and 2 is 7.
7
8
SECTION 0.3
17.
This polynomial cannot be factored; it is prime.
19.
This polynomial cannot be factored; it is prime.
21.
15 + 2 x − x 2 = ( 5 − x )( 3 + x )
23.
3x 2 − 12 x − 36 = 3 ( x 2 − 4 x − 12 )
The product 5 · 3 =15; the sum 5 · 1 + 3 · (– 1) = 2. Factor out the common factor of 3.
= 3 ( x − 6 )( x + 2 )
25.
The product – 6 · 2 = –12; the sum – 6 + 2 = – 4.
y 4 + 11 y 3 + 30 y 2 = y 2 ( y 2 + 11y + 30 )
Factor out the common factor of y 2.
= y 2 ( y + 5 )( y + 6 )
The product 5 · 6 = 30; the sum 5 + 6 = 11.
27.
This polynomial is a perfect square. 2 4 x 2 + 12 x + 9 = ( 2 x + 3)( 2 x + 3) = ( 2 x + 3)
29.
3 x 2 + 4 x + 1 = ( 3 x + 1)( x + 1)
31.
x 4 − 81 = ( x 2 − 9 )( x 2 + 9 )
The products 3 · 1 = 3 and 1 · 1 = 1; the sum 3 · 1 + 1 · 1 = 4. 4
Treat x − 81 as the difference of 2 squares, and factor.
= ( x − 3)( x + 3) ( x 2 + 9 )
33.
Let u = x 3. Then 2 x 6 − 2 x 3 + 1 = u 2 − 2u + 1 = ( u − 1)( u − 1) = ( u − 1) Now substitute back for u. That is, replace u with x 3. x 6 − 2 x 3 + 1 = ( x 3 − 1)
35.
2
x − 9 is the difference of 2 squares.
2
x 7 − x 5 = x 5 ( x 2 − 1)
Factor out the common factor of x 5.
= x 5 ( x − 1)( x + 1)
37.
5 + 16 x − 16 x 2 = (1 + 4 x )( 5 − 4 x )
2
x − 1 is the difference of 2 squares.
The products 4 · (– 4) = – 16 and 1 · 5 = 5; the sum 1 · (– 4) + 4 · 5 = 16.
39.
4 y 2 − 16 y + 15 = ( 2 y − 5 )( 2 y − 3)
The products 2 · 2 = 4 and – 5 · (– 3) = 15; the sum 2 · (– 3) + (– 5) · 2 = – 16.
41.
Let u = x 2, then 1 − 8 x 2 − 9 x 4 = 1 − 8u − 9u 2 = (1 − 9u )(1 + u ) Now substitute back for u. That is, replace u with x 2. 1 − 8 x 2 − 9 x 4 = (1 − 9 x 2 )(1 + x 2 ) 1 − 9x 2 is the difference of 2 squares.
SECTION 0.3
1 − 8 x 2 − 9 x 4 = (1 − 9 x 2 )(1 + x 2 ) = (1 − 3 x )(1 + 3 x ) (1 + x 2 ) 43.
x ( x + 3) − 6 ( x + 3) = ( x + 3)( x − 6 )
45.
( x + 2)
2
Factor out the common factor x + 3.
− 2 ( x + 2 ) = ( x + 2 ) ⎡⎣( x + 2 ) − 5⎤⎦
Factor out the common factor x + 2.
= ( x + 2 )( x − 3)
47.
Simplify.
6 x ( 2 − x ) − 9 x 2 ( 2 − x ) = 3 x ( 2 − x ) ⎡⎣ 2 ( 2 − x ) − 3 x ⎤⎦ 4
3
3
Factor out the common factors.
= 3x ( 2 − x ) ( 4 − 2 x − 3x )
Simplify.
= 3x ( 2 − x ) ( 4 − 5 x )
Simplify.
= 3x ( x − 2 ) ( 5 x − 4 )
Multiply (x – 2) 3 by (– 1) 3 and
3
3
3
(5x – 4) by – 1. [(– 1) 3 · (– 1) = 1].
49.
x 3 + 2 x 2 − x − 2 = ( x 3 + 2 x 2 ) − ( x + 2 ) Group the polynomial into the difference of 2 binomials. = x 2 ( x + 2 ) − 1( x + 2 ) Factor the common factors from each binomial.
= ( x 2 − 1) ( x + 2 )
Factor the common factor x + 2.
= ( x − 1)( x + 1)( x + 2 ) Factor the difference of 2 squares.
51.
x 4 − x 3 + x − 1 = ( x 4 − x 3 ) + ( x − 1) = x 3 ( x − 1) + ( x − 1)
= ( x − 1) ( x 3 + 1)
Group the polynomial into the sum of 2 binomials. Factor the common factors from each binomial. Factor the common factor x – 1.
= ( x − 1)( x + 1) ( x 2 − x + 1) Factor the sum of 2 cubes. 53.
3 x − 6 x 2 − x − 6 3 ( x − 2 ) ( x − 3) ( x + 2 ) ⋅ = ⋅ 5x x2 −4 5x ( x − 2) ( x + 2)
=
3 ( x − 3) 5x
Factor.
Simplify.
55.
4 x 2 − 1 x 2 − 4 x ( 2 x − 1) ( 2 x + 1) x ( x − 4 ) (2 x − 1) x = ⋅ = ⋅ x+4 x 2 − 16 2 x + 1 ( x − 4) ( x + 4) 2x + 1
57.
x x x x − 2 = − x − 7 x + 6 x − 2 x − 24 ( x − 6 )( x − 1) ( x − 6 )( x + 4 ) 2
=
x ( x + 4) x ( x − 1) − ( x − 6 )( x − 1)( x + 4 ) ( x − 6 )( x + 4 )( x − 1)
9
10
59.
SECTION 0.3
=
x 2 + 4x x2 − x − ( x − 6 )( x − 1)( x + 4 ) ( x − 6 )( x + 4 )( x − 1)
=
x 2 + 4x − x 2 + x ( x − 6 )( x − 1)( x + 4 )
=
5x ( x − 6 )( x − 1)( x + 4 )
4 2 4 2 − 2 = − x − 4 x + x − 6 ( x − 2 )( x + 2 ) ( x + 3)( x − 2 ) 2
= =
4 ( x + 3) − 2 ( x + 2 ) ( x − 2 )( x + 2 )( x + 3)
=
4 x + 12 − 2 x − 4 ( x − 2 )( x + 2 )( x + 3)
=
61.
4 ( x + 3) 2 ( x + 2) − ( x − 2 )( x + 2 )( x + 3) ( x + 3)( x − 2 )( x + 2 )
2 ( x + 4) 2x + 8 = ( x − 2 )( x + 2 )( x + 3) ( x − 2 )( x + 2 )( x + 3)
1 2 3 1 2 3 − 2 + 3 = − + 2 2 x x +x x −x x x ( x + 1) x ( x − 1)
=
x ( x + 1)( x − 1) 2 x ( x − 1) 3 ( x + 1) − 2 + 2 2 x ( x + 1)( x − 1) x ( x + 1)( x − 1) x ( x − 1)( x + 1)
=
x ( x + 1)( x − 1) − 2 x ( x − 1) + 3 ( x + 1) x 2 ( x + 1)( x − 1)
=
x 3 − x − 2 x 2 + 2 x + 3x + 3 x 2 ( x + 1)( x − 1)
x 3 − 2x 2 + 4x + 3 = 2 x ( x + 1)( x − 1) 63.
1⎛ 1 1⎞ − ⎟= ⎜ h⎝ x+h x⎠ =
1( x + h ) ⎞ 1⎛ x − ⎜⎜ ⎟ h ⎝ ( x + h ) x x ( x + h ) ⎟⎠ 1⎛ x− x−h⎞ ⎜ ⎟ h ⎜⎝ ( x + h ) x ⎟⎠
⎛ − h ⎞ ⎜⎜ ⎟⎟ ⎝ ( x + h) x ⎠ 1 =− x ( x + h) =
1 h
SECTION 0.3 65.
2 ( 3 x + 4 ) + ( 2 x + 3) ⋅ 2 ( 3 x + 4 ) ⋅ 3 = ( 3 x + 4 ) ⎡⎣ 2 ( 3 x + 4 ) + 6 ( 2 x + 3) ⎤⎦ 2
= ( 3 x + 4 )( 6 x + 8 + 12 x + 18 ) = ( 3 x + 4 )(18 x + 26 )
= 2 ( 3 x + 4 )( 9 x + 13)
67.
2 x ( 2 x + 5) + x 2 ⋅ 2 = 2 x ( 2 x + 5 + x )
= 2 x ( 3x + 5)
69.
2 ( x + 3)( x − 2 ) + ( x + 3) ⋅ 3 ( x − 2 ) = ( x + 3)( x − 2 ) ⎡⎣ 2 ( x − 2 ) + 3 ( x + 3) ⎤⎦ 3
2
2
2
= ( x + 3)( x − 2 )
( 2 x − 4 + 3x + 9 ) = ( x + 3)( x − 2 ) ( 5 x + 5 ) 2 = 5 ( x + 3)( x − 2 ) ( x + 1) 2 2
71.
( 4 x − 3)
2
+ x ⋅ 2 ( 4 x − 3) ⋅ 4 = ( 4 x − 3) ⎡⎣( 4 x − 3) + 8 x ⎤⎦ = ( 4 x − 3)(12 x − 3)
= 3 ( 4 x − 3)( 4 x − 1)
73.
2 ( 3 x − 5 ) ⋅ 3 ( 2 x + 1) + ( 3 x − 5 ) ⋅ 3 ( 2 x + 1) ⋅ 2 = 6 ( 3 x − 5 )( 2 x + 1) ⎡⎣( 2 x + 1) + ( 3 x − 5 ) ⎤⎦ 3
2
2
2
= 6 ( 3x − 5 )( 2 x + 1) ( 5 x − 4 ) 2
75.
77.
( 2 x + 3) ⋅ 3 − ( 3 x − 5 ) ⋅ 2 = 2 ( 3x − 5) x ⋅ 2 x − ( x 2 + 1) ⋅ 1
(x
79.
81.
2
+ 1)
2
=
6 x + 9 − 6 x + 10
( 3x − 5)
2x 2 − x 2 −1
(x
2
+ 1)
2
=
2
+ 1) ⋅ 3 − ( 3x + 4 ) ⋅ 2 x
(x
2
+ 1)
2
=
=
x 2 −1
(x
( 3x + 1) ⋅ 2 x − x 2 ⋅ 3 = 6 x 2 + 2 x − 3x 2 2 2 ( 3x + 1) ( 3x + 1)
(x
2
2
+ 1)
=
2
2
+ 1)
( 3x − 5 ) =
2
( x − 1)( x + 1)
(x
2
+ 1)
2
3x 2 + 2 x
( 3x + 1)
3x 2 + 3 − 6 x 2 − 8 x
(x
19
2
=
2
−3x 2 − 8 x + 3
(x
2
+ 1)
2
11
12
SECTION 0.4
0.4 Solving Equations 1.
3x = 21 x=7
3.
5x + 15 = 0 5x = – 15 x=–3
7.
1 5 x= 3 12 5 3 x= ⋅ 12 4 1 5 x= 4
5.
2x – 3 = 5 2x = 8 x=4
9.
6 – x = 2x + 9 – 3x = 3 x=–1
11.
8 x − ( 2 x + 1) = 3 x − 10
15.
13.
6 x − 1 = 3 x − 10 3x = – 9 x=–3
17.
0.9t = 0.4 + 0.1t 0.8t = 0.4 0.4 t= = 0.5 0.8
21.
( x + 7 )( x − 1) = ( x + 1)
2
x 2 = 9x x 2 − 9x = 0 x ( x − 9) = 0 x = 0 or x − 9 = 0 x=9
1 3 x−4= x 2 4 1 ⎛ ⎞ ⎛3 ⎞ 4 ⋅ ⎜ x − 4⎟ = 4 ⋅ ⎜ x⎟ ⎝2 ⎠ ⎝4 ⎠ 2x – 16 = 3x x = – 16
19.
2 4 + =3 y y 6 =3 y 3y = 6 y=2
23.
z ( z 2 + 1) = 3 + z 3
x 2 − x + 7x − 7 = x 2 + 2x + 1 x 2 + 6x − 7 = x 2 + 2x + 1 6x − 7 = 2x + 1 4x = 8 x=2 25.
2(3 + 2x) = 3(x – 4) 6 + 4x = 3x – 12 x = – 18
z3 + z = 3+ z3 z=3
27.
t 3 − 9t 2 = 0 t 2 (t − 9) = 0 t 2 = 0 or t – 9 = 0 t = 0 or t = 9 The solution set is {0, 9}.
The solution set is {0, 9}.
SECTION 0.4 29.
3 2 = 2x − 3 x + 5 3(x + 5) = 2(2x – 3) 3x + 15 = 4x – 6 x = 21
31.
(x + 2)(3x) = (x + 2)(6) 3x 2 + 6 x = 6 x + 12 3x 2 − 12 = 0 x2 −4 = 0 x2 = 4 x = – 2 or x = 2 The solution set is {– 2, 2}.
33.
2 3 10 = + x − 2 x + 5 ( x + 5 )( x − 2 )
L.C.D.: (x + 5)(x – 2)
2 ( x + 5) 3( x − 2) 10 Write with the common denominator. = + ( x − 2 )( x + 5) ( x + 5)( x − 2 ) ( x + 5)( x − 2 ) 2( x + 5) = 3( x − 2) + 10 Solve the equation formed by the numerators. 2x + 10 = 3x – 6 + 10 x=6 Check the answer for extraneous solutions. The solution set is {6}. 35.
2x = 6
37.
Either 2x = 6 or 2x = – 6 x = 3 or x = – 3 The solution set is {– 3, 3}. 39.
1 − 4t = 5
Either 2x + 3 = 5 or 2x + 3 = – 5 2x = 2 or 2x = – 8 x = 1 or x=–4 The solution set is {– 4, 1}. 41.
Either 1 – 4t = 5 or 1 – 4t = – 5 – 4t = 4 or – 4t = – 6 3 t = – 1 or t= 2 3⎫ ⎧ The solution set is ⎨− 1, ⎬ . 2⎭ ⎩ 43.
−2 x=4
2x = 4 x=2
2x + 3 = 5
− 2x = 8
Either – 2x = 8 or – 2x = – 8 x = – 4 or x=4 The solution set is {– 4, 4}.
45.
1 2 This equation has no solution. Absolute values are always nonnegative. x−2 =−
13
14 47.
SECTION 0.4
x2 −4 = 0
x2 −4 = 0 x2 = 4 x=±2 The solution set is {– 2, 2}. 49.
x 2 − 2x = 3 Either x 2 − 2x = 3 x 2 − 2x − 3 = 0 ( x − 3)( x + 1) = 0
or
x – 3 = 0 or x + 1 = 0 x = 3 or x=–1
x 2 − 2x = − 3 x 2 − 2x + 3 = 0 a = 1, b = – 2, c = 3 The discriminant, b 2 − 4ac = 4 – 12 = – 8 is negative; the equation has no real solutions.
The solution set is {– 1, 3}. 51.
x 2 + x −1 = 1 Either x 2 + x −1 = 1 x2 + x−2 = 0 ( x − 1)( x + 2 ) = 0 x – 1 = 0 or x = 1 or
x+2=0 x=–2
or
or or
x 2 + x −1 = − 1 x2 + x = 0 x ( x + 1) = 0 x = 0 or x + 1 = 0 x = 0 or x = – 1
The solution set is {– 2, – 1, 0, 1}. 53.
57.
x 2 = 4x x 2 − 4x = 0 x(x – 4) = 0 x = 0 or x – 4 = 0 x=4 The solution set is {0, 4}.
2 x 2 − 5x − 3 = 0 (2x + 1)(x – 3) = 0 2x + 1 = 0 or x – 3 = 0 1 x=– or x=3 2 ⎧ 1 ⎫ The solution set is ⎨− , 3⎬ . ⎩ 2 ⎭
55.
z 2 + 4 z − 12 = 0 (z – 2)(z + 6) = 0 z – 2 = 0 or z + 6 = 0 z = 2 or z=–6 The solution set is {– 6, 2}.
59.
x ( x − 7 ) + 12 = 0
x 2 − 7 x + 12 = 0 (x – 3)(x – 4) = 0 x – 3 = 0 or x – 4 = 0 x = 3 or x=4 The solution set is {3, 4}.
SECTION 0.4 61.
4 x 2 + 9 = 12 x 4 x − 12 x + 9 = 0 (2x – 3)(2x – 3) = 0 2x – 3 = 0 3 x= 2 ⎧3⎫ The solution set is ⎨ ⎬ . ⎩2⎭
4( x − 2) 3 −3 + = x−3 x x ( x − 3)
65.
4 x ( x − 2 ) 3 ( x − 3) −3 + = x ( x − 3) x ( x − 3) x ( x − 3) 4 x ( x − 2 ) + 3 ( x − 3) = − 3
4 x − 8 x + 3x − 9 = − 3 4 x 2 − 5x − 6 = 0 (4x + 3)(x – 2) = 0 4x + 3 = 0 or x – 2 = 0 3 x=– or x = 2 4 ⎧ 3 ⎫ The solution set is ⎨− , 2 ⎬ . ⎩ 4 ⎭ 2
67.
71.
x 2 = 25 x = ± 25 x=±5 The solution set is {– 5, 5}.
( 2 x + 3)
2
=9
2x + 3 = ± 9 2x + 3 = ± 3 2 x = − 3 + 3 or 2x = − 3 − 3 2x = 0 or 2x = – 6 x=0 or x=–3 The solution set is {–3, 0}.
63.
6 x 2 6 x − 5x = 6 2 6x − 5x − 6 = 0 (3x + 2)(2x – 3) = 0 3x + 2 = 0 or 2x – 3 = 0 2 3 x=– or x = 3 2 ⎧ 2 3⎫ The solution set is ⎨ − , ⎬ . ⎩ 3 2⎭ 6x − 5 =
The lowest common denominator is x(x – 3).
Write the equation with the common denominator. Consider the equation formed by the numerator. Simplify. Put the quadratic equation in standard form. Factor. Use the Zero-Product Property. Solve; be sure to check for extraneous solutions.
69.
( x − 1)
2
=4
x −1 = ± 4 x −1 = ± 2 x = 2 + 1 or x = – 2 + 1 x=3 or x = – 1 The solution set is {– 1, 3}. 73.
x 2 + 8x 2 ⎛8⎞ Add ⎜ ⎟ = 16. ⎝2⎠ Result x 2 + 8 x + 16
15
SECTION 0.4
16 75.
1 x 2 2 1 ⎛1⎞ Add ⎜ ⎟ = 16 ⎝4⎠ 1 1 Result x 2 + x + 2 16
79.
x 2 + 4 x = 21 x 2 + 4 x + 4 = 21 + 4 2 ( x + 2 ) = 25 x+2=± 5 x = − 2±5 The solution set is {– 7, 3}.
81.
77.
x2 +
1 3 x− =0 2 16 1 3 x2 − x = 2 16 1 1 3 1 x2 − x+ = + 2 16 16 16 2 1⎞ 4 ⎛ ⎜x− ⎟ = 4⎠ 16 ⎝
Add 4 to both sides. Factor. Use the Square Root Method.
x2 −
1 4 =± 4 16 1 2 x= ± 4 4 ⎧ 1 3⎫ The solution set is ⎨− , ⎬ . ⎩ 4 4⎭
x−
83.
2 x 3 2 1 ⎛1⎞ Add ⎜ ⎟ = 9 ⎝3⎠ 2 1 Result x 2 − x + 3 9 x2 −
1 =0 2 1 1 x2 + x− = 0 3 6 1 1 x2 + x = 3 6 1 1 1 1 x2 + x+ = + 3 36 6 36 2 1⎞ 7 ⎛ ⎜x+ ⎟ = 6⎠ 36 ⎝ 3x 2 + x −
x+
1 =± 6
7 36
Add
1 16
to both sides.
SECTION 0.4
x=−
17
1 7 ± 6 6
⎧⎪ − 1 − 7 − 1 + 7 ⎫⎪ The solution set is ⎨ , ⎬. 6 6 ⎪⎭ ⎩⎪ 85.
x 2 − 4x + 2 = 0 a = 1, b = – 4, and c = 2 2 2 The discriminant b – 4ac = ( − 4 ) − 4 (1)( 2 ) = 16 − 8 = 8 is positive, so there are 2 real solutions to the equation. −b ± b 2 − 4ac 4± 8 4±2 2 = =2± 2 = 2 2 2a The solution set is 2 − 2, 2 + 2 .
x=
{
87.
}
x 2 − 5x − 1 = 0 a = 1, b = – 5, and c = – 1 2 2 The discriminant b – 4ac = ( − 5 ) − 4 (1)( − 1) = 25 + 4 = 29 is positive, so there are 2 real solutions to the equation. 5 ± 29 −b ± b 2 − 4ac = 2 2a ⎧⎪ 5 − 29 5 + 29 ⎪⎫ , The solution set is ⎨ ⎬. 2 ⎭⎪ ⎩⎪ 2 x=
89.
2 x 2 − 5x + 3 = 0 a = 2, b = – 5, and c = 3 2 2 The discriminant b – 4ac = ( − 5 ) − 4 ( 2 )( 3) = 25 − 24 = 1 is positive, so there are 2 real solutions to the equation. −b ± b 2 − 4ac 5 ± 1 5 ±1 = = 4 4 2a ⎧ 3⎫ The solution set is ⎨1, ⎬ . ⎩ 2⎭
x=
91.
4y 2 − y + 2 = 0
a = 4, b = – 1, and c = 2
The discriminant b – 4ac = ( − 1) − 4 ( 4 )( 2 ) = 1 − 32 = − 31 is negative, so the equation has no real solution. 2
93.
2
4 x 2 = 1 − 2 x First we rewrite the equation in standard form. 4x 2 + 2x − 1 = 0 a = 4, b = 2, and c = – 1 2 2 The discriminant b – 4ac = ( 2 ) − 4 ( 4 )( −1) = 4 + 16 = 20 is positive, so there are 2 real solutions to the equation. x=
− 2 ± 20 − 2 ± 2 5 − 1 ± 5 −b ± b 2 − 4ac = = = 8 8 4 2a
SECTION 0.4
18
⎧⎪ − 1 − 5 − 1 + 5 ⎫⎪ , The solution set is ⎨ ⎬. 4 4 ⎪⎩ ⎪⎭ 95.
x 2 + 3x − 3 = 0
a = 1, b = 3 , and c = – 3
The discriminant b2 – 4ac =
( 3)
2
− 4 (1)( − 3) = 3 + 12 = 15 is positive, so there are 2 real
solutions to the equation. −b ± b 2 − 4ac − 3 ± 15 = 2 2a ⎧⎪ − 3 − 15 − 3 + 15 ⎫⎪ , The solution set is ⎨ ⎬. 2 2 ⎪⎩ ⎪⎭
x=
97.
a = 1, b = – 5, and c = 7 x 2 − 5x + 7 = 0 2 2 The discriminant b – 4ac = ( − 5 ) − 4 (1)( 7 ) = 25 − 28 = − 3 is negative, so the equation has no real solution.
99.
9 x 2 − 30 x + 25 = 0 a = 9, b = – 30, and c = 25 2 2 The discriminant b – 4ac = ( − 30 ) − 4 ( 9 )( 25 ) = 900 − 900 = 0 , so the equation has a repeated solution, a root of multiplicity 2.
101.
3x 2 + 5 x − 8 = 0 a = 3, b = 5, and c = – 8 2 2 The discriminant b – 4ac = ( 5 ) − 4 ( 3)( − 8 ) = 25 + 96 = 121 is positive, so there are 2 real solutions to the equation.
103.
ax – b = c ax = b + c b+c x= a
105.
x x + =c a b bx ax abc + = ab ab ab bx + ax = abc (b + a)x = abc abc x= a+b
107.
1 1 2 + = x − a x + a x −1
SECTION 0.4
( x + a )( x − 1) + ( x − a )( x − 1) = 2 ( x − a )( x + a ) ( x − a )( x + a )( x − 1) ( x − a )( x + a )( x − 1) ( x − 1)( x − a )( x + a ) ( x + a )( x − 1) + ( x − a )( x − 1) = 2 ( x − a )( x + a ) x 2 + ax − x − a + x 2 − x − ax + a = 2 x 2 + 2ax − 2ax − 2a 2 2 x 2 − 2 x = 2 x 2 − 2a 2 −2 x = −2a 2 x = a2 109.
1 1 1 = + R R1 R 2
R1R 2
=
RR1R 2
RR 2 RR1R 2
+
RR1 RR1R 2
R 1R 2 = RR 2 + RR1
(
R 1R 2 = R R 2 + R1 R=
)
R1R 2 R1 + R 2
111.
mv 2 R FR = mv 2 mv 2 R= F
113.
a 1− r S (1 − r ) = a S − Sr = a Sr = S − a S −a r= S
F=
S=
115. The roots of the quadratic function ax 2 + bx + c = 0 are
x1 =
−b − b 2 − 4ac 2a
The sum x 1 + x 2 =
and
x2 =
−b + b 2 − 4ac 2a
−b − b 2 − 4ac −b + b 2 − 4ac + 2a 2a
− b − b 2 − 4ac + ( − b ) + b 2 − 4ac = 2a − 2b b = =− 2a a
19
20
SECTION 0.5
117. If kx 2 + x + k = 0 has a repeated real solution, then its discriminant is zero. a = k, b = 1, and c = k discriminant: b 2 – 4ac = 1 2 – 4(k)(k) = 0 1 – 4k 2 = 0 4k 2 = 1 2k = ± 1 1 1 So the equation has one repeated root if k = or k = – . 2 2 119. The real solutions of the equation ax 2 + bx + c = 0 are
x1 =
−b − b 2 − 4ac 2a
=−
and x 2 =
b + b 2 − 4ac 2a
−b + b 2 − 4ac 2a
=−
b − b 2 − 4ac 2a
The real solutions of the equation ax 2 – bx + c = 0 are x3 =
b − b 2 − 4ac 2a
and
x4 =
b + b 2 − 4ac 2a
So x 1 = – x 4 and x 2 = – x 3. 121. (a) x 2 = 9 and x = 3 are not equivalent. The solution set of x 2 = 9 is { – 3, 3}, but the solution set of x = 3 is {3}.
(b) x =
9 and x = 3 are equivalent since they both have the same solution set, {3}.
(c) (x – 1)(x – 2) = (x – 1) 2 and x – 2 = x – 1 are not equivalent. The solution of the first equation is {1}, but the second equation has no solution. 123. – 127. Answers will vary.
0.5 Intervals; Solving Inequalities 1.
The graph represents [0, 2] or 0 ≤ x ≤ 2.
3.
The graph represents (– 1, 2) or – 1 < x < 2.
5.
The graph represents [0, 3) or 0 ≤ x < 3.
7.
[0, 4]
9.
[4, 6)
11.
[4, ∞)
20
SECTION 0.5
117. If kx 2 + x + k = 0 has a repeated real solution, then its discriminant is zero. a = k, b = 1, and c = k discriminant: b 2 – 4ac = 1 2 – 4(k)(k) = 0 1 – 4k 2 = 0 4k 2 = 1 2k = ± 1 1 1 So the equation has one repeated root if k = or k = – . 2 2 119. The real solutions of the equation ax 2 + bx + c = 0 are
x1 =
−b − b 2 − 4ac 2a
=−
and x 2 =
b + b 2 − 4ac 2a
−b + b 2 − 4ac 2a
=−
b − b 2 − 4ac 2a
The real solutions of the equation ax 2 – bx + c = 0 are x3 =
b − b 2 − 4ac 2a
and
x4 =
b + b 2 − 4ac 2a
So x 1 = – x 4 and x 2 = – x 3. 121. (a) x 2 = 9 and x = 3 are not equivalent. The solution set of x 2 = 9 is { – 3, 3}, but the solution set of x = 3 is {3}.
(b) x =
9 and x = 3 are equivalent since they both have the same solution set, {3}.
(c) (x – 1)(x – 2) = (x – 1) 2 and x – 2 = x – 1 are not equivalent. The solution of the first equation is {1}, but the second equation has no solution. 123. – 127. Answers will vary.
0.5 Intervals; Solving Inequalities 1.
The graph represents [0, 2] or 0 ≤ x ≤ 2.
3.
The graph represents (– 1, 2) or – 1 < x < 2.
5.
The graph represents [0, 3) or 0 ≤ x < 3.
7.
[0, 4]
9.
[4, 6)
11.
[4, ∞)
SECTION 0.5 13. (– ∞, – 4)
15.
2≤x≤5
17. – 3 < x < – 2
19.
4≤x – 3 (a) 4 + 3 > – 3 + 3 7>0
(b) 4 – 5 > – 3 – 5 –1>–8
(c) (3)(4) > (3)( – 3) 12 > – 9
(d) (– 2)(4) < (– 2)( – 3) –8–4+4 x+4>0 x>6 – 2x < (– 2)(6) – 2x < – 12
2x > 6 2x 6 > 2 2 x>3
21
22 41.
45.
SECTION 0.5
≥
1 x≤3 2 1 ( − 2 ) ⋅ ⎛⎜ − x ⎞⎟ ≥ ( − 2 )( 3) ⎝ 2 ⎠ x≥–6 −
1 – 2x ≤ 3 1 – 2x – 1 ≤ 3 – 1 – 2x ≤ 2 − 2x 2 ≥ −2 −2 x≥–1
43.
The solution set is {x | x < 4} or the interval (– ∞, 4).
47.
3x – 1 ≥ 3 + x 3x – 1 + 1 ≥ 3 + x + 1 3x ≥ 4 + x 3x – x ≥ 4 + x – x 2x ≥ 4 x≥2
51.
The solution set is {x | x ≥ 2} or the interval [2, ∞).
53.
4 − 3 (1 − x ) ≤ 3 4 − 3 + 3x ≤ 3 1 + 3x ≤ 3 1 + 3x – 1 ≤ 3 – 1 3x ≤ 2 2 x≤ 3 ⎧ The solution set is ⎨ x x ≤ ⎩ 2⎤ ⎛ interval ⎜ − ∞, ⎥ . 3⎦ ⎝
– 2(x + 3) < 8 − 2 ( x + 3) 8 > −2 −2 x+3>–4 x+3–3>–4–3 x>–7 The solution set is {x | x > – 7} or the interval (– 7, ∞).
55.
2⎫ ⎬ or the 3⎭
3x – 7 > 2 3x – 7 + 7 > 2 + 7 3x > 9 3x 9 > 3 3 x>3 The solution set is {x | x > 3} or the interval (3, ∞).
The solution set is {x | x ≥ – 1} or the interval [– 1, ∞).
49.
x+1 2x + 16 x – 4 – x > 2x + 16 – x – 4 > x + 16 – 4 – 16 > x + 16 – 16 – 20 > x or x < – 20 The solution set is {x | x < – 20} or the interval (– ∞, – 20).
SECTION 0.5
57.
59.
x x ≥ 1− 2 4 x⎞ ⎛ x⎞ ⎛ 4 ⋅ ⎜ ⎟ ≥ 4 ⋅ ⎜ 1− ⎟ ⎝2⎠ ⎝ 4⎠ 2x ≥ 4 – x 2x + x ≥ 4 – x + x 3x ≥ 4 4 x≥ 3 ⎧ 4⎫ The solution set is ⎨ x x ≥ ⎬ or the 3⎭ ⎩ ⎡4 ⎞ interval ⎢ , ∞ ⎟ . ⎣3 ⎠
0 ≤ 2x – 6 ≤ 4 is equal to the two inequalities 0 ≤ 2x – 6 0 + 6 ≤ 2x – 6 + 6 6 ≤ 2x 3≤x
and
2x – 6 ≤ 4 2x – 6 + 6 ≤ 4 + 6 2x ≤ 10 x≤5
The solution set consists of all x for which x ≥ 3 and x ≤ 5 which is written either as {x | 3 ≤ x ≤ 5} or as the interval [3, 5].
61.
− 5 ≤ 4 − 3x ≤ 2 is equal to the two inequalities − 5 ≤ 4 − 3x – 9 ≤ – 3x 3≥x
and
The solution set consists of all x for which x ≥ ⎧ ⎨x ⎩
⎫ 2 ≤ x ≤ 3 ⎬ or as the interval 3 ⎭
⎡2 ⎤ ⎢⎣ 3 , 3 ⎥⎦ .
4 − 3x ≤ 2 – 3x ≤ – 2 2 x≥ 3
2 and x ≤ 3 which is written either as 3
23
24 63.
SECTION 0.5
2x − 1 < 0 is equal to the two inequalities 4 and 2x − 1 −3< 4 – 12 < 2x – 1 – 11 < 2x 11 − <x 2
−3
0 for – 3 < x < 3, we write the solution set either as {x | – 3 < x < 3} or as the interval (– 3, 3).
75. First we solve the equation x 2 + x = 12 and use the solutions to separate the real number line. x 2 + x = 12
26
SECTION 0.5
x 2 + x − 12 = 0 ( x − 3) ( x + 4 ) = 0 x–3=0 x=3
or or
x+4=0 x=–4
We separate the number line into the intervals (– 4, 3) (3, ∞) (– ∞, – 4) We select a number in each interval and evaluate the expression x 2 + x − 12 at that value. We choose – 5, 0, 5. 2 For x = – 5: ( − 5 ) + ( − 5 ) − 12 = 8 , a positive number. For x = 0: For x = 5:
0 2 + 0 − 12 = − 12 , a negative number. 5 2 + 5 − 12 = 18 , a positive number.
The expression x 2 + x − 12 > 0 for x < – 4 or x > 3. We write the solution set either as {x | x < – 4 or x > 3} or as all x in the interval (– ∞, – 4) or (3, ∞).
77. First we solve the equation x ( x − 7 ) = − 12 and use the solutions to separate the real number line. x ( x − 7 ) = − 12
x 2 − 7 x = − 12 x 2 − 7 x + 12 = 0 ( x − 3) ( x − 4 ) = 0 x–3=0 x=3
or or
x–4=0 x=4
We separate the number line into the intervals (3, 4) (4, ∞) (– ∞, 3) We select a number in each interval and evaluate the expression x 2 − 7 x + 12 at that value. We choose 0, 3.5, and 5. For x = 0: 0 2 − 7 ( 0 ) + 12 = 12 , a positive number. For x = 3.5: For x = 5:
( 3.5 )
2
− 7 ( 3.5 ) + 12 = − 0.25 , a negative number.
5 − 7 ( 5 ) + 12 = 2 , a positive number. 2
The expression x 2 − 7 x + 12 > 0 for x < 3 or x > 4. We write the solution set either as {x | x < 3 or x > 4} or as all x in the interval (– ∞, 3) or (4, ∞).
SECTION 0.5
27
79. First we solve the equation 4 x 2 + 9 = 6 x and use the solutions to separate the real number line. 4x 2 + 9 = 6x 4x 2 − 6x + 9 = 0
This equation has no real solutions. Its discriminant, b 2 – 4ac = 36 – 144 = – 108, is negative. The value of 4 x 2 − 6 x + 9 either is always positive or always negative. To see 2 which is true, we test x = 0. Since 4 ( 0 ) − 6 ( 0 ) + 9 = 9 is positive, we conclude that expression is always positive, and the inequality 4 x 2 + 9 < 6 x has no solution. 81. First we solve the equation ( x − 1) ( x 2 + x + 1) = 0 and use the solutions to separate the
real number line.
( x − 1) ( x 2 + x + 1) = 0
x – 1 = 0 or x 2 + x + 1 = 0 x=1 x = 1 is the only solution, since the equation x 2 + x + 1 = 0 has a negative discriminant. We use x = 1 to separate the number line into two parts: and 1<x 1. The solution set is {x | x > 1}, or for all x in the interval (1, ∞).
83. First we solve the equation ( x − 1) ( x − 2 ) ( x − 3 ) = 0 and use the solutions to separate
the real number line.
( x − 1) ( x − 2 ) ( x − 3 ) = 0
x – 1 = 0 or x – 2 = 0 or x – 3 = 0 x = 1 or x = 2 or x=3
We separate the number line into the following 4 parts, choose a test number in each part, and evaluate the expression ( x − 1) ( x − 2 ) ( x − 3 ) at each test number. 1<x f ⎜ − ⎟ . 2 ⎝ 2⎠
43. (a) The average rate of change from 0 to 2 is 2 2 ∆y f ( 2 ) − f ( 0 ) ⎡⎣ −2 ⋅ 2 + 4 ⎤⎦ − ⎡⎣ −2 ⋅ 0 + 4 ⎤⎦ − 4 − 4 = = = = −4 2−0 2 2 ∆x
(b) The average rate of change from 1 to 3 is 2 2 ∆y f ( 3) − f (1) ⎡⎣ −2 ⋅ 3 + 4 ⎤⎦ − ⎡⎣ −2 ⋅ 1 + 4 ⎤⎦ −16 = = = =−8 3 −1 2 2 ∆x (c) The average rate of change from 1 to 4 is 2 2 ∆y f ( 4 ) − f (1) ⎡⎣ −2 ⋅ 4 + 4 ⎤⎦ − ⎡⎣ −2 ⋅ 1 + 4 ⎤⎦ − 30 = = = = − 10 4 −1 3 3 ∆x 45.
(a)
f ( x ) − f (1) 5 x − 5 (1) 5 ( x − 1) = =5 = x −1 x −1 x −1
(b) Using x = 2 in part (a), we get 5. This is the slope of the secant line containing the points (1, f (1)) and (2, f (2)). (c) msec = 5. Using the point-slope form of the line and the point (1, f (1)) = (1, 5), we get y – y1 = msec(x – x1) y − 5 = 5 ( x − 1) y = 5x – 5 + 5 y = 5x 47. (a)
f ( x ) − f (1) [1 − 3 x ] − [1 − 3] 3 − 3 x 3 (1 − x ) = = = = −3 x −1 x −1 x −1 x −1
80
SECTION 1.3
(b) Using x = 2 in part (a), we get – 3. This is the slope of the secant line containing the points (1, f (1)) and (2, f (2)). (c) msec = – 3. Using the point-slope form of the line and the point (1, f (1)) = (1, – 2), we get y – y1 = msec(x – x1) y − ( − 2 ) = − 3 ( x − 1)
y = − 3x + 3 − 2 y = − 3x + 1
] [ 2 ( )] [ 2 ] [ ] 2 ( ) [ 2 ( ) 49. (a) f x − f 1 = x − 2 x − 1 − 2 1 = x − 2 x − −1 = x − 2 x + 1 x −1 x −1 x −1 x −1 2 ( x − 1) = = x −1 x −1 (b) Using x = 2 in part (a), we get 1. This is the slope of the secant line containing the points (1, f (1)) and (2, f (2)). (c) msec = 1. Using the point-slope form of the line and the point (1, f (1)) = (1, – 1), we get y – y1 = msec(x – x1) y − ( −1) = 1( x − 1) y = x −1−1 y = x−2
( 2 ) ( ) [ 3 ] [ 3 ] [ 3 ] [ ] ( ) 51. (a) f x − f 1 = x − x − 1 − 1 = x − x − 0 = x x − 1 x −1 x −1 x −1 x −1 x ( x − 1) ( x + 1) = = x2 + x x −1 (b) Using x = 2 in part (a), we get 2 2 + 2 = 6. This is the slope of the secant line containing the points (1, f (1)) and (2, f (2)). (c) msec = 6. Using the point-slope form of the line and the point (1, f (1)) = (1, 0), we get y – y1 = msec(x – x1) y − 0 = 6 ( x − 1) y = 6x − 6
⎡ 2 ⎤ ⎡ 2 ⎤ ⎡ 2 ⎤ [] − −1 f ( x ) − f (1) ⎢⎣ x + 1 ⎥⎦ ⎢⎣1 + 1 ⎥⎦ ⎢⎣ x + 1 ⎥⎦ 2 − (1) ( x + 1) 2 − x −1 = = = = 53. (a) ( x − 1)( x + 1) ( x − 1)( x + 1) x −1 x −1 x −1 −1
−x +1 1 =− = ( x − 1) ( x + 1) x +1
SECTION 1.3
81
1 1 = − . This is the slope of the secant line 2 +1 3 containing the points (1, f (1)) and (2, f (2)). (b) Using x = 2 in part (a), we get −
(c) msec = −
1 . Using the point-slope form of the line and the point (1, f (1)) = (1, 1), we 3
get y – y1 = msec(x – x1) 1 y − 1 = − ( x − 1) 3 1 1 y = − x + +1 3 3 1 4 y =− x+ 3 3 55. (a)
f ( x ) − f (1) = x −1
x− 1 = x −1
x −1
(
(b) Using x = 2 in part (a), we get
x − 1) ( x + 1)
=
1 x +1
1 . This is the slope of the secant line containing 2 +1
the points (1, f (1)) and (2, f (2)). 1 2 +1 Using the point-slope form of the line and the point (1, f (1)) = (1, 1), we get y – y1 = msec(x – x1) 1 ( y −1 = x − 1) 2 +1 y ≈ 0.414 x − 0.414 + 1 y ≈ 0.414 x + 0.586 (c) msec =
57. To determine algebraically whether a function is even, odd, or neither we replace x with – x and simplify. 3 f ( − x ) = 4 ( − x ) = 4 ⋅ ( − x3 ) = − 4 x 3 = − f ( x )
Since f ( − x ) = − f ( x ) , the function is odd. 59. To determine algebraically whether a function is even, odd, or neither we replace x with – x and simplify. 2 g ( − x ) = − 3 ( − x ) − 5 = −3 x 2 − 5 = g ( x ) Since g ( − x ) = g ( x ) , the function is even. 61. To determine algebraically whether a function is even, odd, or neither we replace x with – x and simplify.
82
SECTION 1.3
F (−x) = 3 −x = − 3 x = − F ( x) Since F ( − x ) = − F ( x ) , the function is odd. 63. To determine algebraically whether a function is even, odd, or neither we replace x with – x and simplify. f (− x) = − x + − x = − x + x Since f (– x) equals neither f(x) nor – f(x), the function is neither even nor odd. 65. To determine algebraically whether a function is even, odd, or neither we replace x with – x and simplify. 1 1 = 2 = g ( x) g (− x) = 2 (− x) x
Since g ( − x ) = g ( x ) , the function is even. 67. To determine algebraically whether a function is even, odd, or neither we replace x with – x and simplify. 3 − (− x) ⎛ −x 3 ⎞ x3 h (− x) = = 2 = −⎜ 2 ⎟ = − h ( x) 2 ⎝ 3x − 9 ⎠ 3 ( − x ) − 9 3x − 9
Since h ( − x ) = − h ( x ) , the function is odd. 69.
Using 2nd, calculate, we find there is a local maximum at – 1. The local maximum is f (– 1) = 4. There is a local minimum at 1. The local minimum is f (1) = 0. The function is increasing on the intervals (–2, – 1) and (1, 2). The function is decreasing on the interval (– 1, 1).
71.
Using 2nd, CALCULATE, we find there is a local maximum at – 0.77. The local maximum is f (– 0,77) = 0.19. There is a local minimum at 0.77. The local minimum is f (0.77) = – 0.19. The function is increasing on the intervals (–2, – 0.77) and (0.77, 2). The function is decreasing on the interval (– 0.77, 0.77).
73.
Using 2nd, calculate, we find there is a local maximum at 1.77. The local maximum is f (1.77) = – 1.91. There is a local minimum at – 3.77. The local minimum is f (– 3.77) = –18.89.
SECTION 1.3
The function is increasing on the interval (– 3.77, 1.77). The function is decreasing on the intervals (– 6, – 3.77) and (1.77, 4).
75.
Using 2nd, calculate, we find there is a local maximum at 0. The local maximum is f (0) = 3. There is a local minimum at – 1.87. The local minimum is f (– 1.87) = 0.95. There is another local minimum at 0.97. The local minimum is f (0.97) = 2.65. The function is increasing on the intervals (– 1.87, 0) and (0.97, 2). The function is decreasing on the intervals (– 3, – 1.87) and (0, 0.97).
77.
The average rate of change of f from 0 to x is ∆y f ( x ) − f ( 0 ) x 2 − 0 = = =x x≠0 x−0 x ∆x (a) The average rate of change from 0 to 1 is 1. (b) The average rate of change from 0 to 0.5 is 0.5. (c) The average rate of change from 0 to 0.1 is 0.1. (d) The average rate of change from 0 to 0.01 is 0.01. (e) The average rate of change from 0 to 0.001 is 0.001. (g) and (h) Answers will vary.
(f)
83
84
SECTION 1.3
79.
−32 (100 ) (a) h (100 ) = + 100 = 81.07 feet high 1302 2 −32 ( 300 ) (b) h ( 300 ) = + 300 = 129.59 feet high 1302 ( 500 )2 − 32 + 500 = 26.63 feet high (c) h ( 500 ) = 1302 (d) To determine how far the ball was hit, we find h(x) = 0. −32 x 2 +x=0 130 2 −32 x 2 + 130 2 x = 0 x ( −32 x + 130 2 ) = 0 x = 0 or −32 x + 130 2 = 0 x = 0 or x = 528.125 The ball traveled 528.125 feet. 2
(e)
(f) The golf ball is 90 feet high after it has traveled about 115 feet and again when it has traveled about 413 feet. (g)
(h) The ball travels about 275 feet when it reaches its maximum height of about 131.8 feet.
SECTION 1.3
85
(i) The ball actually travels only 264 feet before reaching its maximum height of 132.03 feet.
81. (a) Volume of a box is the product of its length, width, and height. The box has height x inches and length and width equal to 24 – 2x inches. V = V(x) = x(24 -2x) 2 cubic inches
(b) If x = 3 inches, then V = 3(24 – 2 · 3) 2 = 972 cubic inches. (c) If x = 10 inches, then V = 10(24 – 2 · 10) 2 = 10(4) 2 = 160 cubic inches. (d)
83. (a)
(b) The average cost is minimized when 10 riding mowers are produced. (c) The average cost of producing each of the 10 mowers is $239. 85. Best matches are are (a) graph (II), (b) graph (V), (c) graph (IV), (d) graph (III), (e) graph (I). Reasons may vary. 87. Graphs will vary. 89. Explanations and graphs will vary. 91. Descriptions will vary. 93. Functions will vary.
86
SECTION 1.4
1.4 Library of Functions; Piecewise-defined Functions 1. (C) Graphs of square functions are parabolas.
3. (E) Graphs of square root functions are defined for x ≥ 0 and are increasing.
5. (B) Graphs of linear functions are straight lines.
7. (F) Graph of a reciprocal function is not defined at zero.
9.
11.
13.
15.
17. (a) When x = – 2, the equation for f is f (x) = x 2, so f (– 2) = (– 2) 2 = 4
(b) When x = 0, the equation for f is f (x) = 2, so f (0) = 2 (c) When x = 2, the equation for f is f (x) = 2x + 1, so f (2) = 2(2) + 1 = 5 19. (a) f (1.2) = int (2 · 1.2) = int (2.4) = 2
(b) f (1.6) = int (2 · 1.6) = int (3.2) = 3 (c) f (– 1.8) = int (2 · (– 1.8)) = int (– 3.6) = – 4 21. (a) The domain of f is all real numbers or the interval (– ∞, ∞).
(b) To find the y-intercept we let x = 0 and solve. When x = 0, the equation for f is f (0) = 1. So the y-intercept is (0, 1). To find the x-intercept, we let y = 0 and solve for x. f (x) never equals zero, so there is no x-intercept.
SECTION 1.4
(c)
87
(d) The range of f is the set {y | y ≠ 0}.
23. (a) The domain of f is all real numbers or the interval (– ∞, ∞).
(b) To find the y-intercept we let x = 0 and solve. When x = 0, the equation for f is f (x) = – 2x + 3 and f (0) = 3. So the y-intercept is (0, 3). To find the x-intercept, we let y = 0 and solve. If x < 1, then y = – 2x + 3 = 0 or 3 x = = 1.5 . 2 Since x = 1.5 is not less than 1, we ignore this solution. If x ≥ 1, then y = 3x – 2 = 0 2 or x = . 3 2 Since x = < 1, we ignore this solution and conclude f (x) never equals zero and there 3 is no x-intercept. (c) (d) The range of f is the set {y | y ≥ 1} or [1, ∞).
25. (a) The domain of f is the set {x | x ≥ – 2} or the interval [– 2, ∞).
(b) To find the y-intercept we let x = 0 and solve. When x = 0, the equation for f is f (x) = x + 3, and f (0) = 3. So the y-intercept is (0, 3). To find the x-intercept(s), we let y = 0 and solve. If – 2 ≤ x < 1, y = f (x) = x + 3 = 0 or x = – 3. Since x = – 3 is not in the interval [– 2, 1), we ignore this solution. If x > 1, y = – x + 2 = 0 or x = 2. So there is an x-intercept at (2, 0).
88
SECTION 1.4
(c)
(d) The range of f is the set {y | y = 5; y < 4}.
27. (a) The domain of f is all real numbers.
(b) To find the y-intercept we let x = 0 and solve. When x = 0, the equation for f is f (x) = x 2 , and f (0) = 0. So the y-intercept is (0, 0). To find the x-intercept(s), we let y = 0 and solve. If x < 0, y = f (x) = 1 + x = 0 or x = – 1. So there is an x-intercept at (– 1, 0). If x > 0, y = f (x) = x 2 = 0 or x = 0. So there is an x-intercept at (0, 0). (c)
(d) The range of f is the set of all real numbers.
29. (a) The domain of f is the set {x | x ≥ – 2} or the interval [– 2, ∞).
(b) To find the y-intercept we let x = 0 and solve. When x = 0, the equation for f is f (x) = 1, and f (0) = 1. So the y-intercept is (0, 1). To find the x-intercept(s), we let y = 0 and solve. If – 2 ≤ x < 0, y = f (x) = |x| = 0 or x = 0. Since x = 0 is not in the interval [– 2, 0), we ignore this solution. If x > 0, y = x 3 = 0 or x = 0. Since x = 0 is not in the interval (0, ∞), we also ignore this solution, and conclude there is no x-intercept. (c)
(d) The range of f is the set {y | y > 0} or the interval (0, ∞).
SECTION 1.4
89
31. (a) The domain of f is the set of all real numbers.
(b) To find the y-intercept we let x = 0 and solve. When x = 0, the equation for f is f (x) = 2 int (x), and f (0) = 0. So the y-intercept is (0, 0). To find the x-intercept(s), we let y = 0 and solve. y = f (x) = 2 int (x) = 0 for all x in the set {x | 0 ≤ x < 1} or the interval [0, 1). So there are x-intercepts at all points in the interval. (c)
(d) The range of f is the set of even integers.
33. This graph is made of two line segments. We find each. For – 1 ≤ x ≤ 0, we have two points (– 1, 1) and (0, 0). The slope of the line between the points is y −y 1− 0 = −1 m= 2 1 = x2 − x1 −1 − 0 Using the point-slope form of an equation of a line and the point (0, 0), we get y − y1 = m ( x − x1 )
y – 0 = – 1(x – 0) y=–x For 0 < x ≤ 2, we have two points (0, 0) and (2, 1). The slope of the line between the points is y −y 1− 0 1 m= 2 1 = = x2 − x1 2−0 2 Using the point-slope form of an equation of a line and the point (0, 0), we get y − y1 = m ( x − x1 ) 1 (x – 0) 2 1 y= x 2 The piecewise function is ⎧− x ⎪ f (x) = ⎨ 1 ⎪⎩ 2 x y–0=
−1 ≤ x ≤ 0 0< x≤2
90
SECTION 1.4
35. This graph is made of two line segments. We find each. For x ≤ 0, we have two points (– 1, 1) and (0, 0). The slope of the line between the points is y −y 1− 0 m= 2 1 = = −1 x2 − x1 −1 − 0 Using the point-slope form of an equation of a line and the point (0, 0), we get y − y1 = m ( x − x1 )
y – 0 = – 1(x – 0) y=–x For 0 < x ≤ 2, we have two points (0, 2) and (2, 0). The slope of the line between the points is y −y 2−0 m= 2 1 = = −1 x2 − x1 0−2 Using the point-slope form of an equation of a line and the point (0, 2), we get y − y1 = m ( x − x1 )
y – 2 = – 1(x – 0) y=–x+2 The piecewise function is ⎧− x f (x) = ⎨ ⎩− x + 2
x≤0 0< x≤2
37. (a) If x = 200 minutes are used, the equation for C is C(x) = 39.99. So C(200) = $39.99.
(b) If x = 365 minutes are used, the equation for C is C(x) = 0.25x – 47.51. So C(365) = 0.25(365) – 47.51 = $43.74. (c) If x = 351 minutes are used, the equation for C is C(x) = 0.25x – 47.51. So C(351) = 0.25(351) – 47.51 = $40.24. 39. (a) If 50 therms are used, the charge C will be C(50) = 9.45 + 0.36375(50) + 0.6338(50) = $59.33
(b) If 500 therms are used the charge will be C(500) = 9.45 + 0.36375(50) + 0.11445(500 – 50) + 0.6338(500) = $396.04 (c) When 0 ≤ x ≤ 50 the equation that relates the monthly charge for using x therms of gas is C(x) = 9.45 + 0.36375x + 0.6338x = 0.99755x + 9.45 When x > 50 the equation that relates the monthly charge for using x therms of gas is C(x) = 9.45 + 0.36375(50) + 0.11445(x – 50) + 0.6338x = 21.915 + 0.74825 x The function that relates the monthly charge C for x therms of gas is 0 ≤ x ≤ 50 ⎧9.45 + 0.99755 x C(x) = ⎨ x > 50 ⎩21.915 + 0.74825 x
SECTION 1.4
91
(d)
41. (a) When v = 1 meter per second and t = 10º, the equation representing the wind chill is W = t = 10º.
(b) When v = 5 m/sec. and t = 10º, the equation representing the wind chill is (10.45 + 10 v − v ) ( 33 − t ) (10.45 + 10 5 − 5) ( 33 − 10 ) W = 33 − = 33 − 22.04 22.04 ( 5.45 + 10 5 ) ( 23) = 33 − = 3.98 º 22.04 (c) When v = 15 m/sec. and t = 10º, the equation representing the wind chill is (10.45 + 10 v − v ) ( 33 − t ) (10.45 + 10 15 − 15) ( 33 − 10 ) = 33 − W = 33 − 22.04 22.04 (10 15 − 4.55) ( 23) = 33 − = − 2.67 º 22.04 (d) When v = 25 m/sec. and t = 10º, the equation representing the wind chill is W = 33 – 1.5958(33 – t) = 33 – 1.5958(33 – 10) = 33 – 1.5958(23) = – 3.70º (e) Answers may vary. (f) Answers may vary. 43.
0.10 x ⎧ ⎪ 700 + 0.15 x − 7000 ( ) ⎪ ⎪⎪ 3910 + 0.25 ( x − 28, 400 ) y = f ( x) = ⎨ ⎪ 14, 010 + 0.28 ( x − 68,800 ) ⎪34,926 + 0.33 ( x − 143,500 ) ⎪ ⎪⎩90,514 + 0.35 ( x − 311,950 )
0 < x < 7000 7000 ≤ x < 28, 400 28, 400 ≤ x < 68,800 68,800 ≤ x < 143,500 143,500 ≤ x < 311,950 x ≥ 311,950
92
SECTION 1.4
45.
Answers may vary. 47.
Answers may vary. 49.
Answers may vary. 51.
Answers may vary.
SECTION 1.5
93
1.5 Graphing Techniques: Shifts and Reflections 1. (B) This is the graph of a square function reflected around the x-axis and shifted up two units. 3. (H) This is the graph of an absolute value function shifted to the left two units and then reflected over the x-axis. 5. (A) This is the graph of a square function shifted up two units. 7. (F) This is the graph of a square function shifted to the left two units and then reflected over the x-axis. 9. y = (x – 4) 3
11. y = x 3 + 4
13. y = (– x) 3 = – x 3 15. (1) Shift up 2 units.
x +2
Add 2.
y=
(2) Reflect about the x-axis.
Multiply y by – 1.
y=–
(3) Reflect about the y-axis.
Replace x by – x.
y = − −x −2
Multiply y by – 1. Add 2. Replace x by x + 3.
y= – y= – y= –
17. (1) Reflect about the x-axis. (2) Shift up 2 units. (3) Shift left 3 units.
(
x + 2) = − x − 2
x x +2 x+3 + 2
19. Ans: (c) –f (x) changes the sign of y. So if (3, 0) is a point on f, (3, 0) is a point on – f(x). 21.
(a) y = x
2
→ Subtract 1; vertical shift down 1 unit.
(b) y = x 2 − 1
94
SECTION 1.5
23.
(a) y = x
3
→ Add 1; vertical shift up 1 unit.
(b) y = x 3 + 1
25.
(a) y = x
→ Replace x by x – 2; (b) horizontal shift right 2 units.
y = x−2
27.
(a) y = x
3
→ Replace x by x – 1; horizontal shift right 1 unit.
(b)
y = ( x − 1)
3
SECTION 1.5
→ Add 2: vertical shift up 2 units.
3 (c) y = ( x − 1) + 2
29.
(a)
y= x 3
→ Multiply f (x) by – 1; reflect about the x-axis.
(b)
31.
(a)
y= x
→ Replace x by – x; reflect about y-axis.
(b)
y= − x
y=−3 x
95
96
SECTION 1.5
33.
(a) y = x
3
→ Add 2; vertical shift 2 units.
→ Multiply f(x) by – x; reflect about y-axis.
(c)
(b)
y = − x3
y = − x3 + 2
35.
(a)
y= x
→ Replace x by x – 2; horizontal shift right 2 units.
(b)
y = x−2
SECTION 1.5
→ Add 1; vertical shift up 1 unit.
(c)
y = x − 2 +1
37.
(a)
y=
x
→ Subtract 2; vertical down 2 units.
→ Replace x by – x; reflect about y-axis.
(c)
(b)
y = −x
y = −x −2
39.
(a) F ( x ) = f ( x ) + 3
(b)
G ( x ) = f ( x + 2)
97
98
SECTION 1.5
(c)
P ( x) = − f ( x)
(e)
g ( x) = f (− x)
41. (a)
(d)
H ( x ) = f ( x + 1) − 2
(b)
(c)
(d) Answers may vary.
43.
(a)
y = f ( x)
(b)
y= f
(x)
CHAPTER 1 REVIEW
99
Chapter 1 Review TRUE-FALSE ITEMS 1. False
3. False
5. True FILL IN THE BLANKS 1. independent; dependent
3. 5; – 3
5. (– 5, 0), (– 2, 0), (2, 0) REVIEW EXERCISES 1.
x
x2 + 4
−2
( −2 )2 + 4 = 8
−1 0
( −1)
2
+4=5 0 +4=4 2
1
12 + 4 = 5
2
22 + 4 = 8
( x, y ) ( − 2, 8) ( −1, 5) ( 0, 4 ) (1, 5) ( 2, 8)
3. To find the x-intercept(s) we let y = 0 and solve the equation 2x = 3 ⋅ 0 2 x=0 So the x-intercept is (0, 0).
To find the y-intercept(s) we let x = 0 and solve the equation 2 ⋅ 0 = 3y 2 y=0 So the y-intercept is (0, 0). To test the graph of the equation 2x = 3y 2 for symmetry with respect to the x-axis, we replace y by – y in the equation. 2x = 3y 2 2x = 3(– y) 2 2x = 3y 2 Since the resulting equation is equivalent to the original equation, the graph is symmetric with respect to the x-axis. To test the graph of the equation 2x = 3y 2 for symmetry with respect to the y-axis, we replace x by – x in the equation and simplify.
100 CHAPTER 1 REVIEW
2x = 3y 2 2(– x) = 3y 2 – 2x = 3y 2 Since the resulting equation is not equivalent to the original equation, the graph is not symmetric with respect to the y-axis. To test the graph of the equation 2x = 3y 2 for symmetry with respect to the origin, we replace x by – x and y by – y in the equation and simplify. 2(– x) = 3(– y) 2 – 2x = 3y 2 Since the resulting equation is not equivalent to the original equation, the graph is not symmetric with respect to the origin. 5. To find the x-intercept(s) we let y = 0 and solve the equation x 2 + 4 · 0 2 = 16 x 2 = 16 x = – 4 or x = 4 So the x-intercepts are (– 4, 0) and (4, 0).
To find the y-intercept(s) we let x = 0 and solve the equation 0 2 + 4y 2 = 16 4y 2 = 16 y2 = 4 y = – 2 or y = 2 So the y-intercepts are (0, – 2) and (0, 2). To test the graph of the equation x 2 + 4y 2 = 16 for symmetry with respect to the x-axis, we replace y by – y in the equation. x 2 + 4y 2 = 16 2 x + 4(– y) 2 = 16 x 2 + 4y 2 = 16 Since the resulting equation is equivalent to the original equation, the graph is symmetric with respect to the x-axis. To test the graph of the equation x 2 + 4y 2 = 16 for symmetry with respect to the y-axis, we replace x by – x in the equation and simplify. x 2 + 4y 2 = 16 (– x) 2 + 4y 2 = 16 x 2 + 4y 2 = 16 Since the resulting equation is equivalent to the original equation, the graph is symmetric with respect to the y-axis. To test the graph of the equation x 2 + 4y 2 = 16 for symmetry with respect to the origin, we replace x by – x and y by – y in the equation and simplify. x 2 + 4y 2 = 16 2 (– x) + 4(– y) 2 = 16 x 2 + 4y 2 = 16
CHAPTER 1 REVIEW 101
Since the resulting equation is equivalent to the original equation, the graph is symmetric with respect to the origin. 7. To find the x-intercept(s) we let y = 0 and solve the equation x 4 + 2x 2 + 1 = 0 ( x 2 + 1)( x 2 + 1) = 0 x2 + 1 = 0 has no real solution. So there is no x-intercept.
To find the y-intercept(s) we let x = 0 and solve the equation 04 + 2 ⋅ 02 +1 = y y=1 So the y-intercept is (0, 1). To test the graph of the equation x 4 + 2 x 2 + 1 = y for symmetry with respect to the x-axis, we replace y by – y in the equation. x 4 + 2x 2 + 1 = − y Since the resulting equation is not equivalent to the original equation, the graph is not symmetric with respect to the x-axis. To test the graph of the equation x 4 + 2 x 2 + 1 = y for symmetry with respect to the y-axis, we replace x by – x in the equation and simplify. 4 2 (− x) + 2 (− x) +1 = y x 4 + 2x 2 + 1 = y Since the resulting equation is equivalent to the original equation, the graph is symmetric with respect to the y-axis. To test the graph of the equation x 4 + 2 x 2 + 1 = y for symmetry with respect to the origin, we replace x by – x and y by – y in the equation and simplify. 4 2 (− x) + 2 (− x) +1 = − y x 4 + 2x 2 + 1 = − y Since the resulting equation is not equivalent to the original equation, the graph is not symmetric with respect to the origin. 9. To find the x-intercept(s) we let y = 0 and solve the equation x2 + x + 02 + 2 ⋅ 0 = 0 x 2 + x = x ( x + 1) = 0 x = 0 or x + 1 = 0 x=–1 So the x-intercepts are (– 1, 0) and (0, 0).
To find the y-intercept(s) we let x = 0 and solve the equation 02 + 0 + y 2 + 2y = 0 y 2 + 2 y = y ( y + 2) = 0
102 CHAPTER 1 REVIEW
y = 0 or y + 2 = 0 y=–2 So the y-intercepts are (0, – 2) and (0, 0). To test the graph of the equation x 2 + x + y 2 + 2 y = 0 for symmetry with respect to the x-axis, we replace y by – y in the equation. 2 x 2 + x + (− y) + 2(− y) = 0 x2 + x + y 2 − 2y = 0 Since the resulting equation is not equivalent to the original equation, the graph is not symmetric with respect to the x-axis. To test the graph of the equation x 2 + x + y 2 + 2 y = 0 for symmetry with respect to the y-axis, we replace x by – x in the equation and simplify. 2 (− x) + (− x) + y 2 + 2 y = 0 x2 − x + y 2 + 2y = 0 Since the resulting equation is not equivalent to the original equation, the graph is not symmetric with respect to the y-axis. To test the graph of the equation x 2 + x + y 2 + 2 y = 0 for symmetry with respect to the origin, we replace x by – x and y by – y in the equation and simplify. 2 2 (− x) + (− x) + (− y) + 2 (− y) = 0 x2 − x + y 2 − 2y = 0 Since the resulting equation is not equivalent to the original equation, the graph is not symmetric with respect to the origin. 11. (a) We substitute 2 for x in the equation for f to get 3( 2) 6 f ( 2) = 2 = =2 2 −1 3
(b) We substitute – 2 for x in the equation for f to get 3( − 2) −6 = = −2 f ( − 2) = 2 ( − 2) − 1 3 (c) We substitute (– x) for x in the equation for f to get 3( − x) − 3x = 2 f (− x) = 2 (− x) −1 x −1 (d) – f (x) = −
3x x 2 −1
(e) We substitute (x – 2) for x in the equation for f to get 3( x − 2) 3x − 6 3x − 6 = 2 = 2 f ( x − 2) = 2 ( x − 2) − 1 x − 4 x + 4 − 1 x − 4x + 3
CHAPTER 1 REVIEW 103
(f) We substitute 2x for x in the equation for f to get 3( 2x) 6x f (2x) = = 2 2 ( 2x) −1 4x −1 13. (a) We substitute 2 for x in the equation for f to get f ( 2) = 2 2 − 4 = 0
(b) We substitute – 2 for x in the equation for f to get f ( − 2) =
( − 2)
2
−4 =0
(c) We substitute (– x) for x in the equation for f to get f (− x) =
(− x)
2
− 4 = x 2 − 4 = f ( x)
(d) – f (x) = − x 2 − 4 (e) We substitute (x – 2) for x in the equation for f to get f ( x − 2) =
( x − 2)
2
− 4 = x 2 − 4x + 4 − 4 = x 2 − 4x
(f) We substitute 2x for x in the equation for f to get 2 f (2x) = ( 2x) − 4 = 4x 2 − 4 = 2 x 2 −1
15.
(a) We substitute 2 for x in the equation for f to get 22 − 4 ( ) f 2 = =0 22 (b) We substitute – 2 for x in the equation for f to get 2 − 2) − 4 ( f ( − 2) = =0 2 ( − 2) (c) We substitute (– x) for x in the equation for f to get 2 − x) − 4 x 2 − 4 ( f (− x) = = = f ( x) 2 x2 (− x) (d) – f (x) = −
x2 − 4 x2
(e) We substitute (x – 2) for x in the equation for f to get 2 x − 2) − 4 x 2 − 4 x + 4 − 4 ( x 2 − 4x f ( x − 2) = = = 2 2 x 2 − 4x + 4 x − 4x + 4 ( x − 2) (f) We substitute 2x for x in the equation for f to get
104 CHAPTER 1 REVIEW
( 2x) 2 − 4 4x 2 − 4 x 2 −1 f (2x) = = = 4x 2 x2 ( 2x) 2 17. The denominator of the f cannot equal 0, so x 2 – 9 ≠ 0 or x ≠ 3 and x ≠ – 3. The domain is the set {x | x ≠ 3 and x ≠ – 3}. 19. The radicand must be nonnegative, so 2 – x ≥ 0 or x ≤ 2. The domain is the set {x | x ≤ 2} or the interval (– ∞, – 2]. 21. The radicand must be nonnegative and the denominator cannot equal 0. The domain is the set {x | x > 0} or the interval (0, ∞). 23. The denominator of the function f cannot equal 0, so x 2 + 2x – 3 ≠ 0. x 2 + 2x – 3 = 0 (x – 1)(x + 3) = 0 x – 1 = 0 or x + 3 = 0 x = 1 or x = – 3 The domain is the set {x | x ≠ 1 and x ≠ – 3}. 25.
2 f ( x + h ) − f ( x ) − 2 ( x + h ) + ( x + h ) + 1 − [−2 x 2 + x + 1] = h h 2 2 − 2 x − 4 xh − 2h + x + h + 1 + 2 x 2 − x − 1 = h 2 −4 xh − 2h + h = h h ( −4 x − 2h + 1) = h = −4 x − 2 h + 1
27. (a) The domain of f is the set {x | – 4 ≤ x ≤ 3} or the interval [– 4, 3]. The range of f is the set {y | – 3 ≤ y ≤ 3} or the interval [– 3, 3].
(b) The x-intercept is (0, 0), and the y-intercept is (0, 0). (c) f (– 2) = – 1 (d) f (x) = – 3 when x = – 4. (e) f (x) > 0 on the interval (0, 3]. 29. (a) The domain of f is the set of all real numbers or the interval (– ∞, ∞). The range of f is the set {y | y ≤ 1} or the interval (– ∞, 1].
(b) f is increasing on the intervals (– ∞, – 1) and (3, 4); f is decreasing on the intervals (– 1, 3) and (4, ∞).
CHAPTER 1 REVIEW 105
(c) The local maxima are 1 at f (– 1) = 1 and 0 at f (4) = 0. There is a local minimum of – 3 at f (3) = – 3. (d) The graph is not symmetric with respect to the x-axis, y-axis, or the origin. (e) Since the graph of the function has no symmetry, the function is neither even nor odd. (f) The x-intercepts are (– 2, 0), (0, 0) and (4, 0); the y-intercept is (0, 0). 31.
33.
f ( − x ) = (− x) 3 − 4 ( − x ) = − x 3 + 4 x = − f ( x ) Since f ( − x ) = − f ( x ) , the function is odd. h (− x) =
1
(− x)
4
+
1
(− x)
2
+1 =
1 1 + 2 + 1 = h ( x ) Since h ( − x ) = h ( x ) , the function is 4 x x
even. 35.
37.
G ( − x ) = 1 − ( − x ) + ( − x ) = 1 + x − x 3 Since G(– x) is not equal to G(x) or to – G(x), the function is neither even nor odd. 3
(− x) = − x = − f ( x) 2 1+ x 2 1+ (− x) f ( − x ) = − f ( x ) , the function is odd.
f (− x) =
Since
39. There is a local maximum of 4.043 at x = – 0.913, and a local minimum of – 2.043 at x = 0.913.
The function is increasing on the intervals (– 3, – 0.913) and (0.913, 3). The function is decreasing on the interval (– 0.913, 0.913).
41. There is a local maximum of 1.532 at x = 0.414, and local minima of 0.543 at x = – 0.336 and of – 3.565 at x = 1.798.
The function is increasing on the intervals (– 0.336, 0.414) and (1.798, 3). The function is decreasing on the intervals (– 2, – 0.336) and (0.414, 1.798). 43.
(a)
2 2 ∆y f ( 2 ) − f (1) [8 ⋅ 2 − 2] − ⎡⎣8 ⋅ 1 − 1⎤⎦ = = 30 − 7 = 23 = ∆x 2 −1 1
106 CHAPTER 1 REVIEW
(b)
2 2 ∆y f (1) − f ( 0 ) [8 ⋅ 1 − 1] − ⎡⎣8 ⋅ 0 − 0 ⎤⎦ = = 7−0 = 7 = ∆x 1− 0 1
(c)
2 2 ∆y f ( 4 ) − f ( 2 ) [8 ⋅ 4 − 4] − ⎡⎣8 ⋅ 2 − 2 ⎤⎦ 124 − 30 = = = = 47 ∆x 4−2 2 2
45.
∆y f ( x ) − f ( 2 ) [ 2 − 5 x ] − [ 2 − 5 ⋅ 2] 2 − 5 x − 2 + 10 = = = x−2 x−2 ∆x x−2 − 5 x + 10 − 5 ( x − 2 ) = = = −5 x−2 x−2
47.
2 2 ∆y f ( x ) − f ( 2 ) [3x − 4 x ] − ⎡⎣3 ⋅ 2 − 4 ⋅ 2 ⎤⎦ 3x − 4 x 2 − 6 + 16 = = = ∆x x−2 x−2 x−2 2 − 4 x + 3x + 10 − ( 4 x + 5 ) ( x − 2 ) = = = − 4x − 5 x−2 x−2
49. (b), (c), (d), and (e) are graphs of functions because they pass the vertical line test. 51.
53. (a) The domain of f is the set {x | x > – 2} or the interval (– 2, ∞).
(b) To find the y-intercept we let x = 0 and solve. When x = 0, the equation for f is f (x) = 3x, and f (0) = 0. So the y-intercept is (0, 0). To find the x-intercept(s), we let y = 0 and solve. If – 2 ≤ x < 1, y = f (x) = 3x = 0 or x = 0. So there is an x-intercept at (0, 0) If x > 1, y = x + 1 = 0 or x = – 1. Since x = – 1 is not in the interval (1, ∞), we ignore this solution.
CHAPTER 1 REVIEW 107
(d) The range of f is the set {y | y > – 6}.
(c)
55. (a) The domain of f is the set {x | x ≥ – 4} or the interval [– 4, ∞).
(b) To find the y-intercept we let x = 0 and solve. When x = 0, the equation for f is f (x) = 1. So the y-intercept is (0, 1). To find the x-intercept(s), we let y = 0 and solve. If – 4 ≤ x < 0, y = f (x) = x = 0. Since x = 0 is not in the interval [– 4, 0), we ignore this solution. If x > 0, y = 3x = 0 or x = 0. Since x = 0 is not in the interval (0, ∞), we ignore this solution and we conclude that there is no x-intercept on this graph. (d) The range of f is the set {y | y ≥ – 4, but y ≠ 0} or the interval [– 4, 0) and (0, ∞).
(c)
57.
(a) F ( y ) = x
→ Subtract 4; vertical shift down 4 units.
(b)
F ( x) = x − 4
The x-intercepts of F are (−4, 0) and (4, 0); the y-intercept is (0, −4). The domain is all real numbers and the range is {y | y ≥ −4}.
108 CHAPTER 1 REVIEW 59.
→ Replace x with x – 1;
(a) y = x
(b) F ( x ) = x − 1
horizontal shift right 1 unit. The x-intercept of F is (1, 0); there is no y-intercept. The domain of F is the interval [1, ∞) and the range is [0, ∞). 61.
→ Replace x with x – 1;
(a) y = x
(b)
f ( x) = x −1
horizontal shift right 1 unit.
→ Replace x – 1 by – (x – 1);
(c)
f ( x) = 1 − x
reflect about the y-axis. The x-intercept of f is (1, 0); the y-intercept is (0, 1). The domain of f is the set {x | x ≤ 1} or the interval [−∞, 1). The range is the set {y | y ≥ 0} or the interval [0, ∞).
CHAPTER 1 REVIEW 109 63.
(a) y = x2
→ Add – 2; vertical
→ Replace x with x − 1; (b) y = (x − 1)2 horizontal shift right 1 unit.
(c)
f ( x ) = ( x − 1) + 2 2
shift up 2 units. The y-intercept is (0, 3); there is no x-intercept. The domain of f is all real numbers. The range is the set {y | y ≥ 2} or the interval [2, ∞). 65.
(a)
y = f (– x)
(b) y = – f (x)
110 CHAPTER 1 REVIEW
(c) y = f (x + 2)
67.
69.
71.
(d) y = f (x) + 2
Since f is linear, we first find the slope of the function using the 2 given points (4, – 5) and (0, 3). y − y1 − 5 − 3 − 8 = = = −2 m= 2 x2 − x1 4−0 4 Then we get the point-slope form of a line by using m and the point (0, 3). y – y1 = m(x – x1) y – 3 = – 2(x – 0) y = – 2x + 3 So the linear function is f (x) = – 2x + 3. f (1) = 4 means the point (1, 4) satisfies the equation for f. So A (1) + 5 A + 5 4= = 6 (1) − 2 4 16 = A + 5 A = 11 Since the height is twice the radius, we can write h = 2r. Then the volume of the cylinder can be expressed as V = π r 2 ( 2r ) = 2π r 3 .
73. (a) R = R(x) = xp ⎛ 1 ⎞ R(x) = x ⎜ − x + 100 ⎟ ⎝ 6 ⎠ 1 2 = − x + 100 x 6
0 ≤ x ≤ 600
(b) If 200 units are sold, x = 200 and the revenue is 1 2 R ( 200 ) = − ( 200 ) + 100 ( 200 ) = $13,333.33 6 75. (a) To find the revenue function R = R(x), we first solve the demand equation for p. x = −5 p + 100 (1) 5 p = 100 − x 1 p = 20 − x 0 ≤ x ≤ 100 5 We find the domain by using equation (1) and solving
CHAPTER 1 REVIEW 111
when p = 0, x = – 5(0) + 100 = 100 when p = 20, x = – 5(20) + 100 = 0 So the revenue R can be expressed as 1 ⎞ ⎛ R ( x ) = x ⎜ 20 − x ⎟ 5 ⎠ ⎝ 1 0 ≤ x ≤ 100 R ( x ) = 20 x − x 2 5 (b) If 15 units are sold, x = 15, and the revenue is 1 2 R (15 ) = 20 (15 ) − (15 ) = $255.00 5 77. (a) The cost of making the drum is the sum of the costs of making the top and bottom and the side. The amount of material used in the top and bottom is the area of the two circles, Atop + Abottom = π r 2 + π r 2 = 2π r 2 square centimeters. At $0.06 per square centimeter, the cost of the top and the bottom of the drum is C = 0.06(2π r 2) = 0.12π r 2 dollars.
The amount of material used in the side of the drum is the area of the rectangle of material measured by the circumference of the top and the height of the drum. Aside = 2π r h square centimeters. To express the area of the side as a function of r, we use the fact that we are told the volume of the drum is 500 cubic centimeters. V =π r 2h = 500 500 h= πr 2 ⎛ 500 ⎞ 1000 So Aside = 2π r h = 2π r ⎜ 2 ⎟ = square centimeters. r ⎝πr ⎠ At $0.04 per square centimeter, the cost of making the side of the drum is ⎛ 1000 ⎞ 40 C = 0.04 ⎜ dollars. ⎟= ⎝ r ⎠ r 40 dollars. The total cost of making the drum is C = C(r) = 0.12π r 2 + r (b) If the radius is 4 cm, the cost of making the drum is 40 = $16.03 C(4) = 0.12π (4) 2 + 4 (c) If the radius is 8 cm, the cost of making the drum is 40 = $29.13 C(8) = 0.12π (8) 2 + 8
112 CHAPTER 1 REVIEW
(d)
Making the can with a radius of 3.758 centimeters minimizes the cost of making the drum. The minimum cost is $15.97.
Chapter 1 Project 1. Since Avis has unlimited mileage, the cost of driving an Avis car x miles is A = A(x) = 64.99 A is a constant function. 3.
If you drive more than 226 miles, Avis becomes the better choice. 5. If driving fewer than 130 miles, SaveALot Car Rental costs the least. If driving between 130 and 227 miles Enterprise is the most economical. If driving more than 227 miles Avis is the best buy. 7. If driving fewer than 53 miles, USave Car Rental is the cheapest. If driving between 53 and 130 miles, SaveALot Car Rental costs the least. If driving between 130 and 227 miles Enterprise is the most economical. If driving more than 227 miles Avis is the best buy.
MATHEMATICAL QUESTIONS 113
Mathematical Questions from Professional Exams 1. (d) On the open interval (–1, 2), The minimum value of f is 0, and the maximum value of f approaches 4. So the range of f is 0 ≤ y < 4. 3. (c) The domain of f is nonnegative. That is, x3 − x ≥ 0 x ( x 2 − 1) ≥ 0 x ( x − 1)( x + 1) ≥ 0
Solving the related equation for 0 and testing a point in each interval gives x ( x − 1)( x + 1) = 0 x = 0 or
x = 1 or x = – 1
When x = – 2, (– 2) 3 – (– 2) = – 6 < 0, so the interval (– ∞, – 1) is not part of the domain. When x = – 0.5, (– 0.5) 3 – (– 0.5) = 0.375 ≥ 0, so the interval [– 1, 0] is part of the domain. When x = 0.5, (0.5) 3 – ( 0.5) = – 0.375 < 0, so the interval (0, 1) is not part of the domain. When x = 2, (2) 3 – (2) = 6 ≥ 0, so the interval [1, ∞) is part of the domain.
Return to the function editor and enter y = x 3 + x into y1 and graph the equation.
(d)
What do you conclude about the relationship between the graphs of y = f ( x ) and y = f ( x ) ?
The graphs of y = f ( x ) and y = f ( x ) are identical on the x-intervals where f ( x ) ≥ 0 . However, for any x-interval where f ( x ) < 0 , the graph of y = f ( x ) is a reflection of the graph of y = f ( x ) about the x-axis. ________________________________________________________________________
Chapter 1 Review In Problems 39–42, use a graphing utility to graph each function over the indicated interval. Approximate any local maxima and local minima. Determine where the function is increasing and where it is decreasing.
39.
f ( x ) = 2 x3 − 5 x + 1
( −3, 3)
49
Enter the formula for f in the function editor. Go to WINDOW and enter limits for x. We must determine the limits for y so let’s try −10 ≤ y ≤ 10 . If this does not work, then we can go back to WINDOW and adjust the limits on y until we find a good window.
Graph the function.
This window is suitable. Use maximum to find the local maximum. Note that the x-coordinate of the local maximum is between x = −2 and x = 0 .
50
The local maximum is approximately ( −0.913, 4.043) . Use minimum to find the local minimum. Note that the x-coordinate of the local minimum is between x = 0 and x = 2 .
The local minimum is approximately ( 0.913, − 2.043) . Thus, f is increasing on the intervals ( −3, − 0.913) and ( 0.913, 3) , and f is decreasing on
the interval ( −0.913, 0.913) .
41.
f ( x ) = 2 x 4 − 5 x3 + 2 x + 1
( −2, 3)
Enter the formula for f in the function editor. Go to WINDOW and enter limits for x. We must determine limits for y, so try −10 ≤ y ≤ 10 . If this does not work, then we can go back to WINDOW and adjust the limits on y until we find a good window.
51
Graph the function.
This window is suitable. Use maximum to find the local maximum. Note that the x-coordinate of the local maximum is between x = 0 and x = 1 .
The local maximum is approximately ( 0.414, 1.532 ) . Use minimum to find both local minima. Note that the x-coordinate of the first local minimum is between x = −1 and x = 0 .
52
The first local minimum is approximately ( −0.336, 0.543) . The x-coordinate of the second local minimum is between x = 1 and x = 3 .
The second local minimum is approximately (1.798, − 3.565 ) . Thus, f is increasing on the intervals ( −0.336, 0.414 ) and (1.798, 3) , and f is decreasing on the intervals ( −2, − 0.336 ) and ( 0.414,1.798 ) .
77.
Cost of a Drum A drum in the shape of a right circular cylinder is required to have a volume of 500 cubic centimeters. The top and bottom are made of material that costs 6¢ per square centimeter; the sides are made of material that costs 4¢ per square centimeter. Hint: The volume V of a right circular cylinder of height h and radius r is V = π r 2 h .
(d)
Graph C = C ( r ) . For what value of r is the cost C least? 53
40 . Enter the formula for C in the r function editor. Be sure to use X for r when you enter the equation. Go to WINDOW and enter limits for x and y. We are not given any limits for either variable, although for this problem to make sense we must have x > 0 . Let’s try the following window.
The cost function is given by C ( r ) = 0.12π r 2 +
Graph C.
Find the local minimum. Notice that the minimum is between x = 1 and x = 10 .
54
The local minimum is approximately ( 3.76, 15.97 ) . If r is approximately 3.76 centimeters, then the cost C will be minimized.
55
Summary The commands introduced in this chapter are: abs(value) intersect maximum minimum Table
56
Chapter 1 – Functions and Their Graphs Section 1.1 Graphs of Equations In Problems 11–22, the graph of an equation is given. (a) List the intercepts of the graph. (b) Based on the graph, tell whether the graph is symmetric with respect to the x-axis, the y-axis, and/or the origin.
19.
(a)
The x-intercepts are (1, 0 ) and ( −1, 0 ) , and the y-intercept is ( 0, − 1) .
(b)
The graph is symmetric with respect to the y-axis.
21.
17
(a)
This graph does not have any x-intercepts or y-intercepts.
(b)
The graph is symmetric with respect to the origin.
________________________________________________________________________
Section 1.3 Graph of a Function; Properties of Functions In Problems 69–76, use a graphing utility to graph each function over the indicated interval and approximate any local maxima and local minima. Determine where the function is increasing and where it is decreasing. Round answers to two decimal places.
69.
f ( x ) = x3 − 3x + 2
( −2, 2 )
Enter the formula for f in the function editor. Go to WINDOW and enter limits for x. We must determine limits for y, so try −10 ≤ y ≤ 10 . If this does not work, then we can go back to WINDOW and adjust the limits on y until we find a good window.
Graph the function.
This window is suitable.
18
(a)
This graph does not have any x-intercepts or y-intercepts.
(b)
The graph is symmetric with respect to the origin.
________________________________________________________________________
Section 1.3 Graph of a Function; Properties of Functions In Problems 69–76, use a graphing utility to graph each function over the indicated interval and approximate any local maxima and local minima. Determine where the function is increasing and where it is decreasing. Round answers to two decimal places.
69.
f ( x ) = x3 − 3x + 2
( −2, 2 )
Enter the formula for f in the function editor. Go to WINDOW and enter limits for x. We must determine limits for y, so try −10 ≤ y ≤ 10 . If this does not work, then we can go back to WINDOW and adjust the limits on y until we find a good window.
Graph the function.
This window is suitable.
18
You can use your TI-83 Plus to find the x- and y-coordinates of any local maxima and/or local minima using the maximum or minimum functions respectively. Both the maximum and minimum functions require three inputs: an x-value to the left of the maxima or minima; an x-value to the right of the maxima or minima; and an estimate of the x-value of the maxima or minima. Both maximum and minimum can be found under the [CALC] menu. Find the local maximum.
y r
Select maximum. Enter a value for x that is less than (to the left of) the x–coordinate of the local maximum. Notice that the local maximum is between x = −2 and x = 0 , so we can use x = −2 as a left bound.
¶ Ì Á
Input x = 0 as a right bound.
Í Ê
19
Input x = −1 as a guess.
Í Ì À
Press Í to find the local maximum.
Í
Thus the local maximum is ( −1, 4 ) .
Find the local minimum.
y r
Select minimum. Enter a value for x that is less than (to the left of) the x–coordinate of the local minimum. Notice that the local minimum is between x = 0 and x = 2 , so we can use x = 0 as a left bound.
 Ê
20
Input x = 2 as a right bound.
Í Á
Input x = 1 as a guess.
Í À
Press Í to find the local minimum.
Í
Thus the local minimum is (1, 0 ) . Thus, f is increasing on the intervals ( −2, − 1) and (1, 2 ) , and f is decreasing on the
interval ( −1,1) .
71.
f ( x ) = x5 − x3
( −2, 2 )
Enter the formula for f in the function editor. Go to WINDOW and enter limits for x. We must determine limits for y, so try −10 ≤ y ≤ 10 . If this does not work, then we can go back to WINDOW and adjust the limits on y until we find a good window.
21
Graph the function.
This window is not suitable; we cannot see the portion of the graph between x = −1 and x = 1 . We need to reduce the limits for y so let’s try −1 ≤ y ≤ 1 .
This window is much better. Use maximum to find the local maximum. Note that the x-coordinate of the local maximum is between x = −1 and x = 0 .
22
The local maximum is approximately ( −0.77, 0.19 ) . Use minimum to find the local minimum. Note that the x-coordinate of the local minimum is between x = 0 and x = 1 .
The local minimum is approximately ( 0.77, − 0.19 ) . Thus, f is increasing on the intervals ( −2, − 0.77 ) and ( 0.77, 2 ) , and f is decreasing on the interval ( −0.77, 0.77 ) .
73.
f ( x ) = −0.2 x3 − 0.6 x 2 + 4 x − 6
( −6, 4 )
Enter the formula for f in the function editor. Go to WINDOW and enter limits for x. We must determine limits for y, so try −10 ≤ y ≤ 10 . If this does not work, then we can go back to WINDOW and adjust the limits on y until we find a good window.
23
Graph the function.
This window is not suitable, part of the graph goes below the bottom of the screen. We need to change the limits for y so let’s try −20 ≤ y ≤ 2 .
This window is much better. Use maximum to find the local maximum. Note that the x-coordinate of the local maximum is between x = 1 and x = 3 .
24
The local maximum is approximately (1.77, − 1.91) . Use minimum to find the local minimum. Note that the x-coordinate of the local minimum is between x = −5 and x = −3 .
The local minimum is approximately ( −3.77, − 18.89 ) . Thus, f is increasing on the interval intervals ( −6, − 3.77 ) and (1.77, 4 ) .
75.
f ( x ) = 0.25 x 4 + 0.3 x3 − 0.9 x 2 + 3
( −3.77,1.77 ) ,
and f is decreasing on the
( −3, 2 )
Enter the formula for f in the function editor. Go to WINDOW and enter limits for x. We must determine limits for y, so try −10 ≤ y ≤ 10 . If this does not work, then we can go back to WINDOW and adjust the limits on y until we find a good window.
25
Graph the function.
This window may be suitable, but we can obtain a better graph if we use 0 ≤ y ≤ 10 .
This window is much better. Use maximum to find the local maximum. Note that the x-coordinate of the local maximum is between x = −1 and x = 1 .
26
The local maximum is ( 0, 3) . Use minimum to find both local minima. Note that the x-coordinate of the first local minimum is between x = −3 and x = −1 .
The first local minimum is approximately ( −1.87, 0.95 ) . The x-coordinate of the second local minimum is between x = 0 and x = 2 .
27
The second local minimum is approximately ( 0.97, 2.65 ) . Thus, f is increasing on the intervals ( −1.87, 0 ) and ( 0.97, 2 ) , and f is decreasing on the
intervals ( −3, − 1.87 ) and ( 0, 0.97 ) .
77.
For the function f ( x ) = x 2 , compute each average rate of change: (f)
Graph each of the secant lines. Set the viewing rectangle to: Xmin = −0.2 , Xmax = 1.2 , Xscl = 0.1 , Ymin = −0.2 , Ymax = 1.2 , and Yscl = 0.1 .
From parts (a) through (e), the average rates of change are 1, 0.5, 0.1, 0.01, and 0.001, respectively. Each average rate of change is a slope of a secant lines through the points given in each of parts (a) through (e). In each case, one of the two points was ( 0, 0 ) . Using this point and each of the slopes, we can find the equations of the five secant lines. Using m = 1 and ( 0, 0 ) with the point-slope form of the line we obtain y − 0 = 1( x − 0 )
which simplifies to y = x . Using m = 0.5 and ( 0, 0 ) with the point-slope form of the line we obtain y − 0 = 0.5 ( x − 0 )
which simplifies to y = 0.5 x . Using m = 0.1 and ( 0, 0 ) with the point-slope form of the line we obtain y − 0 = 0.1( x − 0 )
which simplifies to y = 0.1x .
28
Using m = 0.01 and ( 0, 0 ) with the point-slope form of the line we obtain y − 0 = 0.01( x − 0 )
which simplifies to y = 0.01x . Using m = 0.001 and ( 0, 0 ) with the point-slope form of the line we obtain y − 0 = 0.001( x − 0 )
which simplifies to y = 0.001x . Enter f ( x ) = x 2 and the equation of the secant line, y = x in the function editor. Go to WINDOW and enter limits for x and y. Graph the two equations.
Graph f ( x ) = x 2 and the secant line y = x .
Replace the equation of the secant line y = x with the next secant line, y = 0.5 x , in the function editor. Graph the two equations.
29
Replace the equation of the secant line y = 0.5 x with the next secant line, y = 0.1x , in the function editor. Graph the two equations.
Replace the equation of the secant line y = 0.1x with the next secant line, y = 0.01x , in the function editor. Graph the two equations.
It looks as though the calculator only graphed f ( x ) = x 2 . The calculator did graph the line, but we cannot distinguish the line from the x-axis. To see the line, you must turn the axes “off.”
yq†††~Í
Regraph the two equations.
30
Replace the equation of the secant line y = 0.01x with the secant line y = 0.001x in the function editor. Graph the two equations.
Turn the axes “on” when you are done.
yq†††Í
79.
Motion of a Golf Ball A golf ball is hit with an initial velocity of 130 feet per second at an inclination of 45 to the horizontal. In physics, it is established that the height h of the golf ball is given by the function −32 x 2 h ( x) = +x 1302 where x is the horizontal distance that the golf ball has traveled. (e)
Graph the function h = h ( x ) .
Enter the formula for h in the function editor. Go to WINDOW and enter limits for x and y. We are not given any limits for either variable. For this problem to make sense, we must have x ≥ 0 . Let’s try the following window.
31
Graph the function.
(f)
Use a graphing utility to determine the distance that the ball has traveled when the height of the ball is 90 feet.
One way to solve this problem is to graph the horizontal line g ( x ) = 90 on the same
graph as h ( x ) , and use intersect to find the point(s) where the two functions intersect. The intersect function requires three inputs: the equation corresponding to the first curve; the equation corresponding to the second curve; and a guess for the xcoordinate of the point of intersection. The intersect function can be found under the [CALC] menu. Enter the function g ( x ) = 90 in the function editor and graph both functions.
There are two points where the two functions intersect. Notice that the x-coordinate of the first (or left) point is close to x = 100 , while the x-coordinate of the second (or right) point is close to x = 450 . Find the first point of intersection.
y r
32
Select intersect.
·
Since the cursor is on first curve (the parabola) we can just press Í to select that curve.
Í
The cursor is now on the second curve (the line), so press Í to select that curve.
Í
Now, input a guess for the x-coordinate of the first intersection point. Let’s use x = 100 .
ÀÊÊ
33
Press Í to find the first intersection point.
Í
The coordinates of the point are approximately (115.07, 90 ) , thus the golf ball has traveled a horizontal distance of approximately 115.07 feet when it first reaches a height of 90 feet. Now, find the second point of intersection.
y r
Select intersect.
·
Since the cursor is on the first intersection point, it is hard to tell which curve it is on. Notice the equation of h ( x ) at the top of the screen, this tells us the cursor is on the parabola, so press Í to select h ( x ) .
34
Í
Again, the cursor is on the first intersection point, but we know that it is on the horizontal line by looking at the equation listed at the top of the screen. Press Í to select that curve.
Í
Now, input a guess for the x-coordinate of the second intersection point. Let’s use x = 450 .
¶·Ê
Press Í to find the second intersection point.
Í
35
The coordinates of the point are approximately ( 413.05, 90 ) , thus the golf ball has traveled a horizontal distance of approximately 413.05 feet when it next reaches a height of 90 feet. NOTE:
(g)
If you have more than two curves graphed, use the } or † keys to move the cursor to curve you wish to mark.
Create a TABLE with TblStart = 0 and ∆ Tbl = 25.
This problem is asking us to generate a table of values for our function h ( x ) . The values for x start with x = 0 , and the rest of the values of x are obtained by adding the increment 25 to the previous x. First, be sure that h ( x ) is the only function in the function editor (so if g ( x ) = 90 is still there from part (f), remove it), or, alternatively, deselect all other functions in the function editor. Go to TBLSET to set up the table.
yp
Set TblStart to zero.
ÊÍ
36
Set ∆ Tbl to 25.
Á·Í
Set Indpnt to Auto.
Í
Generate the table of values.
ys
To scroll through the table, use the † or } keys.
†††††††††††††
(h)
To the nearest 25 feet, how far does the ball travel before it reaches a maximum height? What is the maximum height?
37
Continue to scroll through the table generated in part (g).
†††††††
††
Looking at the table we see that the largest value for h is approximately 132, which occurs when x is 275. In other words, the golf ball reaches a maximum height of approximately 132 feet after it has traveled a horizontal distance of approximately 275 feet. (i)
Adjust the value of ∆ Tbl until you determine the distance, to within 1 foot, that the ball travels before it reaches a maximum height.
Return to TBLSET and set ∆ Tbl to 1 and set TblStart to 250.
38
Return to the table and find the maximum.
Notice that the height h appears to be the same when x is 263, 264, and 265. In order to see a more accurate value for h, move the cursor over to the second column and scroll through the values for h until you find the largest value.
Looking at the table we see that the largest value for h is 132.031, which occurs when x is 264. In other words, the golf ball reaches a maximum height of approximately 132.031 feet after it has traveled a horizontal distance of approximately 264 feet.
81.
Constructing an Open Box An open box with a square base is to be made from a square piece of cardboard 24 inches on a side by cutting our a square from each corner and turning up the sides (see the figure on Page 135).
(d)
Graph V = V ( x ) . For what value of x is V largest?
The function for V is given by V ( x ) = x ( 24 − 2 x ) . Enter the formula for V in the function editor. Go to WINDOW and enter limits for x and y. We are not given any limits for either variable, although for this problem to make sense, we must have 0 ≤ x ≤ 12 . Let’s try the following window. 2
39
Graph V ( x ) .
Find the local maximum. Notice that the maximum is between x = 3 and x = 6 .
The local maximum is ( 4, 1024 ) . Thus, for a maximum volume of 1,024 square inches, cut out squares whose sides are 4 inches.
40
83.
Minimum Average Cost The average cost of producing x riding lawn mowers per hour is given by 2500 C ( x ) = 0.3 x 2 + 21x − 251 + x
(a)
Use a graphing utility to graph C .
Enter the formula for C in the function editor. Go to WINDOW and enter limits for x and y. We are not given any limits for either variable, although for this problem to make sense, we must have x > 0 . Let’s try the following window.
Graph C .
(b)
Determine the number of riding lawn mowers to produce in order to minimize the average cost.
Find the local minimum. Notice that the minimum is between x = 5 and x = 15 .
41
The local minimum is approximately ( 9.66, 238.65 ) . Since they cannot produce 9.66 riding mowers, they should produce either 9 or 10 riding mowers. (c)
What is the minimum average cost?
If they produce 9 riding mowers, the average cost is $240.08, but if they produce 10 riding mowers the average cost is $239.00. They should produce 10 riding mowers for an average cost of $239.00. ________________________________________________________________________
Section 1.4 Library of Functions; Piecewise-Defined Functions 45.
Exploration Graph y = x 2 . Then on the same screen graph y = x 2 + 2 , followed by y = x 2 + 4 , followed by y = x 2 − 2 . What pattern do you observe? Can you predict the graph of y = x 2 − 4 ? Of y = x 2 + 58 .
Enter y = x 2 in the function editor. Go to WINDOW and enter limits for x and y. Let’s use the standard window −10 ≤ x ≤ 10 and −10 ≤ y ≤ 10 .
42
The local minimum is approximately ( 9.66, 238.65 ) . Since they cannot produce 9.66 riding mowers, they should produce either 9 or 10 riding mowers. (c)
What is the minimum average cost?
If they produce 9 riding mowers, the average cost is $240.08, but if they produce 10 riding mowers the average cost is $239.00. They should produce 10 riding mowers for an average cost of $239.00. ________________________________________________________________________
Section 1.4 Library of Functions; Piecewise-Defined Functions 45.
Exploration Graph y = x 2 . Then on the same screen graph y = x 2 + 2 , followed by y = x 2 + 4 , followed by y = x 2 − 2 . What pattern do you observe? Can you predict the graph of y = x 2 − 4 ? Of y = x 2 + 58 .
Enter y = x 2 in the function editor. Go to WINDOW and enter limits for x and y. Let’s use the standard window −10 ≤ x ≤ 10 and −10 ≤ y ≤ 10 .
42
Return to the function editor and enter y = x 2 + 2 into y2 and graph the two equations.
Notice that the new graph is another parabola. The only difference is its position on the yaxis. The graph of y = x 2 + 2 is obtained by shifting the graph of y = x 2 up two units. Return to the function editor and enter y = x 2 + 4 into y2 and graph the two equations.
Notice that the new graph is another parabola. The only difference is its position on the yaxis. The graph of y = x 2 + 4 is obtained by shifting the graph of y = x 2 up four units. Return to the function editor and enter y = x 2 − 2 into y2 and graph the two equations.
Notice that the new graph is another parabola. Again, the only difference is its position on the y-axis. The graph of y = x 2 − 2 is obtained by shifting the graph of y = x 2 down two units.
43
It appears that the graph of y = x 2 + k , k > 0 , is obtained by shifting the graph of y = x 2 up k units, and that the graph of y = x 2 − k , k > 0 , is obtained by shifting the graph of y = x 2 down k units Based on these observations, we predict that the graph of y = x 2 − 4 is obtained by shifting the graph of y = x 2 down four units, while the graph of y = x 2 + 58 is obtained by shifting the graph of y = x 2 up fifty-eight units.
47.
Exploration Graph y = x 2 . Then on the same screen graph y = − x 2 . What pattern do you observe? Now try y = x and y = − x . What do you conclude?
Enter y = x 2 in the function editor. Go to WINDOW and enter limits for x and y. Let’s use the standard window −10 ≤ x ≤ 10 and −10 ≤ y ≤ 10 .
Return to the function editor and enter y = − x 2 into y2 and graph the two equations.
Notice that the new graph is another parabola. The only difference is the parabola is on the opposite side of the x-axis, and it opens down. In other words, the graph of y = − x 2 is obtained by reflecting the graph of y = x 2 about the x-axis. The absolute value function, abs(, is found in the NUM submenu of .
44
Enter the equation y = x into the function editor.
o‘~À„¤Í‘
Graph y = x .
Return to the function editor and enter y = − x into y2 and graph the two equations.
o†Ì~À„¤Í
Graph the two equations.
45
Notice that the new graph is another v shaped graph, the only difference is that the v shape is on the opposite side of the x-axis, and it opens down. In other words, the graph of y = − x is obtained by reflecting the graph of y = x about the x-axis. If we multiply the formula for a function by −1 , the resulting graph is a reflection of the graph of the given function about the x-axis.
49.
Exploration Graph y = x3 . Then on the same screen graph y = ( x − 1) + 2 . 3
Could you have predicted the result? Enter y = x3 in the function editor. Go to WINDOW and enter limits for x and y. Let’s use the standard window −10 ≤ x ≤ 10 and −10 ≤ y ≤ 10 .
Return to the function editor and enter y = ( x − 1) + 2 into y2 and graph the two equations. 3
Yes, based on the results of Problems 45 and 46, we could determine that the graph of 3 y = ( x − 1) + 2 is obtained from the graph of y = x3 by shifting it to the right one unit and up two units.
46
51.
Exploration Graph y = x3 , y = x5 , and y = x 7 on the same screen. What do you notice is the same about each graph? What do you notice that is different?
Enter y = x3 , y = x5 , and y = x 7 in the function editor. Go to WINDOW and enter limits for x and y. Let’s use the window −2 ≤ x ≤ 2 and −2 ≤ y ≤ 2 .
All three graphs have the same general shape and all three graphs pass through ( 0, 0 ) ,
(1,1) , and ( −1, −1) .
As the degree increases, the graph rises faster for values of x that are greater than 1, and falls faster for values of x that are less than −1 . Also, as the degree increases, the graph is closer to the x-axis for values of x that are between −1 and 1. ________________________________________________________________________
Section 1.5 Graphing Techniques: Shifts and Reflections 41.
Exploration (a) Use a graphing utility to graph y = x + 1 and y = x + 1 .
Enter y = x + 1 in the function editor. Go to WINDOW and enter limits for x and y. Let’s use the standard window −10 ≤ x ≤ 10 and −10 ≤ y ≤ 10 .
47
51.
Exploration Graph y = x3 , y = x5 , and y = x 7 on the same screen. What do you notice is the same about each graph? What do you notice that is different?
Enter y = x3 , y = x5 , and y = x 7 in the function editor. Go to WINDOW and enter limits for x and y. Let’s use the window −2 ≤ x ≤ 2 and −2 ≤ y ≤ 2 .
All three graphs have the same general shape and all three graphs pass through ( 0, 0 ) ,
(1,1) , and ( −1, −1) .
As the degree increases, the graph rises faster for values of x that are greater than 1, and falls faster for values of x that are less than −1 . Also, as the degree increases, the graph is closer to the x-axis for values of x that are between −1 and 1. ________________________________________________________________________
Section 1.5 Graphing Techniques: Shifts and Reflections 41.
Exploration (a) Use a graphing utility to graph y = x + 1 and y = x + 1 .
Enter y = x + 1 in the function editor. Go to WINDOW and enter limits for x and y. Let’s use the standard window −10 ≤ x ≤ 10 and −10 ≤ y ≤ 10 .
47
Return to the function editor and enter y = x + 1 into y1 and graph the equation.
(b)
Graph y = 4 − x 2 and y = 4 − x 2 .
Enter y = 4 − x 2 in the function editor. Go to WINDOW and enter limits for x and y. Let’s use the standard window −10 ≤ x ≤ 10 and −10 ≤ y ≤ 10 .
Return to the function editor and enter y = 4 − x 2 into y1 and graph the equation.
(c)
Graph y = x3 + x and y = x3 + x .
Enter y = x3 + x in the function editor. Go to WINDOW and enter limits for x and y. Let’s use the standard window −10 ≤ x ≤ 10 and −10 ≤ y ≤ 10 .
48
Return to the function editor and enter y = x 3 + x into y1 and graph the equation.
(d)
What do you conclude about the relationship between the graphs of y = f ( x ) and y = f ( x ) ?
The graphs of y = f ( x ) and y = f ( x ) are identical on the x-intervals where f ( x ) ≥ 0 . However, for any x-interval where f ( x ) < 0 , the graph of y = f ( x ) is a reflection of the graph of y = f ( x ) about the x-axis. ________________________________________________________________________
Chapter 1 Review In Problems 39–42, use a graphing utility to graph each function over the indicated interval. Approximate any local maxima and local minima. Determine where the function is increasing and where it is decreasing.
39.
f ( x ) = 2 x3 − 5 x + 1
( −3, 3)
49
Chapter 2 Classes of Functions 2.1 Quadratic Functions 1. (C) f is a quadratic function whose graph opens up, and whose vertex is ⎛ b ⎛ b ⎞⎞ ⎜ − , f ⎜ − ⎟ ⎟ = (0, – 1). ⎝ 2a ⎠ ⎠ ⎝ 2a b 0 f ( 0 ) = 0 2 − 1 = −1 − = − = 0; 2a 2 3. (F) f is a quadratic function whose graph opens up, and whose vertex is ⎛ b ⎛ b ⎞⎞ ⎜ − , f ⎜ − ⎟ ⎟ = (1, 0). ⎝ 2a ⎠ ⎠ ⎝ 2a b 2 − = =1 ; f (1) = 1 2 − 2 ⋅ 1 + 1 = 0 2a 2 5. (G) f is a quadratic function whose graph opens up, and whose vertex is ⎛ b ⎛ b ⎞⎞ ⎜ − , f ⎜ − ⎟ ⎟ = (1, 1). ⎝ 2a ⎠ ⎠ ⎝ 2a b 2 − = = 1; f (1) = 1 2 − 2 ⋅ 1 + 2 = 1 2a 2 7. (H) f is a quadratic function whose graph opens up, and whose vertex is ⎛ b ⎛ b ⎞⎞ ⎜ − , f ⎜ − ⎟ ⎟ = (1, – 1). ⎝ 2a ⎠ ⎠ ⎝ 2a b 2 − = =1 ; f ( 1) = 1 2 − 2 ⋅ 1 = −1 2a 2 9. a = 1, b = 2, c = 0. Since a > 0, the parabola opens up. b 2 The x-coordinate of the vertex is − =− = −1 . 2a 2 (1)
The y-coordinate of the vertex is f ( −1) = ( −1) + 2 ( −1) = − 1 . So the vertex is (– 1, – 1) and the axis of symmetry is the line x = – 1. Since f (0) = c = 0, the y-intercept is (0, 0). The x-intercepts are found by solving f (x) = 0. x 2 + 2x = 0 x ( x + 2) = 0 x = 0 or x = – 2 The x-intercepts are (0, 0) and (– 2, 0). The domain is the set of all real numbers or the interval (– ∞, ∞); the range is the set {y | 2
SECTION 2.1 115
y ≥ – 1} or the interval [– 1, ∞). The function is increasing to the right of the axis or on the interval (– 1, ∞), and it is decreasing to the left of the axis or on the interval (– ∞, – 1).
11. a = – 1, b = – 6, c = 0. Since a < 0, the parabola opens down. b −6 The x-coordinate of the vertex is − =− = −3 . 2 ( −1) 2a
The y-coordinate of the vertex is f ( − 3) = − ( − 3) − 6 ( − 3) = 9 . So the vertex is (– 3, 9) and the axis of symmetry is the line x = – 3. Since f (0) = c = 0, the y-intercept is (0, 0). The x-intercepts are found by solving f (x) = 0. − x 2 − 6x = 0 – x ( x + 6) = 0 x = 0 or x = – 6 The x-intercepts are (0, 0) and (– 6, 0). The domain is the set of all real numbers or the interval (– ∞, ∞); the range is the set {y | y ≤ 9} or the interval (– ∞, 9]. The function is increasing to the left of the axis or on the interval (– ∞, – 3), and it is decreasing to the right of the axis or on the interval (– 3, ∞). 2
13. a = 2, b = – 8, c = 0. Since a > 0, the parabola opens up. b −8 The x-coordinate of the vertex is − =− =2. 2a 2 ( 2)
The y-coordinate of the vertex is f ( 2 ) = 2 ( 2 ) − 8 ( 2 ) = − 8 . So the vertex is (2, – 8) and the axis of symmetry is the line x = 2. Since f (0) = c = 0, the y-intercept is (0, 0). The x-intercepts are found by solving f (x) = 0. 2 x 2 − 8x = 0 2x ( x − 4) = 0 x = 0 or x = 4 The x-intercepts are (0, 0) and (4, 0). The domain is the set of all real numbers or the interval (– ∞, ∞); the range is the set {y | y ≥ – 8} or the interval [– 8, ∞). 2
116 SECTION 2.1
The function is increasing to the right of the axis or on the interval (2, ∞), and it is decreasing to the left of the axis or on the interval (– ∞, 2).
15. a = 1, b = 2, c = – 8. Since a > 0, the parabola opens up. b 2 The x-coordinate of the vertex is − =− = −1 . 2 (1) 2a
The y-coordinate of the vertex is f ( −1) = ( −1) + 2 ( −1) − 8 = − 9 . So the vertex is (– 1, – 9) and the axis of symmetry is the line x = – 1. Since f (0) = c = – 8, the y-intercept is (0, – 8). The x-intercepts are found by solving f (x) = 0. x 2 + 2x − 8 = 0 ( x − 2 )( x + 4 ) = 0 x = 2 or x = – 4 The x-intercepts are (2, 0) and (– 4, 0). The domain is the set of all real numbers or the interval (– ∞, ∞); the range is the set {y | y ≥ – 9} or the interval [– 9, ∞). The function is increasing to the right of the axis or on the interval (– 1, ∞), and it is decreasing to the left of the axis or on the interval (– ∞, – 1). 2
17. a = 1, b = 2, c = 1. Since a > 0, the parabola opens up. b 2 The x-coordinate of the vertex is − =− = −1 . 2 (1) 2a
The y-coordinate of the vertex is f ( −1) = ( −1) + 2 ( −1) + 1 = 0 . So the vertex is (– 1, 0) and the axis of symmetry is the line x = – 1. Since f (0) = c = 1, the y-intercept is (0, 1). The x-intercepts are found by solving f (x) = 0. x 2 + 2x + 1 = 0 ( x + 1)( x + 1) = 0 x=–1 The x-intercept is (– 1, 0). The vertex and the x-intercept are the same, so we use symmetry and the y-intercept to obtain a third point (– 2, 1) on the graph. The domain is the set of all real numbers or the interval (– ∞, ∞); the range is the set 2
SECTION 2.1 117
{y | y ≥ 0} or the interval [0, ∞). The function is increasing to the right of the axis or on the interval (– 1, ∞), and it is decreasing to the left of the axis or on the interval (– ∞, – 1).
19. a = 2, b = – 1, c = 2. Since a > 0, the parabola opens up. −1 1 b The x-coordinate of the vertex is − =− = . 2 ( 2) 4 2a 2
15 ⎛1⎞ ⎛1⎞ ⎛1⎞ The y-coordinate of the vertex is f ⎜ ⎟ = 2 ⎜ ⎟ − ⎜ ⎟ + 2 = . 8 ⎝ 4⎠ ⎝ 4⎠ ⎝ 4⎠ ⎛ 1 15 ⎞ So the vertex is ⎜ , ⎟ = (0.25, 1.875) and the axis of symmetry is the line x = 0.25. ⎝4 8 ⎠ Since f (0) = c = 2, the y-intercept is (0, 2). The x-intercepts are found by solving f (x) = 0. Since the discriminant b 2 – 4ac = (– 1) 2 – 4(2)(2) = – 15 is negative, the equation f (x) = 0 has no real solution, and therefore, the parabola has no x-intercept. To graph the function, we choose a point and use symmetry. If we choose x = 1, 2 f (1) = 2 (1) − (1) + 2 = 3 . Using symmetry, we obtain the point (– 0.5, 3).
The domain is the set of all real numbers or the interval (– ∞, ∞); the range is the set {y | y ≥ 1.875} or the interval [1.875, ∞). The function is increasing to the right of the axis or on the interval (0.25, ∞), and it is decreasing to the left of the axis or on the interval (– ∞, 0.25).
21. a = – 2, b = 2, c = – 3. Since a < 0, the parabola opens down. b 2 1 The x-coordinate of the vertex is − =− = = 0.5. 2 ( −2 ) 2 2a 2
5 ⎛1⎞ ⎛1⎞ ⎛1⎞ The y-coordinate of the vertex is f ⎜ ⎟ = − 2 ⎜ ⎟ + 2 ⎜ ⎟ − 3 = − = – 2.5. 2 ⎝ 2⎠ ⎝ 2⎠ ⎝ 2⎠ 5⎞ ⎛1 So the vertex is ⎜ , − ⎟ = (0.5, – 2.5), and the axis of symmetry is the line x = 0.5. 2⎠ ⎝2 Since f (0) = c = – 3, the y-intercept is (0, – 3). The x-intercepts are found by solving f (x) = 0. Since the discriminant
118 SECTION 2.1
b 2 – 4ac = 2 2 – 4(– 2)( – 3) = – 20 is less than zero, the equation f (x) = 0 has no real solution, and therefore, the parabola has no x-intercept. To graph the function, we choose an additional point and use symmetry. If we choose 2 x = 2, f ( 2 ) = − 2 ( 2 ) + 2 ( 2 ) − 3 = − 7 . Using symmetry, we obtain the point (– 1, – 7). The domain is the set of all real numbers or the interval (– ∞, ∞); the range is the set {y | y ≤ – 2.5} or the interval (– ∞, – 2.5]. The function is increasing to the left of the axis or on the interval (– ∞, 0.5), and it is decreasing to the right of the axis or on the interval (0.5, ∞).
23. a = 3, b = 6, c = 2. Since a > 0, the parabola opens up. b 6 The x-coordinate of the vertex is − =− = −1 . 2 ( 3) 2a
The y-coordinate of the vertex is f ( −1) = 3 ( −1) + 6 ( −1) + 2 = − 1 . So the vertex is (– 1, – 1) and the axis of symmetry is the line x = – 1. 2
Since f (0) = c = 2, the y-intercept is (0, 2). The x-intercepts are found by solving f (x) = 0. Using the quadratic formula, we obtain − 6 ± 6 2 − 4 ( 3) ( 2 ) − 6 ± 36 − 24 − 6 ± 12 − 3 ± 3 x= = = = 2 ( 3) 6 6 3 x ≈ – 1.58 or x ≈ – 0.42 The x-intercepts are approximately (– 1.58, 0) and (– 0.42, 0). The domain is the set of all real numbers or the interval (– ∞, ∞); the range is the set {y | y ≥ – 1} or the interval [– 1, ∞). The function is increasing to the right of the vertex or on the interval (– 1, ∞), and it is decreasing to the left of the vertex or on the interval (– ∞, – 1).
25. a = – 4, b = – 6, c = 2. Since a < 0, the parabola opens down. b −6 3 The x-coordinate of the vertex is − =− = − = – 0.75. 2 ( − 4) 4 2a
SECTION 2.1 119 2
17 ⎛ 3⎞ ⎛ 3⎞ ⎛ 3⎞ The y-coordinate of the vertex is f ⎜ − ⎟ = − 4 ⎜ − ⎟ − 6 ⎜ − ⎟ + 2 = . 4 ⎝ 4⎠ ⎝ 4⎠ ⎝ 4⎠ ⎛ 3 17 ⎞ So the vertex is ⎜ − , ⎟ = (– 0.75, 4.25), and the axis of symmetry is the line ⎝ 4 4⎠ x = – 0.75. Since f (0) = c = 2, the y-intercept is (0, 2). The x-intercepts are found by solving f (x) = 0. Using the quadratic formula, we obtain x=
2 6 ± ( −6 ) − 4 ( − 4 ) ( 2 )
=
2 ( − 4) x ≈ – 1.78 or x ≈ 0.28
6 ± 36 + 32 6 ± 68 3 ± 17 = = −8 −8 −4
The x-intercepts are approximately (– 1.78, 0) and (0.28, 0). The domain is the set of all real numbers or the interval (– ∞, ∞); the range is the set {y | y ≤ 4.25} or the interval (– ∞, 4.25]. The function is increasing to the left of the axis or on the interval (– ∞, – 0.75), and it is decreasing to the right of the axis or on the interval (– 0.75, ∞).
27. a = 2, b = 12, c = 0. Since a > 0, the parabola opens up, and the function has a minimum value. The minimum value occurs at b 12 x= − =− = −3 2a 2 ( 2)
The minimum value is f ( − 3) = 2 ( − 3) + 12 ( − 3) = − 18 . 2
29.
a = 2, b = 12, c = – 3. Since a > 0, the parabola opens up, and the function has a minimum value. The minimum value occurs at b 12 x= − =− = −3 2 ( 2) 2a The minimum value is f ( − 3) = 2 ( − 3) + 12 ( − 3) − 3 = − 21 . 2
31. a = – 1, b = 10, c = – 4. Since a < 0, the parabola opens down, and the function has a maximum value. The maximum value occurs at 10 b x= − =− =5 2a 2 ( −1) 2 The maximum value is f ( 5 ) = − ( 5 ) + 10 ( 5 ) − 4 = 21 .
120 SECTION 2.1 33.
a = – 3, b = 12, c = 1. Since a < 0, the parabola opens down and the function has a maximum value. The maximum value occurs at b 12 x= − =− =2 2 ( −3) 2a 2 The maximum value is f ( 2 ) = − 3 ( 2 ) + 12 ( 2 ) + 1 = 13 .
35. If r1 = – 3, and r2 = 1, and f (x) = a(x – r1)(x – r2) (a) Then if a = 1, f (x) = 1(x – (– 3))(x – 1) = (x + 3)(x – 1). If a = 2, f (x) = 2(x – (– 3))(x – 1) = 2(x + 3)(x – 1). If a = – 2, f (x) = – 2(x – (– 3))(x – 1) = – 2(x + 3)(x – 1). If a = 5, f (x) = 5(x – (– 3))(x – 1) = 5(x + 3)(x – 1).
(b) The x-intercepts are found by solving f (x) = 0. a(x – r1)(x – r2) = 0 (x – r1)(x – r2) = 0 So the value of a, a ≠ 0, has no affect on the x-intercept. The y-intercept is found by letting x = 0 and simplifying. f (x) = a(x – r1)(x – r2) f (0) = a(0 – r1)(0 – r2) f (0) = ar1r2 So the y-intercept is (0, ar1r2) is the product of the a and the x-intercepts. (c) The axis of symmetry is the line x = −
b . To determine b, we multiply out the 2a
factors of f. f (x) = a(x – r1)(x – r2) = a(x 2 – r2 x – r1 x + r1r2) = a[x 2 – (r1 + r2) x + r1r2] = ax 2 – a(r1 + r2) x + ar1r2 We find b = – a(r1 + r2). The line of symmetry is − a ( r1 + r2 ) r1 + r2 b =− = x =− 2a 2 2a which does not involve a. So the value of a does not affect the axis of symmetry. ⎛r +r r +r ⎛ r + r ⎞⎞ (d) The vertex is the point ⎜ 1 2 . f ⎜ 1 2 ⎟ ⎟ . If we evaluate f at 1 2 , we get 2 ⎝ 2 ⎠⎠ ⎝ 2 ⎛r +r ⎞ ⎛r +r ⎞⎛ r + r ⎞ f ⎜ 1 2 ⎟ = a ⎜ 1 2 − r1 ⎟⎜ 1 2 − r2 ⎟ ⎝ 2 ⎠ ⎝ 2 ⎠⎝ 2 ⎠ ⎡⎛ r + r ⎞⎛ r + r ⎞⎤ = a ⎢⎜ 1 2 − r1 ⎟ ⎜ 1 2 − r2 ⎟ ⎥ ⎠⎝ 2 ⎠⎦ ⎣⎝ 2 We see the y-value of the vertex is changed by a factor of a. (e) The x-coordinate of the vertex is x =
r1 + r2 . The midpoint of the x-intercepts is 2
SECTION 2.1 121
⎛ x + x y + y2 ⎞ midpoint = ⎜ 1 2 , 1 ⎟ 2 ⎠ ⎝ 2 ⎛ r + r 0 + 0 ⎞ ⎛ r1 + r2 ⎞ = ⎜ 1 2, , 0⎟ ⎟=⎜ 2 ⎠ ⎝ 2 ⎝ 2 ⎠ The x-coordinate of the vertex and the midpoint of the x-intercepts is the same. 37. Since R is a quadratic function with a = – 4 < 0, the vertex will give the maximum revenue. b 4000 The unit price to be charged should be p = − =− = 500 dollars. 2 ( − 4) 2a If the dryers cost $500, the revenue R will be maximized. The maximum revenue will be R(500) = − 4(500) 2 + 4000 ( 500 ) = −1, 000, 000 + 2, 000, 000 = $1,000,000 39. (a) R(x) = xp
1 ⎛ 1 ⎞ R(x) = x ⎜ − x + 100 ⎟ = − x 2 + 100 x 6 ⎝ 6 ⎠
0 ≤ x ≤ 600
(b) If 200 units are sold, x = 200, and the revenue R is 1 − 40, 000 + 120, 000 80, 000 2 R(200) = − ( 200 ) + 100 ( 200 ) = = = $13,333.33 6 6 6 1 < 0, the vertex will give the maximum 6 100 b =− = 300 units are sold. The revenue. Revenue is maximized when x = − 2a ⎛ 1⎞ 2⎜ − ⎟ ⎝ 6⎠ 1( 2 maximum revenue is R(300) = − 300 ) + 100 ( 300 ) = −15, 000 + 30, 000 = $15, 000 6
(c) Since R is a quadratic function with a = –
1 (d) The company should charge p = − ( 300 ) + 100 = − 50 + 100 = $50.00 per unit to 6 maximize revenue. 41. (a) We first solve the demand equation for x. x = − 5 p + 100 0 ≤ p ≤ 20, 5 p = 100 − x 100 − x 1 p= = 20 − x 5 5 Since when p = 0, x = 100 and when p = 20, x = 0, the domain of the function p is 0 ≤ x ≤ 100.
The revenue function R is 1 ⎞ 1 ⎛ R ( x ) = x ⎜ 20 − x ⎟ = 20 x − x 2 5 ⎠ 5 ⎝
0 ≤ x ≤ 100
122 SECTION 2.1
(b) If 15 units are sold, x = 15, and the revenue R is 1 2 R(15) = 20 (15) − (15) = 300 − 45 = $255 5 1 < 0, the vertex will give the maximum 5 b 20 = 50 units are sold. The =− revenue. Revenue is maximized when x = − 2a ⎛ 1⎞ 2⎜ − ⎟ ⎝ 5⎠ 1 2 maximum revenue is R(50) = 20 ( 50 ) − ( 50 ) = 1000 − 500 = $500 5 (c) Since R is a quadratic function with a = –
1 (d) The company should charge p = 20 − ( 50 ) = 20 − 10 = $10 per unit to maximize 5 revenue. 43. If the rectangle is shown at right and we are told the width is x and the perimeter, or the distance around the edge of the rectangle, is 400 yards, then we can find the length of the rectangle. P = 2l + 2w 400 = 2l + 2x 400 − 2 x l= = 200 − x yards 2
(a) A(x) = lw = (200 – x)x = 200x – x 2 square yards. (b) A is a quadratic function with a = – 1, b = 200, and c = 0. Since a < 0, the vertex 200 b = 100 yards. gives the maximum area. The area is maximum when x = − =− 2(−1) 2a (c) The maximum area is A(100) = 200(100) – 100 2 = 10,000 square yards. 45.
From the figure we see that the width of the plot measures x meters and the length of the plot measures 4000 – 2x meters. The area of the plot is
A(x) = lw = (4000 – 2x)x = 4000x – 2x2 meters squared. A is a quadratic function with a = – 2, b = 4000, and c = 0. Since a < 0, the vertex gives b 4000 =− the maximum area. The area is maximum when x = − = 1000 meters. The 2 ( − 2) 2a maximum area is A(1000) = 4000(1000) – 2(1000) 2 = 2,000,000 meters squared. 47.
(a) h is a quadratic function with a = − gives the maximum height.
32 , b = 1, and c = 200. Since a < 0, the vertex 50 2
SECTION 2.1 123
The height is maximum when x = −
b 1 2500 = = 39.06 feet from the cliff. =− 64 2a ⎛ −32 ⎞ 2⎜ 2 ⎟ ⎝ 50 ⎠
(b) The maximum height of the projectile is − 32(39.06) 2 + 39.06 + 200 = 219.53 feet above the water. h(39.06) = 50 2 (c) The projectile will hit the water when h(x) = 0. 32 2 x + x + 200 = 0 502 x = 170.02 feet from the cliff. −
(d)
(e)
When the projectile is 100 feet above the water, it is 135.70 feet from the base of the cliff. 49. If we denote the depth of the rain gutter by x, then the area A of the cross-section is given by A = lw A(x) = (12 – 2x)x = 12x – 2x 2 The function A is quadratic, with a = – 2 < 0, so the vertex is the maximum point. The cross-sectional area is maximum when b 12 x= − =− = 3 inches. 2a 2( − 2) 51.
If x denotes the width of the rectangle (and the diameter of the circle), then the perimeter of the track is ⎛1 ⎞ P = 2l + 2 ⎜ π x ⎟ = 2l + π x = 400 ⎝2 ⎠
1 πx 2
124 SECTION 2.1
We can express l in terms of x by solving the equation for l. 400 − π x l= 2 The area of the rectangle is π 2 ⎛ 400 − π x ⎞ A= ⎜ ⎟ x = 200 x − x 2 2 ⎝ ⎠ A is a maximum when b 200 200 x= − =− = ≈ 63.66 meters, and 2a π ⎛ π⎞ 2 ⋅⎜ − ⎟ ⎝ 2⎠ ⎛ 200 ⎞ ⎤ 1 1⎡ l = ⎢ 400 − π ⎜ ⎟ ⎥ = ( 200 ) = 100 meters. 2⎣ ⎝ π ⎠⎦ 2 53. (a) Since H(x) is a quadratic function, the income level for which there are the most hunters is given by x-value of the vertex. For this function a = – 1.01, b = 114.3, and c = 451. b 114.3 x= − =− = 56.584 2a 2 ( −1.01) The most hunters have an income level of approximately $56,584. There are about H(56.584) = – 1.01(56.584) 2 + 114.3(56.584) + 451 =3684.785 About 3685 hunters have an annual income of $56,584.
(b)
The number of hunters earning between $20,000 and $40,000 is increasing. 55. (a) M ( 23) = 0.76 ( 23) 2 − 107.00 ( 23) + 3854.18 = 1795.2
Approximately 1795 males who are 23 years old are murdered. (b) Using a graphing utility, and finding the intersection of M(x) and the line y = 1456, we find that 1456 males age 28 are murdered. 2500
20
90
SECTION 2.1 125 0
(c)
(d) The number of male murder victims decreases with age until age 70, and then it begins to increase. 57. The reaction rate is modeled by the quadratic function V. V = V(x) = akx − kx 2 where a = −k, b = ak, and c = 0. The reaction rate is maximum at the vertex of V. b ak a x= − =− = 2a 2(− k ) 2 The reaction rate is maximum when half the initial amount of the compound is present. 59.
y = f ( x ) = − 5x 2 + 8
61.
y = f ( x ) = x 2 + 3x + 5
y0 = f ( −1) = − 5 ( −1) + 8 = 3
y0 = f ( − 4 ) = ( − 4 ) + 3 ( − 4 ) + 5 = 9
y1 = f ( 0 ) = − 5 ( 0 ) + 8 = 8
y1 = f ( 0 ) = ( 0 ) + 3 ( 0 ) + 5 = 5
y2 = f ( 1) = − 5 ( 1) + 8 = 3
y2 = f ( 4 ) = ( 4 ) + 3 ( 4 ) + 5 = 33
2
2
2
2
2
h ( y0 + 4 y1 + y2 ) 3 1 = ( 3 + 4 ( 8 ) + 3) 3 38 square units = 3
Area =
2
h ( y0 + 4 y1 + y2 ) 3 4 = ( 9 + 4 ( 5 ) + 33) 3 248 square units = 3
Area =
63. The rectangle is drawn on the right. The width of the rectangle is x units, and the length is y = 10 – x units.
The area A of the rectangle is A = lw A(x) = (10 – x)x = 10x – x 2 A is a quadratic function with a = – 1, b = 10, and c = 0. Since a < 0, the maximum area occurs at the vertex of A. That is, when the width of the rectangle is b 10 x= − =− =5 2a 2 ( −1)
126 SECTION 2.2
The largest area enclosed by the rectangle is A = 10(5) – 52 = 25 square units. 65.
Functions will vary. All answers should have a < 0 and be perfect squares.
67.
Descriptions may vary. 69. Answers will vary.
2.2 Power Functions; Polynomial Functions; Rational Functions 1. Answers will vary. 3. origin 5.
y=x
→ Add 2; vertical shift up 2 units.
f (x) = y + 2 = x 6 + 2
126 SECTION 2.2
The largest area enclosed by the rectangle is A = 10(5) – 52 = 25 square units. 65.
Functions will vary. All answers should have a < 0 and be perfect squares.
67.
Descriptions may vary. 69. Answers will vary.
2.2 Power Functions; Polynomial Functions; Rational Functions 1. Answers will vary. 3. origin 5.
y=x
→ Add 2; vertical shift up 2 units.
f (x) = y + 2 = x 6 + 2
SECTION 2.2 127 7.
y=x
→ Replace y by – y;
5
→ Add 2; vertical shift up 2 units.
– y = – x5
f (x) = – y + 2 = – x 5 + 2
9.
y = x4
→ Replace x with x – 2; horizontal shift to the right 2 units.
f (x) = (x – 2) 4
11.
f is a polynomial. Its degree is 3.
13.
g is a polynomial. Its degree is 2.
15.
f is not a polynomial. The exponent of x is – 1.
17.
g is not a polynomial. One of the exponents is not an integer.
19.
F is a polynomial. Its degree is 4.
21.
G is a polynomial. Its degree is 4.
128 SECTION 2.2 23.
y = 3x 4
25.
y = – 2x 5
27.
2 Multiply the function out. 5 ( x + 1) ( x − 2 ) = 5 ( x 2 + 2 x + 1) ( x − 2 ) = ( 5 x 2 + 10 x + 5 ) ( x − 2 ) = 5 x 3 − 10 x 2 + 10 x 2 − 20 x + 5 x − 10 = 5 x 3 − 15 x − 10 So the power function is y = 5x 3.
29.
R is a rational function. The domain is all real numbers except those for which the denominator is 0. x – 3 ≠ 0 or x ≠ 3 The domain of R is the set {x | x ≠ 3}.
31.
H is a rational function. The domain is all real numbers except those for which the denominator is 0. (x – 2)(x + 4) ≠ 0 x – 2 ≠ 0 or x + 4 ≠ 0 x ≠ 2 or x ≠ – 4 The domain of H is the set {x | x ≠ 2 and x ≠ – 4}.
33. F is a rational function. The domain is all real numbers except those for which the denominator is 0. 2 x 2 − 5x − 3 ≠ 0 ( 2 x + 1)( x − 3) ≠ 0 2 x + 1 ≠ 0 or x – 3 ≠ 0 1 x≠− or x ≠ 3 2 1 The domain of F is the set x | x ≠ − and x ≠ 3 . 2
{
35.
}
R is a rational function. The domain is all real numbers except those for which the denominator is 0. x3 – 8 ≠ 0 2 (x – 2)(x + 2x + 4) ≠ 0 x–2≠0 x 2 + 2x + 4 = 0 has a negative discriminant, and so has no real solutions.
x≠2 The domain of R is the set {x | x ≠ 2}. 37. H is a rational function. The domain is all real numbers except those for which the denominator is 0. x2 + 4 ≠ 0
SECTION 2.3 129
x 2 + 4 = 0 has a negative discriminant, and so has no real solutions. The denominator never equals 0. The domain of H is the set of all real numbers or the interval (– ∞, ∞). 39.
R is a rational function. The domain is all real numbers except those for which the denominator is 0. 4(x 2 – 9) ≠ 0 4(x – 3)(x + 3) ≠ 0 x – 3 ≠ 0 or x + 3 ≠ 0 . x ≠ 3 or x ≠ – 3 The domain of R is the set {x | x ≠ 3 and x ≠ – 3}.
41. (a) The year 2000 is 70 years since the year 1930. (2000 – 1930 = 70) The percentage of union membership in 2000 is u(70). u(70) = 11.93 + 1.9 ( 70 ) − 0.052(70 2 ) + 0.00037 ( 70 3 ) = 17.04% (b) u(75) = 11.93 + 1.9 ( 75) − 0.052(75 2 ) + 0.00037 ( 75 3 ) = 18.02% Answers will vary.
2.3 Exponential Functions 1. (a) 3 2.2 = 11.2116 (c) 3 2.236 = 11.6639
(b) 3 2.23 = 11.5873 (d) 3 5 = 11.6648
3. (a) 2 3.14 = 8.8152 (c) 2 3.1415 = 8.8244
(b) 2 3.141 = 8.8214 (d) 2 π = 8.8250
5. (a) 3.1 2.7 = 21.2166 (c) 3.1412.718 = 22.4404
(b) 3.14 2.71 = 22.2167 (d) π e = 22.4592
7. e 1.2 = 3.3201 9. e – 0.85 = 0.4274
SECTION 2.3 129
x 2 + 4 = 0 has a negative discriminant, and so has no real solutions. The denominator never equals 0. The domain of H is the set of all real numbers or the interval (– ∞, ∞). 39.
R is a rational function. The domain is all real numbers except those for which the denominator is 0. 4(x 2 – 9) ≠ 0 4(x – 3)(x + 3) ≠ 0 x – 3 ≠ 0 or x + 3 ≠ 0 . x ≠ 3 or x ≠ – 3 The domain of R is the set {x | x ≠ 3 and x ≠ – 3}.
41. (a) The year 2000 is 70 years since the year 1930. (2000 – 1930 = 70) The percentage of union membership in 2000 is u(70). u(70) = 11.93 + 1.9 ( 70 ) − 0.052(70 2 ) + 0.00037 ( 70 3 ) = 17.04% (b) u(75) = 11.93 + 1.9 ( 75) − 0.052(75 2 ) + 0.00037 ( 75 3 ) = 18.02% Answers will vary.
2.3 Exponential Functions 1. (a) 3 2.2 = 11.2116 (c) 3 2.236 = 11.6639
(b) 3 2.23 = 11.5873 (d) 3 5 = 11.6648
3. (a) 2 3.14 = 8.8152 (c) 2 3.1415 = 8.8244
(b) 2 3.141 = 8.8214 (d) 2 π = 8.8250
5. (a) 3.1 2.7 = 21.2166 (c) 3.1412.718 = 22.4404
(b) 3.14 2.71 = 22.2167 (d) π e = 22.4592
7. e 1.2 = 3.3201 9. e – 0.85 = 0.4274
130 SECTION 2.3 11.
13.
15.
x
f ( x)
−1
3
0
6
1
12
2
18
3
30
x
H ( x)
−1
1 4
0
1
1
4
2
16
3
64
x
f ( x)
−1
3 2
0
3
1
6
2
12
3
24
f ( x) f ( x − 1)
The ratio of consecutive outputs is not constant for unit increases in inputs. So the function f is not exponential.
6 =2 3 12 =2 6 18 3 = 12 2 30 5 = 18 3 H ( x) H ( x − 1)
The ratio of consecutive outputs is constant for unit increases in inputs. So the function H is exponential. The base a = 4.
1 =4 1 4 4 =4 1 16 =4 4 64 =4 16 f ( x) f ( x − 1)
3 =2 3 2 6 =2 3 12 =2 6 24 =2 12
The ratio of consecutive outputs is constant for unit increases in inputs. So the function f is exponential. The base a = 2.
SECTION 2.3 131 17.
x
H ( x)
−1
2
0
4
1
6
2
8
3
10
H ( x) H ( x − 1)
The ratio of consecutive outputs is not constant for unit increases in inputs. So the function H is not exponential.
4 =2 2 6 3 = 4 2 8 4 = 6 3 10 5 = 8 4 x
–x
19.
B. This is the graph of y = 3 reflected about the y-axis. It is the graph of y = 3
21.
D. This is the graph of y = 3 reflected over both the x- and the y-axes. It is the graph –x of y = – 3 .
23.
A.
This is the graph of y = 3 .
25.
E.
This is the graph of y = 3 vertically shifted down one unit. It is the graph x of y = 3 – 1.
.
x
x
x
27.
y=2
x
→ Add 1; vertical shift up 1 unit.
f (x) = 2x + 1
The domain of f is all real numbers or the interval (– ∞, ∞); the range is the set {y | y > 1} or the interval (1, ∞). The horizontal asymptote is the line y = 1.
132 SECTION 2.3 29.
y = 3x
→ Subtract 2; vertical shift down 2 units.
→ Replace x by – x; reflect about the y-axis.
y = 3–x
f (x) = 3 – x – 2
The domain of f is all real numbers or the interval (– ∞, ∞); the range is the set {y | y > –2} or the interval (– 2, ∞). The horizontal asymptote is the line y = – 2. 31.
y = ex
→ Replace x by – x; reflect about the y-axis.
y = e–x
The domain of f is all real numbers or the interval (– ∞, ∞); the range is the set {y | y > 0} or the interval (0, ∞). The horizontal asymptote is the line y = 0.
SECTION 2.3 133 33.
y = ex
→ Subtract 1; vertical shift 1 unit down.
→ Replace x by x – 2; horizontal shift 2 units to the right.
y = ex-2
y = ex-2 – 1
The domain of f is all real numbers or the interval (– ∞, ∞); the range is the set {y | y > – 1} or the interval (– 1, ∞). The horizontal asymptote is the line y = – 1. 35.
2 2x+ 1 = 4 = 2 2 Since the bases are the same, we obtain 2x + 1 = 2 2x = 1 1 x= 2 1 2
37.
{}
39.
8x
2
−2x
=
2 3 x −2x
(2 )
2 3( x
2
− 2 x)
1 2
= 2 −1 = 2 −1
Since the bases are the same, we obtain 3(x 2 – 2x) = – 1 3x 2 – 6x + 1 = 0, where a = 3, b = – 6, c = 1
3 x = 9 x = (3 2 ) = 3 2x Since the bases are the same, we obtain x 3 = 2x x 3 – 2x = 0 x(x 2 – 2) = 0 x = 0 or x 2 – 2 = 0 x= ± 2
{−
41.
x
3
2, 0, 2}
2 x ⋅ 8 − x = 4x 2 x ⋅ (23 ) = (2 2 ) 2 x ⋅ 2 −3x = 2 2 x 2 x −3 x = 2 2 x −x
x
Since the bases are the same, we obtain x – 3x = 2x – 4x = 0 x=0 {0}
134 SECTION 2.3
Using the quadratic formula we find x=
6 ± (−6) 2 − 4 ( 3) (1) 6 ± 36 − 12 = 6 2 ( 3) =
6 ± 24 6 ± 2 6 3 ± 6 = = 6 6 3
⎧3 − 6 3 + 6 ⎫ , ⎨ ⎬ 3 ⎭ ⎩ 3 43.
45.
2− x
⎛1⎞ = 25 ⎜ ⎟ ⎝5⎠ 5 − ( 2 − x) = 5 2
Since the bases are the same, we obtain 2x = 3
Since the bases are the same, we obtain – (2 – x) = 2 x=4 {4}
47.
e x = (e 3x ) ⋅ 2
x
4 =8 2 2x = 2 3
x=
{} 3 2
49.
1 e2
3 2
4 – 2x = (4 x) – 2 So if 4 x = 7, 4 – 2x = 7 – 2 =
e x = e 3x − 2 2
1 1 = 2 7 49
Since the bases are the same, we obtain x 2 = 3x – 2 x 2 – 3x + 2 = 0 (x – 2)(x – 1) = 0 x – 2 = 0 or x – 1 = 0 x = 2 or x=1 {2, 1} 51. If 3 – x = 2, then 3 x = 2 – 1, and 3 2x = (2 – 1) 2 1 = 2–2 = 4 53. The graph is increasing and ⎛⎜ −1, ⎝ x function f (x) = 3 .
1⎞ ⎟ , (0, 1), and (1, 3) are points on the graph. So the 3⎠
SECTION 2.3 135
1 55. The graph is decreasing, negative, and ⎛⎜ −1, − ⎞⎟ , (0, – 1), and (1, – 6) are points on the 6⎠ ⎝ x graph. So the function f (x) = – 6 . 57. (a) The percent of light passing through 10 panes of glass is p(10) = 100e – 0.03(10) = 74.08%
(b) The percent of light passing through 25 panes of glass is p(25) = 100e – 0.03(25) = 47.24% 59. (a) After 30 days there will be w(30) = 50e – 0.004(30) = 44.35 watts of power.
(b) After 365 days there will be w(365) = 50e – 0.004(365) = 11.61 watts of power. 61. After 1 hour, there will be D(1) = 5e – 0.4(1) = 3.35 milligrams of drug will be in the patient’s bloodstream.
After 6 hours, there will be D(6) = 5e – 0.4(6) = 0.45 milligrams of drug will be in the patient’s bloodstream. 63. (a) The probability that a car will arrive within 10 minutes of 12:00 PM is F(10) = 1 – e – 0.1(10) = 0.632
(b) The probability that a car will arrive within 40 minutes of 12:00 PM is F(40) = 1 – e – 0.1(40) = 0.982 (c) As t becomes unbounded in the positive direction, e – 0.1t = approaches 1. (d)
1 e
− 0.1t
approaches 0. So F
136 SECTION 2.3
(e)
1.2
0
60 0
It takes 6.9 minutes for the probability to reach 50%. 65. (a) If 15 cars arrive, then x = 15 and 20 15 e − 20 = 0.0516 P ( x ) = P (15 ) = 15! The probability that 15 cars arrive between 5:00 PM and 6:00 PM is 0.0516.
(b) The probability that 20 cars arrive between 5:00 PM and 6:00 PM is given by 20 20 e − 20 = 0.0888 P ( x ) = P ( 20 ) = 20! 67. (a) If the Civic is 3 years old x = 3, and its cost is 3 p ( x ) = p ( 3) = 16,630 ( 0.90 ) = $12,123.27
(b) If the Civic is 9 years old x = 9, and its cost is 9 p ( x ) = p ( 9 ) = 16,630 ( 0.90 ) = $6442.80 69. (a) If E = 120 volts, R = 10 ohms, and L = 5 henrys, then the amperage is given by the function 120 ⎡1 − e − (10 / 5) t ⎤⎦ = 12 ⎡⎣1 − e − 2t ⎤⎦ I (t ) = 10 ⎣
After 0.3 second t = 0.3 and I ( t ) = I ( 0.3) = 12 ⎡⎣1 − e − 2(0.3) ⎤⎦ = 5.414 amps. After 0.5 second t = 0.5 and I ( t ) = I ( 0.5 ) = 12 ⎡⎣1 − e − 2(0.5) ⎤⎦ = 7.585 amps. After 1 second t = 1 and I ( t ) = I (1) = 12 ⎡⎣1 − e − 2 ⎤⎦ = 10.375 amps. (b)
SECTION 2.3 137
(c) The maximum current of I 1(t) approaches 12 amps as t becomes unbounded in the positive direction. (d) If E = 120 volts, R = 5 ohms, and L = 10 henrys, then the amperage is given by the function 120 ⎡⎣1 − e − (5 /10) t ⎤⎦ = 24 ⎡1 − e − (1/ 2) t ⎤ I (t ) = ⎣ ⎦ 5 After 0.3 second t = 0.3 and I ( t ) = I ( 0.3) = 24 ⎡1 − e ⎣
− (1/ 2 )( 0.3)
⎤ = 3.343 amps. ⎦
After 0.5 second t = 0.5 and I ( t ) = I ( 0.5 ) = 24 ⎡1 − e ⎣
− (1/ 2 )( 0.5)
⎤ = 5.309 amps. ⎦
After 1 second t = 1 and I ( t ) = I (1) = 24 ⎡⎣1 − e − 1/ 2 ⎤⎦ = 9.443 amps. (e)
(f) The maximum current of I 2(t) approaches 24 amps as t becomes unbounded in the positive direction. 71. Using a calculator, we get 1 1 1 2 + + + = 2.70833333 n = 4, 2! 3! 4! 1 1 1 1 1 n = 6, 2 + + + + + = 2.71805556 2! 3! 4! 5! 6!
n = 8,
2+
1 1 1 1 1 1 1 + + + + + + = 2.71827877 2! 3! 4! 5! 6! 7! 8!
n = 10
2+
1 1 1 1 1 1 1 1 1 + + + + + + + + = 2.718281801 2! 3! 4! 5! 6! 7! 8! 9! 10!
and e = 2.718281828 Comparisons might differ.
138 SECTION 2.3 73.
f ( x + h ) − f ( x ) a ( x + h) − a x = h h x h a a −ax = h x a a h −1 ⎛ a h −1⎞ = =ax⎜ ⎟ h ⎝ h ⎠
(
Use the exponential property a r · a s = a r + s.
)
Factor.
75. If f ( x ) = a x , then f ( − x ) = a − x
1 ax 1 = f ( x) =
Use the exponential property a – r =
1 a
r
.
Substitute.
77. (a) If F = 50° and D = 41°, then 4221
R = 10 T + 459.4
−
4221 +2 D + 459.4
4221
= 10 50 + 459.4 4221
= 10 509.4 = 70.95 The relative humidity is 70.95%.
−
−
4221 +2 41+ 459.4
4221 +2 500.4
(b) If F = 68° and D = 59°, then 4221
R = 10 T + 459.4
−
4221 +2 D + 459.4
4221
= 10 68 + 459.4 4221
= 10 527.4 = 72.62 The relative humidity is 72.62%. (c) If F = D, then
−
−
4221 +2 59 + 459.4
4221 +2 518.4
4221 4221 4221 4221 − = 0 , and = . So T + 459.4 D + 459.4 T + 459.4 D + 459.4
R = 10 0 + 2 = 100 When the temperature and the dew point are equal, the relative humidity is 100%. 79.
(a) To show f (x) = sinh x is an odd function, we evaluate f (– x) and simplify. 1 − −x f (− x) = e − x − e ( ) 2 1 = (e − x − e x ) Simplify. 2 1 Rearrange the terms. = ( −e x + e − x ) 2 1 = − (e x − e − x ) Factor out – 1. 2 = – f (x)
(
)
SECTION 2.4 139
So f is an odd function.
(b)
81. It takes 59 minutes. Explanations will vary. 83. There is no power function that increases more rapidly than an exponential function whose base is greater than 1. Explanations will vary. 85. Answers will vary.
2.4 Logarithmic Functions 1. We use the fact that y = log a x and x = a y, a > 0, a ≠ 1 are equivalent. If 9 = 32, then log 3 9 = 2. 3. We use the fact that y = log a x and x = a y, a > 0, a ≠ 1 are equivalent. If a2 = 1.6, then log a 1.6 = 2. 5. We use the fact that y = log a x and x = a y, a > 0, a ≠ 1 are equivalent. If 1.12 = M, then log 1.1 M = 2. 7. We use the fact that y = log a x and x = a y, a > 0, a ≠ 1 are equivalent. If 2x = 7.2, then log 2 7.2 = x. 9. We use the fact that y = log a x and x = a y, a > 0, a ≠ 1 are equivalent. If x 2 = π , then log x π = 2 . 11. We use the fact that y = log a x and x = a y, a > 0, a ≠ 1 are equivalent. If e x = 8, then log e 8 = x or ln 8 = x.
SECTION 2.4 139
So f is an odd function.
(b)
81. It takes 59 minutes. Explanations will vary. 83. There is no power function that increases more rapidly than an exponential function whose base is greater than 1. Explanations will vary. 85. Answers will vary.
2.4 Logarithmic Functions 1. We use the fact that y = log a x and x = a y, a > 0, a ≠ 1 are equivalent. If 9 = 32, then log 3 9 = 2. 3. We use the fact that y = log a x and x = a y, a > 0, a ≠ 1 are equivalent. If a2 = 1.6, then log a 1.6 = 2. 5. We use the fact that y = log a x and x = a y, a > 0, a ≠ 1 are equivalent. If 1.12 = M, then log 1.1 M = 2. 7. We use the fact that y = log a x and x = a y, a > 0, a ≠ 1 are equivalent. If 2x = 7.2, then log 2 7.2 = x. 9. We use the fact that y = log a x and x = a y, a > 0, a ≠ 1 are equivalent. If x 2 = π , then log x π = 2 . 11. We use the fact that y = log a x and x = a y, a > 0, a ≠ 1 are equivalent. If e x = 8, then log e 8 = x or ln 8 = x.
140 SECTION 2.4 13. We use the fact that x = a y, a > 0, a ≠ 1 and y = log a x are equivalent. If log 2 8 = 3, then 23 = 8. 15. We use the fact that x = a y, a > 0, a ≠ 1 and y = log a x are equivalent. If log a 3 = 6, then a 6 = 3. 17. We use the fact that x = a y, a > 0, a ≠ 1 and y = log a x are equivalent. If log 3 2 = x, then 3 x = 2. 19. We use the fact that x = a y, a > 0, a ≠ 1 and y = log a x are equivalent. If log 2 M = 1.3, then 21.3 = M. 21. We use the fact that x = a y, a > 0, a ≠ 1 and y = log a x are equivalent.
If log 2 π = x , then
( 2)
x
=π .
23. We use the fact that x = a y, a > 0, a ≠ 1 and y = log a x are equivalent. If ln 4 = x, then e x = 4. 25. To find the exact value of the logarithm, we change the expression to its equivalent exponential expression and simplify. y = log 2 1
2 y =1 2 y = 20 y=0 Therefore, log 2 1 = 0
Write in exponential form. 1 = 20 Equate exponents.
27. To find the exact value of the logarithm, we change the expression to its equivalent exponential expression and simplify. y = log 5 25
5 y = 25 5 y = 52 y=2 Therefore, log 5 25 = 2 .
Write in exponential form. 25 = 5 2 Equate exponents.
29. To find the exact value of the logarithm, we change the expression to its equivalent exponential expression and simplify. y = log 1/ 2 16 y
⎛1⎞ ⎜ ⎟ = 16 ⎝2⎠ y −4 ⎛1⎞ ⎛1⎞ = ⎜ ⎟ ⎜ ⎟ ⎝2⎠ ⎝2⎠ y=–4 Therefore, log 1/ 2 16 = − 4 .
Write in exponential form. 16 =
⎛1⎞ ⎜ ⎟ ⎝2⎠
−4
Equate exponents.
SECTION 2.4 141 31. To find the exact value of the logarithm, we change the expression to its equivalent exponential expression and simplify. y = log 10 10
10 y = 10 y
10 = 10 1 y= 2
Write in exponential form.
1/ 2
Therefore, log 10 10 =
10 = 10
1/ 2
Equate exponents.
1 . 2
33. To find the exact value of the logarithm, we change the expression to its equivalent exponential expression and simplify. y = log 2 4
( 2) = 4 ( 2) = ( 2) y
y
Therefore, log
2
Write in exponential form. 4
y=4 4 = 4.
4=2 = 2
⎡ ⎣
( ) 2
2
2
⎤ = ⎦
( 2)
4
Equate exponents.
35. To find the exact value of the logarithm, we change the expression to its equivalent exponential expression and simplify. y = ln e
ey = e y
e =e 1 y= 2 1 Therefore, ln e = . 2
1/ 2
Write in exponential form. e =e
1/ 2
Equate exponents.
37. The domain of a logarithmic function is limited to all positive real numbers, so for f ( x ) = ln ( x − 3) , x – 3 > 0. The domain is all x > 3, or using interval notation (3, ∞). 39. The domain of a logarithmic function is limited to all positive real numbers, so for F ( x ) = log 2 x 2 , x 2 > 0. x 2 is positive except when x = 0, meaning the domain is the
set {x | x ≠ 0}. 41.
The domain of f (x) = 3 − 2log 4
x x is restricted by log 4 which is defined only when 2 2
x is positive. So the domain of f is all x > 0, or using interval notation (0, ∞). 2 43.
There are two restrictions on the domain of f ( x ) = ln x .
142 SECTION 2.4
First, y = x is defined only when x is nonnegative, that is when x ≥ 0. Second, y = ln x is defined only when x is positive, that is when x > 0. However, on the interval (0, 1), ln x < 0, so f ( x ) = ln x is not defined. So the domain of f is all x ≥ 1, or using interval notation [1, ∞). 45.
ln
47.
5 = 0.511 3
10 3 = 30.099 0.04
ln
49. If the graph of f contains the point (2, 2), then x = 2 and y = 2 must satisfy the equation y = f ( x ) = log a x
2 = log a 2 a2 = 2 a= 2 51. The graph of y = log 3 x has the properties:
The domain is (0, ∞). The range is (–∞, ∞). The x-intercept is (1, 0). The y-axis is a vertical asymptote. The graph is increasing since a = 3 > 1. ⎛1 ⎞ The graph contains the points (3, 1) and ⎜ , − 1⎟ . ⎝3 ⎠ The graph is continuous.
53. The graph of y = log 1/ 5 x has the properties:
The domain is (0, ∞). The range is (– ∞, ∞). The x-intercept is (1, 0). The y-axis is a vertical asymptote. 1 The graph is decreasing since a = < 1. 5 ⎛1 The graph contains the points (5, – 1) and ⎜ , ⎝5 The graph is continuous.
⎞ 1⎟ . ⎠
55. B. This graph is reflected over the y-axis. 57. D. This graph was reflected about both the x- and y-axes. 59. A. This graph has not been shifted or reflected. 61. E. This graph was shifted vertically down 1 unit.
SECTION 2.4 143 63.
y = ln x
→ Replace x by x + 4; horizontal shift 4 units to the left.
f (x) = ln(x + 4)
The domain of f is all x > – 4, or in interval notation, (– 4, ∞). The range is all real numbers or (– ∞, ∞). The vertical asymptote is the line x = – 4. 65.
y = ln x
→ Add 2; vertical shift up 2 units.
f (x) = 2 + ln x
The domain of f is all x > 0, or in interval notation, (0, ∞). The range is all real numbers or (– ∞, ∞). The vertical asymptote is the y-axis, that is the line x = 0. 67.
→
y = log x
Replace x with x − 4; horizontal shift right 4 units.
f (x) = log (x − 4)
The domain of f is all x > 4, or in interval notation, (4, ∞). The range is all real numbers or (– ∞, ∞). The vertical asymptote is the line x = 4.
144 SECTION 2.4 69.
→
y = log x
Add 2; vertical shift up 2 units.
f (x) = log x + 2
The domain of f is all x > 0, or in interval notation, (0, ∞). The range is all real numbers or (– ∞, ∞). The vertical asymptote is the y-axis, that is the line x = 0. 71.
log 3 x = 2
x = 32 = 9 73.
Change to exponential form and simplify.
log 2 ( 2 x + 1) = 3
2x + 1 = 2 3 2x + 1 = 8 2x = 7 7 x= 2 75.
log x 4 = 2 4 = x2 x=2
77.
79.
Change to exponential form. Solve the linear equation.
Change to exponential form. Solve using the Square Root Method. (x ≠ – 2, the base is positive.)
ln e x = 5 ex= e5 x=5 log 4 64 = x 64 = 4 x 43 = 4x 3=x
81.
Change to exponential form. Since the bases are equal, the exponents are equal.
Change to exponential form. Since the bases are equal, the exponents are equal.
log 3 243 = 2 x + 1 3 2x + 1 = 243 3 2x + 1 = 3 5 2x + 1 = 5 2x = 4 x=2
Change to exponential form. Since the bases are equal, the exponents are equal. Solve the linear equation.
SECTION 2.4 145 83.
85.
87.
e 3 x = 10 ln 10 = 3x ln 10 x= 3 ≈ 0.768
Change to a logarithmic expression. Exact solution. Approximate solution.
e 2 x +5 = 8 ln 8 = 2x + 5 2x = – 5 + ln 8 − 5 + ln 8 x= 2 ≈ – 1.460
=±2 2
Approximate solution.
Change to an exponential expression.
Use the Square Root Method. Simplify.
log 2 8 x = − 3
8x= 2–3 (2 3) x = 2 – 3 2 3x = 2 – 3 3x = – 3 x=–1 91.
Exact solution.
log 3 ( x 2 + 1) = 2
x2 + 1 = 32 x2 + 1 = 9 x2 = 8 x=± 8
89.
Change to a logarithmic expression.
Change to an exponential expression.
Since the bases are equal, the exponents are equal.
(a) When H + = 0.1, pH = – log (0.1) = 1. (b) When H + = 0.01, pH = – log (0.01) = 2. (c) When H + = 0.001, pH = – log (0.001) = 3. (d) As the hydrogen ion concentration decreases, pH increases. (e) If pH = 3.5, then 3.5 = – log x log x = – 3.5 10 – 3.5 = x x = 0.000316 (f) If pH = 7.4, then 7.4 = – log x log x = – 7.4 10 – 7.4 = x x = 3.981 × 10 – 8
93.
(a) If p = 320 mm, 320 = 760 e − 0.145 h
146 SECTION 2.4
320 8 = 760 19 ⎛ 8⎞ – 0.145h = ln ⎜ ⎟ ⎝ 19 ⎠ 1000 ⎛ 8 ⎞ h= − ln ⎜ ⎟ 145 ⎝ 19 ⎠ ≈ 5.965
e − 0.145 h =
Divide both sides by 760. Change to a logarithmic expression. 0.145 =
145 1000
The aircraft is at an altitude of approximately 5.965 kilometers above sea level. (b) If p = 667 mm, 667 = 760 e − 0.145 h 667 e − 0.145 h = 760 ⎛ 667 ⎞ – 0.145h = ln ⎜ ⎟ ⎝ 760 ⎠ 1000 ⎛ 667 ⎞ h= − ln ⎜ ⎟ 145 ⎝ 760 ⎠ ≈ 0.900
Divide both sides by 760. Change to a logarithmic expression. 0.145 =
145 1000
The mountain is approximately 0.9 kilometer above sea level. 95. (a) We want F(t) = 0.50. 0.50 = 1 − e − 0.1t e − 0.1t = 0.50 – 0.1t = ln 0.5 t = – 10 ln 0.5 ≈ 6.93 After approximately 6.9 minutes the probability that a car arrives at the drive-thru reaches 50%.
(b) We want F(t) = 0.80. 0.80 = 1 − e − 0.1t e − 0.1t = 0.20 – 0.1t = ln 0.2 t = – 10 ln 0.2 ≈ 16.09 After approximately 16.1 minutes the probability that a car arrives at the drive-thru reaches 80%. 97. We want D = 2. 2 = 5 e − 0.4 h 2 e − 0.4 h = = 0.4 5 – 0.4h = ln 0.4
SECTION 2.4 147
5 ln 0.4 2 ≈ 2.29 The drug should be administered every 2.3 hours (about 2 hours 17 minutes).
h=–
99. We want I = 0.5 12 ⎡ − 10 / 5 t 0.5 = 1 − e ( ) ⎤ = 1.2 (1 − e − 2t ) ⎦ 10 ⎣ 0.5 5 = 1 − e − 2t = 1.2 12 7 e − 2t = 12 ⎛ 7⎞ – 2t = ln ⎜ ⎟ ⎝ 12 ⎠ 1 ⎛7⎞ t = − ln ⎜ ⎟ 2 ⎝ 12 ⎠ ≈ 0.269 It takes about 0.27 seconds to obtain a current of 0.5 ampere.
Next we want I = 1.0 1.0 = 1.2 (1 − e − 2t ) 1.0 5 = 1.2 6 1 e − 2t = 6 ⎛1⎞ – 2t = ln ⎜ ⎟ ⎝6⎠ 1 ⎛1⎞ t = − ln ⎜ ⎟ 2 ⎝6⎠ ≈ 0.896 It takes about 0.9 seconds to obtain a current of 1ampere. 1 − e − 2t =
101. We are interested in the population when t = 11, so we want P(11). P(10) = 298,710,000 + 10,000,000 log 11 = 309,123,926.9 On January 1, 2020 the population (to the nearest thousand) will be 309,124,000. 103. Normal conversation: x = 10 – 7 watt per square meter, 10 − 7 L (10 − 7 ) = 10 log − 12 = 50 decibels 10 105. Amplified rock music: x = 10 – 1 watt per square meter, 10 − 1 L (10 − 1 ) = 10 log − 12 = 110 decibels 10
148 SECTION 2.5 107. Mexico City, 1985: x = 125,892 mm. ⎛ 125,892 ⎞ M(125,892) = log ⎜ ⎟ = 8.1 = 8.1 −3 ⎝ 10 ⎠ 109. (a) Since R = 10 when x = 0.06, we can find k. 10 = 3 e 0.06k ⎛ 10 ⎞ 0.06k = ln ⎜ ⎟ ⎝ 3⎠ 100 ⎛ 10 ⎞ 50 ⎛ 10 ⎞ k= ln ⎜ ⎟ = ln ⎜ ⎟ ≈ 20.066 6 ⎝ 3⎠ 3 ⎝ 3⎠
(b) If x = 0.17, then the risk of having a car accident is ⎛ 50 ⎛ 10 ⎞ ⎞ ln ⎜ ⎟ ⎟ ( 0.17 ) ⎜ ⎝ 3 ⎠⎠ ⎝ 3
R = 3e = 90.91% (c) If the risk is 100%, then R = 100. ⎛ 50 ⎛ 10 ⎞ ⎞ ln ⎜ ⎟ ⎟ x ⎜ ⎝ 3 ⎠⎠
3e ⎝ 3
⎛ 50 ⎛ 10 ⎞ ⎞ ln ⎜ ⎟ ⎟ x ⎜ ⎝ 3 ⎠⎠ ⎝ 3
= 100
100 3 50 ⎛ 10 ⎞ 100 ln ⎜ ⎟ x = ln 3 3 ⎝ 3⎠ x ≈ 0.1747
e
=
R = 15 = 3 e
(d) e
⎛ 50 ⎛ 10 ⎞ ⎞ ln ⎜ ⎟ ⎟ x ⎜ ⎝ 3 ⎠⎠ ⎝ 3
⎛ 50 ⎛ 10 ⎞ ⎞ ln ⎜ ⎟ ⎟ x ⎜ ⎝ 3 ⎠⎠ ⎝ 3
=5
50 ⎛ 10 ⎞ ln ⎜ ⎟ x = ln 5 3 ⎝ 3⎠ x ≈ 0.080 A driver with a blood alcohol concentration greater than or equal to 0.08 should be arrested and charged with DUI. (e) Answers will vary. 111. Answers may vary.
2.5 Properties of Logarithms 1. log 3 371 = 71
log a a r = r
5.
a
2 log 2 7 = 7
log a M
=M
3.
ln e – 4 = – 4
log a a r = r
148 SECTION 2.5 107. Mexico City, 1985: x = 125,892 mm. ⎛ 125,892 ⎞ M(125,892) = log ⎜ ⎟ = 8.1 = 8.1 −3 ⎝ 10 ⎠ 109. (a) Since R = 10 when x = 0.06, we can find k. 10 = 3 e 0.06k ⎛ 10 ⎞ 0.06k = ln ⎜ ⎟ ⎝ 3⎠ 100 ⎛ 10 ⎞ 50 ⎛ 10 ⎞ k= ln ⎜ ⎟ = ln ⎜ ⎟ ≈ 20.066 6 ⎝ 3⎠ 3 ⎝ 3⎠
(b) If x = 0.17, then the risk of having a car accident is ⎛ 50 ⎛ 10 ⎞ ⎞ ln ⎜ ⎟ ⎟ ( 0.17 ) ⎜ ⎝ 3 ⎠⎠ ⎝ 3
R = 3e = 90.91% (c) If the risk is 100%, then R = 100. ⎛ 50 ⎛ 10 ⎞ ⎞ ln ⎜ ⎟ ⎟ x ⎜ ⎝ 3 ⎠⎠
3e ⎝ 3
⎛ 50 ⎛ 10 ⎞ ⎞ ln ⎜ ⎟ ⎟ x ⎜ ⎝ 3 ⎠⎠ ⎝ 3
= 100
100 3 50 ⎛ 10 ⎞ 100 ln ⎜ ⎟ x = ln 3 3 ⎝ 3⎠ x ≈ 0.1747
e
=
R = 15 = 3 e
(d) e
⎛ 50 ⎛ 10 ⎞ ⎞ ln ⎜ ⎟ ⎟ x ⎜ ⎝ 3 ⎠⎠ ⎝ 3
⎛ 50 ⎛ 10 ⎞ ⎞ ln ⎜ ⎟ ⎟ x ⎜ ⎝ 3 ⎠⎠ ⎝ 3
=5
50 ⎛ 10 ⎞ ln ⎜ ⎟ x = ln 5 3 ⎝ 3⎠ x ≈ 0.080 A driver with a blood alcohol concentration greater than or equal to 0.08 should be arrested and charged with DUI. (e) Answers will vary. 111. Answers may vary.
2.5 Properties of Logarithms 1. log 3 371 = 71
log a a r = r
5.
a
2 log 2 7 = 7
log a M
=M
3.
ln e – 4 = – 4
log a a r = r
SECTION 2.5 149 7.
log 8 2 + log 8 4 = log 8 ( 2 ⋅ 4 ) = log 8 8 = 1
9.
⎛ 18 ⎞ log 6 18 − log 6 3 = log 6 ⎜ ⎟ = log 6 6 = 1 ⎝ 3⎠
11. Use a change of base formula, ln 6 ln 4 ln 4 log 2 6 ⋅ log 6 4 = ⋅ = ln 2 ln 6 ln 2
Next simplify ln 4,
ln 4 ln 2 2 2 ln 2 = = =2 ln 2 ln 2 ln 2 So, log 2 6 ⋅ log 6 4 = 2 13.
3
log 3 5 − log 3 4
=3
5 log 3 ⎛⎜ ⎞⎟ ⎝4⎠
=
5 4
15. First we write the exponent as y = log e 2 16 , and express it as an exponential.
y = log e 2 16 e 2y = 16 Then, e
log
e2
16
or
e y = 4. So y = ln 4.
= e y = e ln 4 = 4.
17. ln 6 = ln (2 · 3) = ln 2 + ln 3 = a + b
19.
ln 1.5 = ln
21. ln 8 = ln 23 = 3 ln 2 = 3a 23.
ln
5
6=
1 1 1 1 ln 6 = ln (2 · 3) = (ln 2 + ln 3) = (a + b) 5 5 5 5
25.
log 5 ( 25 x ) = log 5 ( 5 2 x ) = log 5 52 + log 5 x = 2log 5 5 + log 5 x
27.
log 2 z 3 = 3log 2 z
29. ln (ex) = ln e + ln x = 1 + ln x 31. ln (xe x) = ln x + ln e x = ln x + x ln e = ln x + x 33.
log a ( u 2v 3 ) = log a u 2 + log a v 3 = 2log a u + 3log a v
35.
ln x 2 1 − x = ln x 2 + ln (1 − x )
(
)
1/ 2
1 = 2ln x + ln (1 − x ) 2
3 = ln 3 – ln 2 = b – a 2
150 SECTION 2.5 37.
⎛ x3 ⎞ 3 log 2 ⎜ ⎟ = log 2 x − log 2 ( x − 3) = 3log 2 x − log 2 ( x − 3) x − 3 ⎝ ⎠
39.
⎡ x ( x + 2) ⎤ 2 log ⎢ = log ⎡⎣ x ( x + 2 ) ⎤⎦ − log ( x + 3) = log x + log ( x + 2 ) − 2log ( x + 3) 2 ⎥ ⎣⎢ ( x + 3) ⎦⎥
41.
⎡ x 2 − x − 2⎤ ln ⎢ 2 ⎥ ⎣⎢ ( x + 4 ) ⎦⎥
43.
45. 47.
ln
5 x 1 + 3x
( x − 4)
3
1/ 3
1 ⎡ x 2 − x − 2⎤ 1 ⎡ 2 = ln ⎢ = ln ( x 2 − x − 2) − ln ( x + 4 ) ⎤ 2 ⎥ ⎣ ⎦ 3 ⎢⎣ ( x + 4 ) ⎦⎥ 3 1 2 = ⎡ ln ⎡⎣ ( x − 2 )( x + 1) ⎤⎦ − ln ( x + 4 ) ⎤ ⎦ 3⎣ 1 = ⎡⎣ ln ( x − 2 ) + ln ( x + 1) − 2 ln ( x + 4 ) ⎤⎦ 3 1 1 2 = ln ( x − 2 ) + ln ( x + 1) − ln ( x + 4 ) 3 3 3
1 3 = ln ⎡⎣5 x 1 + 3x ⎤⎦ − ln ( x − 4 ) = ln 5 + ln x + ln (1 + 3x ) − 3ln ( x − 4 ) 2
3 log 5 u + 4 log 5 v = log 5 u 3 + log 5 v 4 = log 5 ( u 3 v 4 ) log 3
⎛ x⎞ 5 x − log 3 x 3 = log 3 ⎜ 3 ⎟ = log 3 x − 5 / 2 = − log 3 x 5 / 2 or − log 3 x 2 ⎝x ⎠
49.
⎡ x 2 −1 ⎤ 5 log 4 ( x 2 − 1) − 5 log 4 ( x + 1) = log 4 ( x 2 − 1) − log 4 ( x + 1) = log 4 ⎢ 5⎥ ⎢⎣ ( x + 1) ⎥⎦ ⎡ ( x − 1) ( x + 1) ⎤ ⎡ x −1 ⎤ ⎥ log = = log 4 ⎢ ⎢ 4 4⎥ ⎢ ( x + 1) 5 4 ⎥ 1 + x ( ) ⎢ ⎥⎦ ⎣ ⎣ ⎦
51.
⎡ x ⎤ x +1 1 ⎛ x ⎞ ⎛ x +1⎞ 2 ⎢ ln ⎜ ln ln x 1 ln ⋅ + − − = ⋅ ( ) ⎢ x − 1 x x − 1 x + 1 ⎥⎥ ⎟ ⎜ ⎟ ⎝ x −1⎠ ⎝ x ⎠ ( )( )⎦ ⎣ ⎡ 1 ⎤ −2 = ln ⎢ = ln ( x − 1) = − 2 ln ( x − 1) 2⎥ ⎣⎢ ( x − 1) ⎦⎥
53.
⎛4⎞ 8log 2 3x − 2 − log 2 ⎜ ⎟ + log 2 4 = log 2 ⎝x⎠
(
3x − 2
= log 2 ( 3 x − 2 )
4
)
⎛4⎞ − log 2 ⎜ ⎟ + log 2 4 ⎝ x⎠ ⎛4⎞ − log 2 ⎜ ⎟ + log 2 4 ⎝ x⎠ 8
SECTION 2.5 151
⎡ ⎤ ⎢ ( 3x − 2 ) 4 ⋅ 4 ⎥ 4 = log 2 ⎢ ⎥ = log 2 ⎡⎣ x ( 3x − 2 ) ⎤⎦ 4 ⎢ ⎥ ⎢⎣ ⎥⎦ x 55.
2 1 2log a ( 5 x 3 ) − log a ( 2 x + 3) = log a ( 5 x 3 ) − log a 2 x + 3 2
(5x ) 3
= log a 57.
2
2x + 3
= log a
25 x 6 2x + 3
2log 2 ( x + 1) − log 2 ( x + 3) − log 2 ( x − 1) = log 2 ( x + 1) − log 2 ( x + 3) − log 2 ( x − 1) 2
( x + 1) = log ( x + 1) = log 2 2 x 2 + 2x − 3 ( x + 3)( x − 1) 2
59.
63.
67.
71.
log 3 21 =
log
2
7=
ln 21 = 2.771 ln 3
61.
ln 7 = 5.615 ln 2
65.
y = log 4 x =
ln x ln 4
y = log x −1 ( x + 1) =
ln ( x + 1) ln ( x − 1)
log 1/ 3 71 =
log π e =
2
ln 71 = − 3.880 1 ln 3
ln e = 0.874 ln π
69.
y = log 2 ( x + 2 ) =
73.
ln y = ln x + ln C ln y = ln ( Cx ) y = Cx
ln ( x + 2 ) ln 2
152 SECTION 2.5
75.
ln y = ln x + ln ( x + 1) + ln C
77.
ln y = ln ⎡⎣ x ( x + 1) C ⎤⎦ y = x ( x + 1) C 79.
ln y = 3x + ln C ln y = ln e 3 x + ln C ln y = ln Ce 3 x y = Ce 3 x
ln ( y − 3) = − 4 x + ln C ln ( y − 3) = ln e − 4 x + ln C
ln ( y − 3) = ln ( Ce − 4 x ) y − 3 = Ce − 4 x y = Ce − 4 x + 3 81.
1 1 ln ( 2 x + 1) − ln ( x + 4 ) + ln C 2 3 1/ 2 1/ 3 ln y 3 = ln ( 2 x + 1) − ln ( x + 4 ) + ln C
3 ln y =
1/ 2 1/ 3 ln y 3 = ln ⎡C ( 2 x + 1) ⎤ − ln ( x + 4 ) ⎣ ⎦ ⎡ C ( 2 x + 1) 1/ 2 ⎤ 3 ln y = ln ⎢ 1/ 3 ⎥ ⎢⎣ ( x + 4 ) ⎥⎦
y = 3
C ( 2 x + 1)
( x + 4)
1/ 2
1/ 3
⎡ C ( 2 x + 1) 1/ 2 ⎤ y =⎢ 1/ 3 ⎥ ⎢⎣ ( x + 4 ) ⎥⎦
1/ 3
=
C ( 2 x + 1)
( x + 4)
1/ 6
C 1/ 3 is still a positive constant. We write C.
1/ 9
83. We use the change of base formula and simplify. log 2 3 ⋅ log 3 4 ⋅ log 4 5 ⋅ log 5 6 ⋅ log 6 7 ⋅ log 7 8
= =
log 3 log 2
⋅
log 4 log 3
log 8 log 23 = log 2 log 2
⋅
log 5 log 4
⋅
log 6 log 7 log 8 ⋅ ⋅ log 5 log 6 log 7
SECTION 2.5 153
=
3 log 2 log 2
=3
85. We use the change of base formula and simplify. log 2 3 ⋅ log 3 4 ⋅ log 4 5 ⋅ log 5 6 ⋅ log 6 7 ⋅ log 7 8 ⋅ … ⋅ log n ( n + 1) ⋅ log n +1 2
Noticing that the first n – 1 factors follow the pattern of problem 83, we get ⎡⎣ log 2 3 ⋅ log 3 4 ⋅ log 4 5 ⋅ log 5 6 ⋅ log 6 7 ⋅ log 7 8 ⋅ … ⋅ log n ( n + 1) ⎤⎦ ⋅ log n +1 2 ⎡ log 3 log 4 log 5 log ( n + 1) ⎤ log 2 ⋅ ⋅ ... ⋅ ⋅ =⎢ ⎥⋅ log n ⎦⎥ log ( n + 1) ⎣⎢ log 2 log 3 log 4 =1 87.
(
)
)
(
(
)( x − ) ⎞⎟⎠
log a x + x 2 − 1 + log a x − x 2 − 1 = log a ⎡ x + x 2 − 1 ⎢⎣ ⎛ = log a ⎜ x 2 − x 2 − 1 ⎝
(
(
= log a x 2 − x 2 − 1
)
)
x 2 −1 ⎤ ⎥⎦
2
= log a ( x 2 − x 2 + 1) = log a 1 = 0 89.
ln (1 + e 2 x ) = 2 x + ln (1 + e − 2 x ) We work from the complicated side (the right) and simplify to get to the left side of the equation. 2 x + ln (1 + e − 2 x ) = ln e 2 x + ln (1 + e − 2 x ) = ln ⎡⎣e 2 x ⋅ (1 + e − 2 x ) ⎤⎦ = ln ⎡⎣e 2 x + e 2 x ⋅ e − 2 x ⎤⎦
= ln ⎡⎣e 2 x + 1⎤⎦ = ln (1 + e 2 x )
91. We show that − f ( x ) = log 1/ a x by using the Change of Base Formula. f ( x ) = log a x so − f ( x ) = − log a x = −
ln x ln a
ln x − ln a ln x ln x = = −1 1 ln a ln a
=
154 SECTION 2.6
Using the Change of base formula in reverse, the last expression becomes ln 1/ a x . So − f ( x ) = log 1/ a x
93. If f ( x ) = log a x , then
⎛1⎞ ⎛1⎞ f ⎜ ⎟ = log a ⎜ ⎟ = log a x −1 = − log a x = − f ( x) ⎝ x⎠ ⎝ x⎠ 95.
If A = log a M and B = log a N , then (writing each expression as an exponential) we get a A = M and a B = N . ⎛aA⎞ ⎛M ⎞ log a ⎜ ⎟ = log a ⎜ B ⎟ ⎝N⎠ ⎝a ⎠ = log a a A − B
= A− B = log a M − log a N 5
5
97. –6
6
–5
Y1 = log ( x
2
)
–6
6
–5
Y2 = 2log ( x )
Explanations may vary.
2.6 Continuously Compounded Interest 1. If $1000 is invested at 4% compounded continuously, the amount A after 3 years is 0.04 3 A = Pe rt = 1000e ( )( ) = 1000e 0.12 = $1127.50 3. If $500 is invested at 5% compounded continuously, the amount A after 3 years is 0.05 3 A = Pe rt = 500e ( )( ) = 500e 0.15 = $580.92 5. The present value of $100 that will be invested at 4% compounded continuously for 6 months is − 0.04 0.5 P = Ae – rt = 100e ( )( ) = 100e − 0.02 = $98.02 7. The present value of $500 that will be invested at 7% compounded continuously for 1 year is − 0.07 1 P = Ae – rt = 500e ( )( ) = 500e − 0.07 = $466.20
154 SECTION 2.6
Using the Change of base formula in reverse, the last expression becomes ln 1/ a x . So − f ( x ) = log 1/ a x
93. If f ( x ) = log a x , then
⎛1⎞ ⎛1⎞ f ⎜ ⎟ = log a ⎜ ⎟ = log a x −1 = − log a x = − f ( x) ⎝ x⎠ ⎝ x⎠ 95.
If A = log a M and B = log a N , then (writing each expression as an exponential) we get a A = M and a B = N . ⎛aA⎞ ⎛M ⎞ log a ⎜ ⎟ = log a ⎜ B ⎟ ⎝N⎠ ⎝a ⎠ = log a a A − B
= A− B = log a M − log a N 5
5
97. –6
6
–5
Y1 = log ( x
2
)
–6
6
–5
Y2 = 2log ( x )
Explanations may vary.
2.6 Continuously Compounded Interest 1. If $1000 is invested at 4% compounded continuously, the amount A after 3 years is 0.04 3 A = Pe rt = 1000e ( )( ) = 1000e 0.12 = $1127.50 3. If $500 is invested at 5% compounded continuously, the amount A after 3 years is 0.05 3 A = Pe rt = 500e ( )( ) = 500e 0.15 = $580.92 5. The present value of $100 that will be invested at 4% compounded continuously for 6 months is − 0.04 0.5 P = Ae – rt = 100e ( )( ) = 100e − 0.02 = $98.02 7. The present value of $500 that will be invested at 7% compounded continuously for 1 year is − 0.07 1 P = Ae – rt = 500e ( )( ) = 500e − 0.07 = $466.20
SECTION 2.6 155 9. If $1000 is invested at 2% compounded continuously, the amount A after 1 year is 0.02 1 A = Pe rt = 1000e ( )( ) = 1000e 0.02 = $1020.20
A – P = $20.20 interest was earned. 11. We need to deposit the present value of $5000 that will be invested at 3% compounded continuously for 4 years. − 0.03 4 P = Ae – rt = 5000e ( )( ) = 5000e − 0.12 = $4434.60
We need to deposit the present value of $5000 that will be invested at 3% compounded continuously for 8 years. − 0.03 8 P = Ae – rt = 5000e ( )( ) = 5000e − 0.24 = $3933.14 13. If P is invested, it will double when amount A = 2P. A = 2 P = Pe rt 2 = e 3r 3r = ln 2 ln 2 r= ≈ 0.2310 3 It will require an interest rate of about 23.1% compounded continuously to double an investment in 3 years. 15. If P is invested, it will triple when amount A = 3P. A = 3P = Pe rt 3 = e 0.10t 0.10t = ln 3 ln 3 t= ≈ 10.986 0.10 It will take about 11 years for an investment to triple at 10 % compounded continuously. 17. We need the present value of $1000 that will be invested at 9% compounded continuously for 1 year. − 0.09 1 P = Ae – rt = 1000e ( )( ) = 1000e − 0.09 = $913.93
We need the present value of $1000 that will be invested at 9% compounded continuously for 2 years. − 0.09 2 P = Ae – rt = 1000e ( )( ) = 1000e − 0.18 = $835.27 19. Tami and Todd need to invest the present value of $40,000 that will earn 3% compounded continuously for 4 years. − 0.03 4 P = Ae – rt = 40,000e ( )( ) = 40,000e − 0.12 = $35,476.82 21. If P is invested, it will triple when amount A = 3P. A = 3P = Pe rt 3 = e 5r
156 CHAPTER 2 REVIEW
5r = ln 3 ln 3 r= ≈ 0.2197 5 It will require an interest rate of about 22% compounded continuously to triple an investment in 5 years. 23. The rule of 70: (a) The actual time it takes to double an investment if r = 1% is ln 2 t= = 69.3147 years. 0.01 The estimated time it takes to double an investment using the Rule of 70 if r = 1% is 0.70 t= = 70.0 years. 0.01 The Rule of 70 overestimates the time to double money by 0.6853 year, about 8 months.
(b) The actual time it takes to double an investment if r = 5% is ln 2 t= = 13.8629 years. 0.05 The estimated time it takes to double an investment using the Rule of 70 if r = 5% is 0.70 t= = 14 years. 0.05 The Rule of 70 overestimates the time to double money by 0.1371 year, about 1.6 months. (c) The actual time it takes to double an investment if r = 10% is ln 2 t= = 6.9315 years. 0.10 The estimated time it takes to double an investment using the Rule of 70 if r = 10% is 0.70 t= = 7.0 years. 0.10 The Rule of 70 overestimates the time to double money by 0.0685 year, almost 1 month.
Chapter 2 Review TRUE-FALSE ITEMS 1. True
3. True
7. False
9. False
5. False
156 CHAPTER 2 REVIEW
5r = ln 3 ln 3 r= ≈ 0.2197 5 It will require an interest rate of about 22% compounded continuously to triple an investment in 5 years. 23. The rule of 70: (a) The actual time it takes to double an investment if r = 1% is ln 2 t= = 69.3147 years. 0.01 The estimated time it takes to double an investment using the Rule of 70 if r = 1% is 0.70 t= = 70.0 years. 0.01 The Rule of 70 overestimates the time to double money by 0.6853 year, about 8 months.
(b) The actual time it takes to double an investment if r = 5% is ln 2 t= = 13.8629 years. 0.05 The estimated time it takes to double an investment using the Rule of 70 if r = 5% is 0.70 t= = 14 years. 0.05 The Rule of 70 overestimates the time to double money by 0.1371 year, about 1.6 months. (c) The actual time it takes to double an investment if r = 10% is ln 2 t= = 6.9315 years. 0.10 The estimated time it takes to double an investment using the Rule of 70 if r = 10% is 0.70 t= = 7.0 years. 0.10 The Rule of 70 overestimates the time to double money by 0.0685 year, almost 1 month.
Chapter 2 Review TRUE-FALSE ITEMS 1. True
3. True
7. False
9. False
5. False
CHAPTER 2 REVIEW 157 FILL IN THE BLANKS 1. parabola
3. x = −
7. (0, ∞)
9. one
b 2a
5. one
REVIEW EXERCISES 1.
First we expand the function. 2 f ( x ) = ( x − 2) + 2 = x 2 − 4x + 4 + 2 = x 2 − 4 x + 6 and find that a = 1, b = – 4, c = 6.
a > 0; the parabola opens up.
( − 4 ) = 2 , and the y-coordinate is b =− 2 (1) 2a 2 f ( 2) = ( 2) − 4 ( 2) + 6 = 2 . So the vertex is (2, 2) and the axis of symmetry is the line x = 2. The x-coordinate of the vertex is −
Since f (0) = c = 6, the y-intercept is (0, 6). The x-intercepts are found by solving f (x) = 0. Since the discriminant of f b 2 – 4ac = (– 4) 2 – 4(1)(6) = – 8 is negative, the equation f (x) = 0 has no real solution, and therefore, the parabola has no x-intercept. To graph the function, we use symmetry. If we choose the y-intercept, we obtain its symmetric point (4, 6).
3.
1 , b = 0, c = – 16 4 a > 0, the parabola opens up.
a=
The x-coordinate of the vertex is −
b = 0, and the y-coordinate is 2a
1 2 f ( 0 ) = ( 0 ) − 16 = −16 . 4 So the vertex is (0, – 16), and the axis of symmetry is the line x = 0. Since f (0) = c = – 16, the y-intercept is also (0, – 16) The x-intercepts are found by solving f (x) = 0.
158 CHAPTER 2 REVIEW
1 2 x − 16 = 0 4 x 2 − 64 = 0 x 2 = 64 x=±8 The x-intercepts are (8, 0) and (– 8, 0).
5. a = – 4, b = 4, c = 0 a < 0, the parabola opens down.
The x-coordinate of the vertex is −
b 4 1 =− = , and the y-coordinate is 2 ( − 4) 2 2a 2
⎛1⎞ ⎛1⎞ f ⎜ ⎟ = − 4⎜ ⎟ + 4x = 1 ⎝2⎠ ⎝2⎠ 1 ⎛1 ⎞ So the vertex is ⎜ , 1⎟ , and the axis of symmetry is the line x = . 2 ⎝2 ⎠ Since f (0) = c = 0, the y-intercept is (0, 0). The x-intercepts are found by solving f (x) = 0. − 4 x 2 + 4x = 0 − 4 x ( x − 1) = 0
x = 0 or x = 1 The x-intercepts are (0, 0) and (1, 0).
7.
a=
9 , b = 3, c = 1 2
a > 0; the parabola opens up. The x-coordinate of the vertex is −
b 3 1 =− = − , and the y-coordinate is 3 ⎛9⎞ 2a 2⎜ ⎟ ⎝2⎠ 2
1 ⎛ 1⎞ 9⎛ 1⎞ ⎛ 1⎞ f ⎜ − ⎟ = ⎜ − ⎟ + 3⎜ − ⎟ + 1 = 2 ⎝ 3⎠ 2⎝ 3⎠ ⎝ 3⎠ 1 ⎛ 1 1⎞ So the vertex is ⎜ − , ⎟ , and the axis of symmetry is the line x = − . 3 ⎝ 3 2⎠
CHAPTER 2 REVIEW 159
Since f (0) = c = 1, the y-intercept is (0, 1). The x-intercepts are found by solving f (x) = 0. Since the discriminant of f ⎛9⎞ b 2 – 4ac = 3 2 − ( 4 ) ⎜ ⎟ (1) = − 9 ⎝2⎠ is negative, the equation f (x) = 0 has no real solution, and therefore, the parabola has no x-intercept. To graph the function, we use symmetry. If we choose the y-intercept, we obtain its ⎛ 2 ⎞ symmetric point ⎜ − , 1⎟ . Because this point is so close to the vertex, we also use the ⎝ 3 ⎠ 5⎞ ⎛ point ⎜ −1, ⎟ . 2⎠ ⎝
9. a = 3, b = 4, c = – 1
a > 0; the parabola opens up. The x-coordinate of the vertex is −
b 4 2 =− = − , and the y-coordinate is 2a 2 ( 3) 3 2
7 ⎛ 2⎞ ⎛ 2⎞ ⎛ 2⎞ f ⎜ − ⎟ = 3⎜ − ⎟ + 4 ⎜ − ⎟ − 1 = − 3 ⎝ 3⎠ ⎝ 3⎠ ⎝ 3⎠ 7⎞ 2 ⎛ 2 So the vertex is ⎜ − , − ⎟ and the axis of symmetry is the line x = − . 3 3⎠ ⎝ 3
Since f (0) = c = – 1, the y-intercept is (0, – 1). The x-intercepts are found by solving f (x) = 0. 3x 2 + 4 x − 1 = 0 x=
x=
− 4 ± 4 2 − 4 ( 3)( − 1) 2 ( 3)
=
− 4 ± 28 6
− 4±2 7 − 2± 7 = 6 3
The x-intercepts are difficult to graph, so we choose another point and use symmetry. ⎛2 ⎞ We chose (– 2, 3) and by symmetry ⎜ , 3 ⎟ . ⎝3 ⎠
160 CHAPTER 2 REVIEW
11. a = 3, b = – 6, c = 4 The quadratic function has a minimum since a = 3 is greater than zero. b −6 x= − =− =1 2a 2 ( 3)
f (1) = 3 (1) − 6 (1) + 4 = 1 2
The minimum value is 1, and it occurs at x = 1. 13. a = – 1, b = 8, c = – 4 The quadratic function has a maximum since a = – 1 is less than zero. b 8 x= − =− =4 2a 2 ( −1)
f ( 4 ) = − ( 4 ) + 8 ( 4 ) − 4 = 12 2
The maximum value is 12, and it occurs at x = 4. 15. a = – 3, b = 12, c = 4 The quadratic function has a maximum since a = – 3 is less than zero. b 12 x= − =− =2 2a 2 ( −3)
f ( 2 ) = − 3 ( 2 ) + 12 ( 2 ) + 4 = 16 2
The maximum value is 16, and it occurs at x = 2. 17. Answers will vary. 19.
y = x4
→ Add 2; vertical shift 2 units up.
f ( x) = x 4 + 2
CHAPTER 2 REVIEW 161 21.
y = x5
→ Add 1; vertical shift up 1 unit.
→ Multiply y by – 1; reflect about x-axis.
– y = – x5
f ( x) = – x 5 + 1
23. f is a polynomial function. Its degree is 5. 25. f is not a polynomial function. The exponent on the middle term is not a positive integer. 27. The power function that models the end behavior of the function f is p(x) = – 2x 4. 29. The domain of a rational function is all real numbers except those that make the denominator zero. The denominator of R(x), x 2 – 9 = 0 when x = – 3 or when x = 3. The domain of R is {x | x ≠ – 3 or x ≠ 3}. 31. The domain of a rational function is all real numbers except those that make the denominator zero. The denominator of R(x), (x + 2) 2 = 0, when x = – 2. The domain of R is {x | x ≠ – 2}. 33. (a) f (4) = 3 4 = 81
(c) f (– 2) = 3
35.
52 = z log 5 z = 2
−2
1 = 9
(b) g(9) = log 3 9 = log 3 3 2 = 2 ⎛ 1 ⎞ ⎛ 1 ⎞ ⎛ 1 ⎞ g ⎜ ⎟ = log 3 ⎜ ⎟ = log 3 ⎜ 3 ⎟ (d) ⎝ 27 ⎠ ⎝ 27 ⎠ ⎝3 ⎠ −3 = log 3 3 = − 3 37.
log 5 u = 13
5 13 = u
162 CHAPTER 2 REVIEW 39.
41.
2 The domain of f consists of all x for which 3x – 2 > 0, that is for all x > , or using 3 2 ⎛ ⎞ interval notation, ⎜ , ∞ ⎟ . ⎝3 ⎠ The domain of H consists of all x for which – 3x + 2 > 0, that is for all x
0}. The x-axis is a horizontal asymptote as f becomes unbounded in the negative direction. 69.
→ y=e
x
→
Multiply y by – 1;
–y=– e
reflect about the x-axis.
x
Add 1; vertical
f ( x) = 1 − e x
shift up 1 unit.
The domain of the function f is all real numbers, or in interval notation (– ∞, ∞); the range is all {y | y < 1}. The line y = 1 is a horizontal asymptote as f becomes unbounded in the negative direction. 71.
→ y = ln x
Add 3; vertical shift up 3 units.
f (x) = 3 + ln x
The domain of the function f is {x | x > 0}, or in interval notation (0, ∞); the range is all real numbers. The y-axis is a vertical asymptote. 73.
4 1− 2 x = 2
(2 )
2 1− 2 x
=2
2 (1− 2 x )
2 = 21 2 − 4x = 1 4x = 1 1 x= 4 75.
3x
2
+x
= 3 = 3 1/ 2 1 x2 + x = 2
Set exponents equal.
CHAPTER 2 REVIEW 165
1 =0 2 2x 2 + 2x − 1 = 0 x2 + x−
x=
−2 ± 2 2 − 4 ( 2 )( −1) 2 ( 2)
=
−2 ± 4 + 8 −2 ± 12 − 2 ± 2 3 −1 ± 3 = = = 4 4 2 42
⎧⎪ −1 + 3 −1 − 3 ⎫⎪ x=⎨ , ⎬ 2 ⎪⎭ ⎩⎪ 2 77.
log x 64 = −3
79.
x − 3 = 64 = 4 3
9 2 x = 27 3 x − 4
(3 ) 2
3
−3 ⎡⎛ 1 ⎞ − 1 ⎤ ⎛1⎞ x = ⎢⎜ ⎟ ⎥ = ⎜ ⎟ ⎝4⎠ ⎣⎢⎝ 4 ⎠ ⎦⎥ 1 x= 4 −3
81.
2x
= (3 3 )
3x −4
3 4x = 3 ( ) 4 x = 3 ( 3x − 4 ) 3 3x−4
4 x = 9 x − 12 5x = 12 12 x= 5
log 3 ( x − 2 ) = 2
32 = x − 2 9=x–2 x = 11 83. If $100 is invested at 10% compounded continuously, the amount A after 2.25 years (2 years and 3 months) is 0.1 2.25 A = Pe rt = 100e ( )( ) = 100e 0.225 = $125.23 85. I need to invest the present value of $1000 at 4% compounded continuously for 2 years. − 0.04 2 P = Ae – rt = 1000e ( )( ) = 1000e − 0.08 = $923.12 87. If P is invested, it will double when amount A = 2P. A = 2 P = Pe rt 2 ( 220,000 ) = 220,000e 0.06t
2 = e 0.06t 0.06t = ln 2 ln 2 ≈ 11.55 t= 0.06 It will take about 11 and one half years for an investment to double at 6 % compounded continuously.
166 CHAPTER 2 REVIEW 89. When T = 0° and P 0 = 760 mm. of mercury, ⎛ 760 ⎞ h ( x ) = 8000log ⎜ ⎟ ⎝ x ⎠ ⎛ 760 ⎞ h ( 300 ) = 8000log ⎜ ⎟ = 3229.54 ⎝ 300 ⎠ The Piper Cub is flying at an altitude of approximately 3230 meters above sea level. 91. The perimeter of the pond is P = 200 = 2l + 2w 200 − 2w l= = 100 − w 2
If x denotes the width of the pond, then A = lw = (100 – x)x =100x – x 2 A = A(x) is a quadratic function with a = – 1, so the maximum value of A is found at the vertex of A. b 100 x= − =− = 50 2a 2 ( −1) A ( 50 ) = 100(50) − 502 = 2500 square feet.
The dimensions should be 50 feet by 50 feet. 93. The perimeter of the window is 100 feet and can be expressed as P = 2l + πd = 2l + π x = 100
Solving for l in terms of x, we find 100 − π x πx l= = 50 − 2 2 The area A of the rectangle is a quadratic function given by πx⎞ π x2 ⎛ A ( x ) = ⎜ 50 − x x 50 = − ⎟ 2 ⎠ 2 ⎝ The maximum area is found at the vertex of A. The width of the rectangle with maximum area is 50 50 b =− = feet x=− 2a ⎛ π⎞ π 2⎜ − ⎟ ⎝ 2⎠ The length of the rectangle is ⎛ 50 ⎞ π⎜ ⎟ πx π l = 50 − = 50 − ⎝ ⎠ = 50 − 25 = 25 feet 2 2 95. (a) A 3.5 inch telescope has a limiting magnitude of L = 9 + 5.1 log 3.5 = 11.77
CHAPTER 2 REVIEW 167
(b) If the star’s magnitude is 14, the telescope must have a lens with a diameter of 9 + 5.1 log d = 14 5.1 log d = 5 5 log d = 5.1 d = 10 5 / 5.1 = 9.56 inches. 97.
(a) If $620.17 grows to $5000 in 20 years when interest is compounded continuously, the interest rate is 10.436% A = Pe rt 5000 = 620.17 e 20 r 5000 e 20 r = 620.17 ⎛ 5000 ⎞ 20r = ln ⎜ ⎟ ⎝ 620.17 ⎠ 1 ⎛ 5000 ⎞ r= ln ⎜ ⎟ = 0.10436 20 ⎝ 620.17 ⎠ (b) An investment of $4000 will have a value A in 20 years if it is invested at 10.436% compounded continuously. A = Pe rt = 4000 e (0.10436)(20) = $32, 249.24
99. (a) The Calloway Company will minimize marginal cost if it produces ( − 617.4 ) = 63 golf clubs. b x=− =− 2a 2 ( 4.9 )
(b) The marginal cost of making the 64 golf club is 2 C ( 63) = 4.9 ( 63) − 617.4 ( 63) + 19,600 = $151.90
168 CHAPTER 2 PROJECT CHAPTER 2 PROJECT 1.
3. (a) t = 150 represents the year 2000. The projected population at t = 150 is P (150 ) = 2.4 ⋅ 1.59 150 /10 = 2.4 ⋅ 1.59 15 = 2518.7 thousand persons.
(b) The projected population overestimates the actual population by 565.1 thousand persons, or by 28.9%. (c) Explanations will vary. (d) t = 160 represents the year 2010. Using the exponential growth function, the predicted population is P (160 ) = 2.4 ⋅ 1.59 160 /10 = 2.4 ⋅ 1.59 16 = 4004.7 thousand persons. (e) t = 200 represents the year 2050. Using the exponential growth function, the predicted population is P ( 200 ) = 2.4 ⋅ 1.59 200 /10 = 2.4 ⋅ 1.59 20 = 25,594 The predicted population is 25.594 million people. 5.
The growth rate for Houston is P ( t ) = 2.4 e
ln1.59 ( t /10)
7. Answers will vary. 9. Answers will vary. 11. 3. To do this problem with the result from above, use TBLSET Indpnt: Ask, TABLE. (a) The population of Houston in 2000 (t = 50) is predicted to be 2159.1 thousand.
(b) The prediction overestimates the actual population by 205.5 thousand, or 10.5%.
MATHEMATICAL QUESTIONS 169
(c) Answers will vary. (d) The population of Houston in 2010 (t = 60) is predicted to be 2702.0 thousand. (e) The population of Houston in 2050 (t = 100) is predicted to be 6627.0 thousand. 4. If P = P0 a t /10 = P0 e k t /10 , where P0 = 703.4597 and a = 1.0227 then a t /10 = e k t /10 k = ln a = ln 1.0227 = 0.022446 5. The growth rate for Houston is P ( t ) = 703.4597 e
ln1.0227 ( t /10 )
Explanations will vary. MATHEMATICAL QUESTIONS FROM PROFESSIONAL EXAMS 1.
(a) 4 · 27 = 2 2 · 3 3 = 6 2 · 3 =
62 ⋅ 3 ⋅ 2 62 ⋅ 6 63 = = 2 2 2
⎛ 63 ⎞ log 6 ( 4 ⋅ 27 ) = log 6 ⎜ ⎟ = log 6 ( 6 3 ) − log 6 ( 2 ) = 3 − b ⎝ 2 ⎠
3.
(c)
e 2x
e x − e −x y= 2 x 2y = e − e −x 2 ye x = e x ⋅ e x − e − x ⋅ e x 2 ye x = e 2 x − 1 − 2 ye x − 1 = 0
Multiply both sides by e x. a = 1, b = – 2y, c = – 1
2y ± 4y 2 + 4 e = 2 x
2y ±2 y 2 +1 = 2 = y ± y 2 +1 Since
e x − e −x > 0 , y > 0, so e x = y + y 2 + 1 . 2
(
x = ln y + y 2 + 1 5.
(b)
log b log a ⋅ =1 ( log a b ) ( log b a ) = log a log b
)
170 MATHEMATICAL QUESTIONS 7. (a)
e
2 ln ( x −1) ln ( x −1)
=4
2
=4 ( x − 1) = 4
e
2
x–1=±2 x=3
Chapter 2 – Classes of Functions Section 2.1 Quadratic Functions 47.
Analyzing the Motion of a Projectile A projectile is fired from a cliff 200 feet above the water at an inclination of 45 to the horizontal, with a muzzle velocity of 50 feet per second. The height h of the projectile above the water is given by −32 x 2 h ( x) = + x + 200 2 ( 50 ) where x is the horizontal distance of the projectile from the base of the cliff. (d)
Using a graphing utility, graph the function h, 0 ≤ x ≤ 200 .
Enter the formula for h into the function editor. Go to WINDOW and enter limits for x. We must determine limits for y, so try 0 ≤ y ≤ 300 . If this does not work, then we can go back to WINDOW and adjust the limits on y until we find a good window.
(e)
When the height of he projectile is 100 feet above the water, how far is it from the cliff?
Return to the function editor, input the equation y = 100 , and graph the two equations.
57
The point where the graph of h and the horizontal line y = 100 intersect represents the point where the projectile is 100 feet above the water. Recall that you can find the point of intersection on your calculator using the intersect function, which can be found in the CALC menu. The x-coordinate of the point of intersection appears to be close to x = 125 .
Thus, the point of intersection is approximately (135.698,100 ) . When the projectile is 100 feet above the water it is approximately 135.698 feet from the cliff.
58
53.
Hunting The function H ( x ) = −1.01x 2 + 114.3 x + 451.0 models the number of individuals who engage in hunting activities whose annual income is x thousand dollars.
(b)
Using a graphing utility, graph H = H ( x ) . Are the number of hunters increasing or decreasing for individuals earning between $20,000 and $40,000?
Enter the formula for H into the function editor. Since x is in thousands, the earnings range of $20,000 to $40,000 corresponds to 20 ≤ x ≤ 40 . Go to WINDOW and enter limits 20 ≤ x ≤ 40 for x. We must determine limits for y, so try 0 ≤ y ≤ 1000 . If this does not work, then we can go back to WINDOW and adjust the limits on y until we find a good window.
There is no graph in our viewing rectangle, so we will need to find a better window. Lets try 0 ≤ y ≤ 5000 .
59
This time we can see the graph. The graph is increasing on this interval, so the number of hunters is increasing for individuals earning between $20,000 and $40,000.
55.
Male Murder Victims The function M ( x ) = 0.76 x 2 − 107.00 x + 3854.18 models
the number of male murder victims who are x years of age ( 20 ≤ x ≤ 90 ). (c)
Using a graphing utility, graph M = M ( x ) .
Enter the formula for h into the function editor. Go to WINDOW and enter limits for x. We must determine limits for y, so try 0 ≤ y ≤ 4000 . If this does not work, then we can go back to WINDOW and adjust the limits on y until we find a good window.
________________________________________________________________________
Section 2.3 Exponential Functions 63.
Exponential Probability Between 12:00 PM and 1:00 PM, cars arrive at Citibank’s drive-thru at the rate of 6 cars per hour (0.1 car per minute). The following formula from probability can be used to determine the probability that a car will arrive within t minutes of 12:00 PM: F ( t ) = 1 − e −0.1t
60
This time we can see the graph. The graph is increasing on this interval, so the number of hunters is increasing for individuals earning between $20,000 and $40,000.
55.
Male Murder Victims The function M ( x ) = 0.76 x 2 − 107.00 x + 3854.18 models
the number of male murder victims who are x years of age ( 20 ≤ x ≤ 90 ). (c)
Using a graphing utility, graph M = M ( x ) .
Enter the formula for h into the function editor. Go to WINDOW and enter limits for x. We must determine limits for y, so try 0 ≤ y ≤ 4000 . If this does not work, then we can go back to WINDOW and adjust the limits on y until we find a good window.
________________________________________________________________________
Section 2.3 Exponential Functions 63.
Exponential Probability Between 12:00 PM and 1:00 PM, cars arrive at Citibank’s drive-thru at the rate of 6 cars per hour (0.1 car per minute). The following formula from probability can be used to determine the probability that a car will arrive within t minutes of 12:00 PM: F ( t ) = 1 − e −0.1t
60
(d)
Graph F using your graphing utility.
Enter the formula for F into the function editor using x instead of t. Since x is in minutes, the time range 12:00 PM to 1:00 PM corresponds to 0 ≤ x ≤ 60 . Go to WINDOW and enter the limits 0 ≤ x ≤ 60 for x. We must determine limits for y, so try 0 ≤ y ≤ 1 . If this does not work, then we can go back to WINDOW and adjust the limits on y until we find a good window.
(e)
Using TRACE, determine how many minutes are needed for the probability to reach 50%.
The TRACE feature uses the points on the curve that the calculator generated to draw the curve. As you TRACE along the curve, the distance between successive x-coordinates of points is the same. It is determined by ∆x =
Xmax − Xmin 94
Because of this, as we trace along the curve, we may not get “nice” values for x (with the window used for the problem we will not get nice values). We must trace until we find the value of x that corresponds to the value y that is closest to 0.5. Select the TRACE feature.
61
r
Note that your cursor may not be in the same spot as shown above. Use the | and ~ keys to move the cursor along the curve. Move the cursor until the value of y changes from more than 0.5 to less than 0.5.
Since y is less than 0.5 when x ≈ 6.383 , and y is more than 0.5 when x ≈ 7.021 , then y is equal to 0.5 when 6.383 < x < 7.021 . We can improve our approximation if we TRACE over a smaller viewing rectangle. We can obtain a smaller viewing window by “zooming in” on the graph using Zoom In.
q
Select Zoom In.
Í
62
Be sure the y-value of the cursor is close to 0.5. Zoom in on the graph.
Á
Use TRACE and move the cursor so that the y-value is close 0.5.
Since y is less than 0.5 when x ≈ 6.861 , and y is more than 0.5 when x ≈ 7.021 , then y is equal to 0.5 when 6.383 < x < 7.021 . We can continue to improve our solution by zooming in on our graph. Zoom In again, then TRACE.
. Since y is less than 0.5 when x ≈ 6.902 , and y is more than 0.5 when x ≈ 6.914 , then y is equal to 0.5 when 6.902 < x < 6.914 . We can continue to improve our solution by zooming in on our graph.
63
Zoom In again, then TRACE.
. Since y is less than 0.5 when x ≈ 6.921 , and y is more than 0.5 when x ≈ 6.932 , then y is equal to 0.5 when 6.921 < x < 6.932 . We can continue to improve our solution by zooming in on our graph. Zoom In again, then TRACE.
. Since y is less than 0.5 when x ≈ 6.929 , and y is more than 0.5 when x ≈ 6.932 , then y is equal to 0.5 when 6.929 < x < 6.932 . We can continue to improve our solution by zooming in on our graph. Zoom In again, then TRACE.
. Since y is less than 0.5 when x ≈ 6.9309 , and y is more than 0.5 when x ≈ 6.9315 , then y is 0.5 when x is approximately 6.931. Thus, the probability will reach 50% after 6.931 minutes.
64
79.
The hyperbolic sine function, designated by sinh x, is defined as 1 sinh x = ( e x − e − x ) 2 (b)
Graph f ( x ) = sinh x using a graphing utility.
Enter the formula for f into the function editor. Go to WINDOW and enter limits for x and y. Since we are not given limits for x or y, use the standard window, −10 ≤ x ≤ 10 and −10 ≤ y ≤ 10 .
________________________________________________________________________
Section 2.5 Properties of Logarithms In Problems 67–72, graph each function using a graphing utility and the Change-of-Base formula. 67.
y = log 4 x
log b M . Because we will log b a use our calculator to graph the equation, we must use 10 or e for the new base b. Use e as the new base. Thus
Recall that the Change-of-Base formula states that log a M =
65
79.
The hyperbolic sine function, designated by sinh x, is defined as 1 sinh x = ( e x − e − x ) 2 (b)
Graph f ( x ) = sinh x using a graphing utility.
Enter the formula for f into the function editor. Go to WINDOW and enter limits for x and y. Since we are not given limits for x or y, use the standard window, −10 ≤ x ≤ 10 and −10 ≤ y ≤ 10 .
________________________________________________________________________
Section 2.5 Properties of Logarithms In Problems 67–72, graph each function using a graphing utility and the Change-of-Base formula. 67.
y = log 4 x
log b M . Because we will log b a use our calculator to graph the equation, we must use 10 or e for the new base b. Use e as the new base. Thus
Recall that the Change-of-Base formula states that log a M =
65
y = log 4 x =
log e x ln x = log e 4 ln 4
ln x into the function editor. Go to WINDOW and enter limits for x and y. Since ln 4 we are not given limits for x or y, use the viewing rectangle 0 ≤ x ≤ 10 and −10 ≤ y ≤ 10 .
Enter y =
69.
y = log 2 ( x + 2 )
Using the Change-of-Base formula we obtain y = log 2 ( x + 2 ) =
ln ( x + 2 ) ln 2
ln ( x + 2 ) into the function editor. Go to WINDOW and enter limits for x and y. ln 2 Since we are not given limits for x or y, use the viewing rectangle −3 ≤ x ≤ 10 and −10 ≤ y ≤ 10 .
Enter y =
66
71.
y = log x −1 ( x + 1)
Using the Change-of-Base formula we obtain y = log x −1 ( x + 1) =
ln ( x + 1) ln ( x − 1)
ln ( x + 1) into the function editor. Go to WINDOW and enter limits for x and y. ln ( x − 1) Since we are not given limits for x or y, use the viewing rectangle 0 ≤ x ≤ 10 and −10 ≤ y ≤ 10 .
Enter y =
67
97.
Graph Y1 = log ( x 2 ) and Y2 = 2 log ( x ) on your graphing utility. Are they equivalent? What might account for any differences in the two functions?
Enter Y1 = log ( x 2 ) into the function editor. Go to WINDOW and enter limits for x and y.
Since we are not given limits for x or y, use the viewing rectangle −10 ≤ x ≤ 10 and −10 ≤ y ≤ 10 .
Return to the function editor, clear the first function, enter Y2 = 2 log ( x ) and graph on the same view rectangle.
68
No, they are not equivalent, because the graphs are not identical. The difference is due to the range of x and x 2 . The range of x is all real numbers, but the domain of log ( x ) is
just x > 0 . The range of x 2 is x ≥ 0 , so the domain of log ( x 2 ) is all real numbers except 0.
________________________________________________________________________
Chapter 2 Review 99.
Minimizing Marginal Cost The marginal cost of a product can be thought of as the cost of producing one additional unit of output. For example, if the marginal cost of producing the 50th product is $6.20, then it cost $6.20 to increase production from 49 to 50 units of output. Callaway Golf Company has determined that the marginal cost C of manufacturing x Big Bertha golf clubs may be expressed by the quadratic function C ( x ) = 4.9 x 2 − 617.4 x + 19, 600
(a)
How many golf clubs should be manufactured to minimize the marginal cost?
Enter the formula for C into the function editor. Go to WINDOW and enter limits for x and y. Since we are not given limits for x or y, use the viewing rectangle 0 ≤ x ≤ 150 and 0 ≤ y ≤ 30000 .
69
No, they are not equivalent, because the graphs are not identical. The difference is due to the range of x and x 2 . The range of x is all real numbers, but the domain of log ( x ) is
just x > 0 . The range of x 2 is x ≥ 0 , so the domain of log ( x 2 ) is all real numbers except 0.
________________________________________________________________________
Chapter 2 Review 99.
Minimizing Marginal Cost The marginal cost of a product can be thought of as the cost of producing one additional unit of output. For example, if the marginal cost of producing the 50th product is $6.20, then it cost $6.20 to increase production from 49 to 50 units of output. Callaway Golf Company has determined that the marginal cost C of manufacturing x Big Bertha golf clubs may be expressed by the quadratic function C ( x ) = 4.9 x 2 − 617.4 x + 19, 600
(a)
How many golf clubs should be manufactured to minimize the marginal cost?
Enter the formula for C into the function editor. Go to WINDOW and enter limits for x and y. Since we are not given limits for x or y, use the viewing rectangle 0 ≤ x ≤ 150 and 0 ≤ y ≤ 30000 .
69
Recall that we can use the minimum function to find the local minimum. Note that the xcoordinate of the local minimum is between x = 50 and x = 75 .
The local minimum is approximately ( 63.0, 151.9 ) . Thus, Callaway should produce 63 Big Bertha clubs. (b)
At this level of production, what is the marginal cost?
The marginal cost for producing 63 Big Bertha clubs is $151.90.
70
Summary The commands introduced in this chapter were: TRACE Zoom In
71
72
Chapter 3 The Limit of a Function 3.1 Finding Limits Using Tables and Graphs 1. Here f (x) = 2x, and c = 1. We complete the table by evaluating the function f at each value of x. x 0.9 0.99 0.999 f ( x) = 2x 1.8 1.98 1.998 x 1.1 1.01 1.001 f ( x) = 2x 2.2 2.02 2.002 We infer from the table that lim f ( x ) = lim 2 x = 2 . x →1
x →1
3. Here f (x) = x 2 + 2, and c = 0. We complete the table by evaluating the function f at each value of x. x −0.1 −0.01 −0.001 f ( x ) = x2 + 2 x f ( x) = x + 2 2
2.01 0.1
2.0001 0.01
2.0000 0.001
2.01
2.0001
2.0000
We infer from the table that lim f ( x ) = lim ( x 2 + 2 ) = 2. x →0
x →0
x2 − 4 , and c = – 2. We complete the table by evaluating the function f at x+2 each value of x. x −2.1 −2.01 −2.001
5. Here f (x) =
f ( x) =
x2 − 4 x+2
x f ( x) =
x2 − 4 x+2
−4.1
−4.01
−4.001
−1.9
−1.99
−1.999
−3.9
−3.99
−3.999
⎛ x2 − 4 ⎞ We infer from the table that lim f ( x) = lim ⎜ ⎟ = −4 . x →−2 x →−2 ⎝ x+2 ⎠
x3 + 1 , and c = – 1. We complete the table by evaluating the function f at x +1 each value of x.
7. Here f (x) =
172 SECTION 3.1 −1.1
x f ( x) =
x +1 x +1
3.31 −0.9
x f ( x) =
−1.01
−1.001
3
3.0301 −0.99
3.0030 −0.999
3
x +1 x +1
2.71
2.9701
2.9970
⎛ x3 + 1 ⎞ We infer from the table that lim f ( x) = lim ⎜ ⎟ = 3. x →−1 x →−1 ⎝ x +1 ⎠
9. Here f (x) = 4x 3 , and c = 2. We choose values of x close to 2, starting at 1.99. Then we select additional numbers that get closer to 2, but remain less than 2. Next we choose values of x greater than 2, starting with 2.01, that get closer to 2. Finally we evaluate the function f at each choice to obtain the table: x 1.99 1.999 1.9999 → ← 2.0001 2.001 2.01 f ( x) = 4 x3
31.522
31.952
31.995
→ ←
32.005
32.048
32.482
We infer that as x gets closer to 2, f gets closer to 32. That is, lim f ( x ) = lim ( 4 x3 ) = 32 x →2
x→2
11. Here f (x) = x2 + 1 , and c = 0. We choose values of x close to 0, starting at – 0.01. Then x +1 we select additional numbers that get closer to 0, but remain less than 0. Next we choose values of x greater than 0, starting with 0.01, that get closer to 0. Finally we evaluate the function f at each choice to obtain the table: x −0.01 −0.001 −0.0001 → ← 0.0001 0.001 0.01 x +1 f ( x) = 2 0.9899 0.9990 0.9999 → ← 1.0001 1.001 1.0099 x +1
We infer that as x gets closer to 0, f gets closer to 1. That is, ⎛ x +1 ⎞ lim f ( x ) = lim ⎜ 2 ⎟ =1 x →0 x →0 x + 1 ⎝ ⎠ 13.
x2 − 4 x , and c = 4. We choose values of x close to 4, starting at 3.99. Then x−4 we select additional numbers that get closer to 4, but remain less than 4. Next we choose values of x greater than 4, starting with 4.01, that get closer to 4. Finally we evaluate the function f at each choice to obtain the table: x 3.99 3.999 3.9999 → ← 4.0001 4.001 4.01
Here f (x) =
f ( x) =
x2 − 4 x x−4
3.99
3.999
3.9999
→ ←
4.0001
4.001
4.01
SECTION 3.1 173
We infer that as x gets closer to 4, f gets closer to 4. That is, ⎛ x2 − 4 x ⎞ lim f ( x ) = lim ⎜ ⎟=4 x →4 x→4 x 4 − ⎝ ⎠ 15. Here f (x) = ex + 1, and c = 0. We choose values of x close to 0, starting at – 0.01. Then we select additional numbers that get closer to 0, but remain less than 0. Next we choose values of x greater than 0, starting with 0.01, that get closer to 0. Finally we evaluate the function f at each choice to obtain the table: x −0.01 −0.001 −0.0001 → ← 0.0001 0.001 0.01 f ( x) = e x + 1 1.9900 1.9990 1.9999 → ← 2.0001 2.0010 2.0101
We infer that as x gets closer to 0, f gets closer to 2. That is, lim f ( x ) = lim ( e x + 1) = 2 x →0
17.
x →0
To determine the lim f ( x ) we observe that as x gets closer to 2, f (x) gets closer to 3. So x →2
We conclude that
lim f ( x ) = 3 x →2
19.
To determine the lim f ( x ) we observe that as x gets closer to 2, f (x) gets closer to 4. x →2
We conclude that
lim f ( x ) = 4 x →2
21.
To determine the lim f ( x ) we observe that as x gets closer to 3, but remains less than 3, x→ 3
the value of f gets closer to 3. However, we see that as x gets closer to 3, but remains greater than 3, the value of f gets closer to 6. Since there is no single number that the values of f are close to when x is close to 3, we conclude that lim f ( x ) does not exist. x→ 3
23.
We conclude from the graph that lim f ( x ) = lim ( 3 x + 1) = 13 x →4
x →4
174 SECTION 3.1 25.
We conclude from the graph that lim f ( x ) = lim 1 − x 2 = – 3 x →2
x →2
(
)
27. We conclude from the graph that lim f ( x ) = lim ( x − 2 ) = 1
x→−3
x→−3
29. We conclude from the graph that
lim f ( x ) = lim e x = 1
x→0
x→0
31. We conclude from the graph that 1 lim f ( x ) = lim = −1 x → −1 x → −1 x
33. We conclude from the graph that
lim f ( x ) = 0
x→0
Notice that lim− f ( x ) = lim− x 2 = 0 x→0
and
x→0
lim f ( x ) = lim+ ( 2 x ) = 0
x→0+
x→0
35. We conclude from the graph that lim f ( x ) does not exist. x →1
Notice that lim− f ( x ) = lim− 3 x = 3 x →1
x →1
but lim+ f ( x ) = lim+ ( x + 1) = 2 x →1
x →1
SECTION 3.1 175 37. We conclude from the graph that lim f ( x ) = 0 x→0
Notice that lim− f ( x ) = lim− x = 0 x→0
x→0
and lim+ f ( x ) = lim+ ( 3 x ) = 0 x→0
x→0
39. We conclude from the graph that lim f ( x ) = 0 x→0
Notice that lim− f ( x ) = lim− ( e x − 1) = 1 − 1 = 0 x→0
and
x→0
lim+ f ( x ) = lim+ x 2 = 0
x→0
x→0
41. To find the limit we create the tables shown below and conclude x 3 − x 2 + x −1 2 lim 4 = 3 x→ 1 x − x + 2x − 2 3
43. To find the limit we create the tables shown below and conclude x3 − 2 x 2 + 4 x − 8 lim = 1.6 x →−2 x2 + x − 6
45. To find the limit we create the tables shown below and conclude x3 + 2 x 2 + x =0 lim 4 x →−1 x + x 3 + 2 x + 2
176 SECTION 3.2
3.2 Techniques for Finding Limits of Functions 3. Using formula (2) (p. 243), we find lim x = 4
1. Using formula (1) (p. 243), we find lim 5 = 5
x →4
x→1
5.
lim ( 3 x + 2 ) = lim ( 3x ) + lim 2 = ⎡⎢ lim 3⎤⎥ ⎡⎢ lim x ⎤⎥ + lim 2 x →2 x →2 x →2 ⎣ x →2 ⎦ ⎣ x →2 ⎦ x →2 = (3) · (2) + 2 = 6 + 2 = 8
7.
lim 3x 2 − 5 x = 3 ⋅ ( −1) − 5 ⋅ ( −1) = 3 ⋅ 1 + 5 = 3 + 5 = 8
x → −1
9.
(
)
2
(
)
lim 5 x 4 − 3 x 2 + 6 x − 9 = 5 · 14 – 3 · 12 + 6 · 1 – 9 x →1
=5·1–3·1+6–9 = 5 – 3 – 3 = –1 11.
3 3 lim x 2 + 1 = ⎡⎢lim x 2 + 1 ⎤⎥ = ( 2 ) = 8 x →1 ⎣ x→1 ⎦
13.
lim 5 x + 4 = lim ( 5 x + 4 ) = 9 = 3
(
)
(
x →1
)
3
x →1
15. The limit we seek is the limit of a rational function whose domain is {x | x ≠ –2, x ≠ 2}. Since 0 is in the domain, we use formula (12). x 2 − 4 02 − 4 −4 lim 2 = 2 = = −1 x→ 0 x + 4 0 +4 4 17.
lim ( 3x − 2 ) x →2
5 2
5
5
5 = ⎡⎢ lim ( 3x − 2 ) ⎤⎥ 2 = ( 4 ) 2 = ( 2 ) = 32 ⎣ x →2 ⎦
19. The domain of the rational function R(x) =
x2 − 4 is {x | x ≠ 0, x ≠ 2}. Since 2 is not x2 − 2x
in the domain, we cannot evaluate R(2), but we notice that the function can be factored as x 2 − 4 ( x + 2 )( x − 2 ) = x ( x − 2) x2 − 2x
176 SECTION 3.2
3.2 Techniques for Finding Limits of Functions 3. Using formula (2) (p. 243), we find lim x = 4
1. Using formula (1) (p. 243), we find lim 5 = 5
x →4
x→1
5.
lim ( 3 x + 2 ) = lim ( 3x ) + lim 2 = ⎡⎢ lim 3⎤⎥ ⎡⎢ lim x ⎤⎥ + lim 2 x →2 x →2 x →2 ⎣ x →2 ⎦ ⎣ x →2 ⎦ x →2 = (3) · (2) + 2 = 6 + 2 = 8
7.
lim 3x 2 − 5 x = 3 ⋅ ( −1) − 5 ⋅ ( −1) = 3 ⋅ 1 + 5 = 3 + 5 = 8
x → −1
9.
(
)
2
(
)
lim 5 x 4 − 3 x 2 + 6 x − 9 = 5 · 14 – 3 · 12 + 6 · 1 – 9 x →1
=5·1–3·1+6–9 = 5 – 3 – 3 = –1 11.
3 3 lim x 2 + 1 = ⎡⎢lim x 2 + 1 ⎤⎥ = ( 2 ) = 8 x →1 ⎣ x→1 ⎦
13.
lim 5 x + 4 = lim ( 5 x + 4 ) = 9 = 3
(
)
(
x →1
)
3
x →1
15. The limit we seek is the limit of a rational function whose domain is {x | x ≠ –2, x ≠ 2}. Since 0 is in the domain, we use formula (12). x 2 − 4 02 − 4 −4 lim 2 = 2 = = −1 x→ 0 x + 4 0 +4 4 17.
lim ( 3x − 2 ) x →2
5 2
5
5
5 = ⎡⎢ lim ( 3x − 2 ) ⎤⎥ 2 = ( 4 ) 2 = ( 2 ) = 32 ⎣ x →2 ⎦
19. The domain of the rational function R(x) =
x2 − 4 is {x | x ≠ 0, x ≠ 2}. Since 2 is not x2 − 2x
in the domain, we cannot evaluate R(2), but we notice that the function can be factored as x 2 − 4 ( x + 2 )( x − 2 ) = x ( x − 2) x2 − 2x
SECTION 3.2 177
Since x is near 2, but x ≠ 2, we can cancel the (x – 2)’s. Formula (11), can then be used to find the limit of the function as x approaches 2. ( x + 2) 4 ( x + 2 ) ( x − 2 ) lim x2 − 4 = x →2 = =2 lim 2 = lim x →2 x − 2 x x →2 lim ( x ) 2 x ( x − 2) x →2 x 2 − x − 12 is {x | x ≠ –3, x ≠ 3}. Since –3 is x2 − 9 not in the domain, we cannot evaluate R(–3), but we notice that the function can be factored as x 2 − x − 12 ( x − 4 )( x + 3) = x2 − 9 ( x − 3)( x + 3) Since x is near –3, but x ≠ –3, we can cancel the (x + 3)’s. Formula (11), can then be used to find the limit of the function as x approaches –3. ( x − 4 ) ( −3) − 4 −7 7 ( x − 4 ) ( x + 3) xlim x 2 − x − 12 → −3 lim = lim = = = = x → −3 x → −3 x − 3 x2 − 9 ( ) ( x + 3) lim ( x − 3) ( −3) − 3 −6 6
21. The domain of the rational function R(x) =
x → −3
x3 − 1 is {x | x ≠ 1}. Since 1 is not in the x −1 domain, we cannot evaluate R(1), but we notice that the function can be factored as 2 x 3 − 1 ( x − 1) x + x + 1 = x −1 x −1 Since x is near 1, but x ≠ 1, we can cancel the (x – 1)’s. Formula (11), can then be used to find the limit of the function as x approaches 1. x 2 + x + 1 12 + 1 + 1 ( x − 1) x 2 + x + 1 lim x3 − 1 x →1 = = =3 lim = lim x →1 x − 1 x →1 lim (1) 1 x −1
23. The domain of the rational function R(x) =
(
)
(
)
(
)
x →1
( x + 1)
2
is {x | x ≠ –1, x ≠ 1}. Since –1 is not x2 − 1 in the domain, we cannot evaluate R(–1), but we notice that the function can be factored as 2 2 ( x + 1) ( x + 1) = x 2 − 1 ( x − 1)( x + 1) Since x is near –1, but x ≠ –1, we can cancel (x + 1)’s. Formula (11), can then be used to find the limit of the function as x approaches –1.
25. The domain of the rational function R(x) =
lim
x → −1
( x + 1)
2
x2 − 1
( x + 1) = lim x → −1 x − 1 ( ) ( x + 1) 2
( x + 1) ( −1) + 1 0 = = =0 x → −1 ( x − 1) ( −1) − 1 −2
= lim
178 SECTION 3.2 27.
The limit of the denominator of this function as x approaches 1 is zero, so formula (11), cannot be used directly. We first factor the function by grouping. 2 x 2 ( x − 1) + 1 ⋅ ( x − 1) ( x − 1) x + 1 x3 − x 2 + x − 1 = = x 4 − x3 + 2 x − 2 x3 ( x − 1) + 2 ⋅ ( x − 1) ( x − 1) x3 + 2
( (
) )
Since x is near 1, but x ≠ 1, we can cancel (x – 1)’s. Then using formula (11), we get x2 + 1 ( x − 1) x 2 + 1 lim x3 − x 2 + x − 1 2 x →1 lim 4 = lim = = 3 3 3 x →1 x − x + 2 x − 2 x →1 x − 1 ( ) x + 2 lim x + 2 3
( (
) )
x →1
( (
) )
29. The limit of the denominator of this function as x approaches 2 is zero, so formula (11), cannot be used directly. We first factor the function, using grouping to factor the numerator.
(
)
2 x2 + 4 ( x − 2) x3 − 2 x 2 + 4 x − 8 x ( x − 2 ) + 4 ⋅ ( x − 2 ) = = x2 + x − 6 ( x + 3)( x − 2 ) ( x + 3)( x − 2 ) Since x is near 2, but x ≠ 2, we can cancel (x – 2)’s. Then using formula (11), we get x2 + 4 x 2 + 4 ( x − 2 ) xlim 22 + 4 8 x3 − 2 x 2 + 4 x − 8 →2 = = = lim = lim x→ 2 x → 2 ( x + 3) ( x − 2 ) lim ( x + 3) 2+3 5 x2 + x − 6
(
(
)
)
x→ 2
31. The limit of the denominator of this function as x approaches –1 is zero, so formula (11), cannot be used directly. We first factor the function, using grouping to factor the denominator. x x2 + 2 x + 1 x ( x + 1)( x + 1) x3 + 2 x 2 + x = 3 = 3 4 3 x + x + 2 x + 2 x ( x + 1) + 2 ( x + 1) x + 2 ( x + 1)
(
)
(
)
Since x is near –1, but x ≠ –1, we can cancel (x + 1)’s. Then using formula (11), we get x ( x + 1) ( −1) ⋅ ⎡( −1) + 1⎤ x ( x + 1) ( x + 1) xlim x3 + 2 x 2 + x → −1 ⎣ ⎦ = ( −1) ⋅ ( 0 ) = 0 = = lim 4 = lim 3 3 x → −1 x + x + 2 x + 2 x→ −1 x 3 + 2 1 ( x + 1) lim x3 + 1 ( −1) + 2
(
)
x → −1
(
)
33. The average rate of change of f from 2 to x is ∆y f ( x ) − f ( 2 ) ( 5 x − 3) − ( 5 ⋅ 2 − 3) 5 x − 3 − 7 5 x − 10 5 ( x − 2 ) = = = = = ∆x x−2 x−2 x−2 x−2 x−2 The limit of the average rate of change as x approaches 2 is 5 ( x − 2) f ( x ) − f ( 2) ( 5 x − 3) − 7 5 x − 10 =5 lim = lim = lim = lim x→ 2 x→ 2 x→ 2 x − 2 x→ 2 x−2 x−2 x−2 35. The average rate of change of f from 3 to x is ∆y f ( x ) − f ( 3) x 2 − 32 ( x + 3)( x − 3) = = = ∆x x−3 x−3 x−3
SECTION 3.2 179
The limit of the average rate of change as x approaches 3 is ( x + 3) ( x − 3) f ( x ) − f ( 3) x 2 − 32 =6 lim = lim = lim x→ 3 x→ 3 x − 3 x→ 3 x −3 x−3 37. The average rate of change of f from –1 to x is 2 x 2 + 2 x − ⎡( −1) + 2 ⋅ ( −1) ⎤ x 2 + 2 x − ( −1) x 2 + 2 x + 1 ( x + 1)2 ∆y f ( x ) − f ( −1) ⎣ ⎦ = = = = = ∆x x − ( −1) x +1 x +1 x +1 x +1 The limit of the average rate of change as x approaches –1 is 2 x 2 + 2 x − ( −1) f ( x ) − f ( −1) ( x + 1) x2 + 2 x + 1 = ( −1) + 1 = 0 lim = lim = lim = lim x→ −1 x → −1 x→ −1 x → −1 x − ( −1) x +1 x +1 x +1
(
)
(
(
)
)
39. The average rate of change of f from 0 to x is 3 2 2 ∆y f ( x ) − f ( 0 ) 3 x − 2 x + 4 − ( 4 ) 3 x 3 − 2 x 2 x ( 3 x − 2 ) = = = = ∆x x−0 x x x The limit of the average rate of change as x approaches 0 is f ( x ) − f ( 0) x 2 ( 3x − 2 ) 3x3 − 2 x 2 lim = lim = lim =0 x→ 0 x→ 0 x→ 0 x−0 x x
(
)
41. The average rate of change of f from 1 to x is 1 1 1 x 1− x − − ( −1) ⋅ ( x − 1) ∆y f ( x ) − f (1) x 1 x x 1− x = = = = x = = x −1 x −1 x − 1 x − 1 x ( x − 1) x ( x − 1) ∆x ↑ Find a common denominator
↑ Simplify
↑ Factor out (–1)
The limit of the average rate of change as x approaches 1 is ( −1) ⋅ ( x − 1) f ( x ) − f (1) 1− x = −1 lim = lim = lim x →1 x →1 x ( x − 1) x →1 x −1 x ( x − 1) 43. The average rate of change of f from 4 to x is ∆y f ( x ) − f ( 4 ) x− 4 x −2 = = = ∆x x−4 x−4 x +2 x −2
(
)(
)
↑ Factor the denominator
The limit of the average rate of change as x approaches 4 is f ( x ) − f ( 4) x −2 1 1 x− 4 = = lim = lim = lim x→ 4 x → 4 x → 4 x−4 x−4 4 +2 4 x +2 x −2
(
)(
)
180 SECTION 3.3 ⎞⎛ lim f x ⎞ = 2 ⋅ 5 = 10 45. Since lim f ( x ) = 5 and lim g ( x ) = 2 , lim ⎣⎡ 2 f ( x ) ⎦⎤ = ⎜⎛ lim 2 ⎟⎜ ( )⎟ x→ c x → c x→ c x→ c ⎝ ⎠⎝ x→ c ⎠ 3
3 47. Since lim f ( x ) = 5 and lim g ( x ) = 2 , lim ⎡ g ( x ) ⎤ = ⎡ lim g ( x ) ⎤ = 23 = 8 ⎣ ⎦ ⎢⎣ x→ c ⎥⎦ x→ c x→ c x→ c
49. Since lim f ( x ) = 5 and lim g ( x ) = 2 , lim x→ c
x→ c
x→ c
lim 4 4 4 x→ c = = f ( x ) lim f ( x ) 5 x→ c
51. Since lim f ( x ) = 5 and lim g ( x ) = 2 , x→ c
x→ c
lim ⎡⎣ 4 f ( x ) − 5 g ( x ) ⎤⎦ = lim ⎡⎣ 4 f ( x ) ⎤⎦ − lim ⎡⎣5 g ( x )⎤⎦ x→ c x→ c x→ c
= ⎛⎜ lim 4 ⎞⎟ ⋅ ⎛⎜ lim f ( x ) ⎞⎟ − ⎛⎜ lim 5 ⎞⎟ ⋅ ⎛⎜ lim g ( x ) ⎞⎟ ⎝ x→ c ⎠ ⎝ x→ c ⎠ ⎝ x→ c ⎠ ⎝ x→ c ⎠ = 4 · 5 – 5 · 2 = 20 – 10 = 10
3.3 One-Sided Limits; Continuous Functions 1. The domain of f is {x | –8 ≤ x < – 3 or – 3 < x < 4 or 4 < x ≤ 6} or the intervals [8, – 3) or (– 3 , 4) or (4, 6]. 3. The x-intercepts of the graph of f are (– 8, 0) and (– 5, 0). At these points the graph of f either crosses or touches the x-axis. 5. f (– 8) = 0 and f (– 4) = 2 7. To find lim f ( x ) , we look at the values of f when x is close to – 6, but less than – 6. − x → −6
Since the graph of f is approaching y = 3 for these values, we have lim − f ( x ) = 3 . x → −6
9. To find lim f ( x ) , we look at the values of f when x is close to – 4, but less than – 4. − x→ −4
Since the graph of f is approaching y = 2 for these values, we have lim − f ( x ) = 2 . x→ −4
11. To find lim f ( x ) , we look at the values of f when x is close to 2, but less than 2. Since − x→ 2
the graph of f is approaching y = 1 for these values, we have lim− f ( x ) = 1 . x→ 2
180 SECTION 3.3 ⎞⎛ lim f x ⎞ = 2 ⋅ 5 = 10 45. Since lim f ( x ) = 5 and lim g ( x ) = 2 , lim ⎣⎡ 2 f ( x ) ⎦⎤ = ⎜⎛ lim 2 ⎟⎜ ( )⎟ x→ c x → c x→ c x→ c ⎝ ⎠⎝ x→ c ⎠ 3
3 47. Since lim f ( x ) = 5 and lim g ( x ) = 2 , lim ⎡ g ( x ) ⎤ = ⎡ lim g ( x ) ⎤ = 23 = 8 ⎣ ⎦ ⎢⎣ x→ c ⎥⎦ x→ c x→ c x→ c
49. Since lim f ( x ) = 5 and lim g ( x ) = 2 , lim x→ c
x→ c
x→ c
lim 4 4 4 x→ c = = f ( x ) lim f ( x ) 5 x→ c
51. Since lim f ( x ) = 5 and lim g ( x ) = 2 , x→ c
x→ c
lim ⎡⎣ 4 f ( x ) − 5 g ( x ) ⎤⎦ = lim ⎡⎣ 4 f ( x ) ⎤⎦ − lim ⎡⎣5 g ( x )⎤⎦ x→ c x→ c x→ c
= ⎛⎜ lim 4 ⎞⎟ ⋅ ⎛⎜ lim f ( x ) ⎞⎟ − ⎛⎜ lim 5 ⎞⎟ ⋅ ⎛⎜ lim g ( x ) ⎞⎟ ⎝ x→ c ⎠ ⎝ x→ c ⎠ ⎝ x→ c ⎠ ⎝ x→ c ⎠ = 4 · 5 – 5 · 2 = 20 – 10 = 10
3.3 One-Sided Limits; Continuous Functions 1. The domain of f is {x | –8 ≤ x < – 3 or – 3 < x < 4 or 4 < x ≤ 6} or the intervals [8, – 3) or (– 3 , 4) or (4, 6]. 3. The x-intercepts of the graph of f are (– 8, 0) and (– 5, 0). At these points the graph of f either crosses or touches the x-axis. 5. f (– 8) = 0 and f (– 4) = 2 7. To find lim f ( x ) , we look at the values of f when x is close to – 6, but less than – 6. − x → −6
Since the graph of f is approaching y = 3 for these values, we have lim − f ( x ) = 3 . x → −6
9. To find lim f ( x ) , we look at the values of f when x is close to – 4, but less than – 4. − x→ −4
Since the graph of f is approaching y = 2 for these values, we have lim − f ( x ) = 2 . x→ −4
11. To find lim f ( x ) , we look at the values of f when x is close to 2, but less than 2. Since − x→ 2
the graph of f is approaching y = 1 for these values, we have lim− f ( x ) = 1 . x→ 2
SECTION 3.3 181 13. The lim f ( x) exists because both lim− f ( x) = 0 and lim+ f ( x) = 0. Since both onex→ 4
x→ 4
x→ 4
sided limits exist and are equal, the limit of f as x approaches 4 exists and is equal to the one-sided limits. That is, lim f ( x) = 0. x→ 4
15. The function f is not continuous at x = – 6, because lim f ( x ) does not exist. (The onex → −6
sided limits are not equal. See Problems 7 and 8.) 17. The function f is continuous at x = 0. The function is defined at zero, f (0) = 3, and lim f ( x) = f (0) = 3. (See Problem 14.) x→ 0
19. The function f is not continuous at x = 4. The function is not defined at 4. That is, 4 is not part of the domain of f . 21. To find the one-sided limit we look at values of x close to 1, but greater than 1. Since f (x) = 2x + 3 for such numbers, we conclude that lim+ f ( x ) = lim+ ( 2 x + 3) = 5 x →1
x →1
23. To find the one-sided limit we look at values of x close to 1, but less than 1. Since f (x) = 2x 3 + 5x for such numbers, we conclude that lim− f ( x ) = lim− 2 x3 + 5 x = 7 x →1
x →1
(
)
25. To find the one-sided limit we look at values of x close to 0, but less than 0. Since f (x) = ex for such numbers, we conclude that lim− f ( x ) = lim− e x = 1 x→ 0
x→ 0
( )
27. To find the one-sided limit we look at values of x close to 2, but greater than 2. Since x2 − 4 for such numbers, we first factor the function. f (x) = x−2 ( x + 2 )( x − 2 ) x2 − 4 = f (x) = x−2 x−2 We can then conclude that ( x + 2) ( x − 2) x2 − 4 =4 lim+ f ( x ) = lim+ = lim+ x→ 2 x→ 2 x − 2 x→ 2 x−2 29. To find the one-sided limit we look at values of x close to –1, but less than –1. Since x2 −1 for such numbers, we first factor the function. f (x) = 3 x +1 ( x + 1)( x − 1) x2 −1 f (x) = 3 = x + 1 ( x + 1) x 2 − x + 1
(
)
182 SECTION 3.3
We can then conclude that lim − f ( x ) = lim −
x → −1
x → −1
( x + 1) ( x − 1) −2 2 x2 − 1 = =− = lim − 3 2 x → − 1 3 x +1 ( x + 1) x − x + 1 3
(
)
31. To find the one-sided limit we look at values of x close to –2, but greater than –2. Since x2 + x − 2 for such numbers, we first factor the function. f (x) = 2 x + 2x x 2 + x − 2 ( x + 2 )( x − 1) = f (x) = 2 x + 2x x ( x + 2) We can then conclude that ( x + 2 ) ( x − 1) x2 + x − 2 x −1 1 lim+ f ( x) = lim+ 2 = lim+ = lim+ = x→2 x→2 x→2 x→2 x + 2x 2 x x ( x + 2) 33. f (x) = x 3 – 3x 2 + 2x – 6 is continuous at c = 2 because f is a polynomial, and polynomials are continuous at every number.
x2 + 5 is a rational function whose domain is {x | x ≠ 6}. f is continuous at c = 3 x−6 since f is defined at 3.
35. f (x) =
x+3 is a rational function whose domain is {x | x ≠ 3}. f is not continuous at c = x−3 3 since f is not defined at 3.
37. f (x) =
39.
x3 + 3x f (x) = 2 is a rational function whose domain is {x | x ≠ 0, x ≠ 3}. f is not x − 3x continuous at c = 0 since f is not defined at 0.
⎧ x3 + 3x ⎪ 41. To determine whether f ( x ) = ⎨ x 2 − 3x ⎪1 ⎩ f when x = 0. f (0) = 1
if x ≠ 0
is continuous at c = 0, we investigate
if x = 0
(
)
(
)
x x2 + 3 3 x3 + 3x = = −1 lim− f ( x ) = lim− 2 = lim− x→ 0 x→ 0 x − 3 x x→ 0 x ( x − 3) −3 x x2 + 3 3 x3 + 3x = = −1 lim+ f ( x ) = lim+ 2 = lim+ x→ 0 x→ 0 x − 3 x x→ 0 x ( x − 3) −3
Since lim f ( x ) = −1 ≠ f (0) = 1, the function f is not continuous at c = 0. x→ 0
SECTION 3.3 183
⎧ x3 + 3x ⎪ 43. To determine whether f ( x ) = ⎨ x 2 − 3x ⎪ −1 ⎩ f when x = 0. f (0) = – 1
if x ≠ 0
is continuous at c = 0, we investigate
if x = 0
(
)
(
)
x x2 + 3 3 x3 + 3x = = −1 lim f ( x ) = lim− 2 = lim− x→ 0 − x→ 0 x − 3 x x→ 0 x ( x − 3) −3 x x2 + 3 3 x3 + 3x = = −1 lim+ f ( x ) = lim+ 2 = lim+ x→ 0 x→ 0 x − 3 x x→ 0 x ( x − 3) −3
Since lim f ( x ) = −1 = f (0), the function f is continuous at c = 0. x→ 0
⎧ x3 − 1 ⎪ 2 ⎪⎪ x − 1 45. To determine whether f ( x ) = ⎨2 ⎪ 3 ⎪ ⎪⎩ x + 1
if x < 1 if x = 1 is continuous at c = 1, we investigate f if x > 1
when x = 1. f (1) = 2
(
)
( x − 1) x 2 + x + 1 1 + 1 + 1 3 x3 − 1 = = lim f ( x ) = lim− 2 = lim− x→ 1 − x→ 1 x − 1 x→ 1 1+1 2 ( x − 1) ( x + 1) lim+ f ( x ) = lim+
x→ 1
Since lim f ( x ) = x→ 1
x→ 1
3 3 = x +1 2
3 ≠ f (1), the function f is not continuous at c = 1. 2
⎧ ⎪2e x ⎪⎪ 47. To determine whether f ( x ) = ⎨2 ⎪ 3 2 ⎪ x + 2x ⎪⎩ x 2 investigate f when x =0. f (0) = 2 lim− f ( x ) = lim− 2e x = 2 x→ 0
if x < 0 if x = 0 is continuous at c = 0, we if x > 0
x→ 0
x2 ( x + 2) x3 + 2 x 2 =2 lim f ( x ) = lim+ = lim+ x→ 0 + x→ 0 x→ 0 x2 x2 The lim f ( x ) exists, and lim f ( x ) = f (0) = 2. So we conclude that the function f is x→ 0
continuous at c = 0.
x→ 0
184 SECTION 3.3 49.
f (x) = 2x + 3 is a first degree polynomial function. Polynomial functions are continuous at all real numbers.
51.
f (x) = 3x 2 + x is a second degree polynomial function. Polynomial functions are continuous at all real numbers.
53. f (x) = 4 ln x is the product of a constant function h(x) = 4, which is continuous at every number, and the logarithmic function g(x) = ln x, which is continuous for every number in the domain (0, ∞). So f (x) = 4 ln x is continuous for all values x > 0. 55.
57.
59.
61.
f (x) = 3e x is the product of a constant function h(x) = 3, which is continuous at every number, and the exponential function g(x) = e x, which is continuous for every number in the domain (– ∞, ∞). So f (x) = 3e x is continuous for all real numbers. 2x + 5 is a rational function. Rational functions are continuous at every number in x2 − 4 the domain. The domain of f is {x | x ≠ – 2, x ≠ 2}, and f is continuous at all those numbers. f is discontinuous at x = – 2 and x = 2.
f (x) =
x−3 is the quotient of a polynomial function, which is continuous at all real ln x numbers and the logarithmic function, which is continuous at all numbers in the domain (0, ∞). So f is continuous at all positive numbers or for x > 0. f (x) =
1 x − 5 , are each continuous for 2 every number since they are polynomials. So we only need to investigate x = 0 and x = 2, the two points at which the pieces change. For x = 0: f (0) = 3(0) + 1 = 1 lim− f ( x ) = lim− ( 3x + 1) = 1
The “pieces” of f , that is, y = 3x + 1, y = – x 2, and y =
x→ 0
x→ 0
lim f ( x ) = lim+ ( − x 2 ) = 0
x→ 0 +
x→ 0
Since lim+ f ( x ) ≠ f (0), we conclude that the function f is discontinuous at x = 0. x→ 0
For x = 2:
f (2) = – 2 2 = – 4 lim− f ( x ) = lim− ( − x 2 ) = −4
x→ 2
x→ 2
⎛1 ⎞ lim+ f ( x ) = lim+ ⎜ x − 5 ⎟ = −4 x→ 2 x→ 2 ⎝ 2 ⎠ ( ) ( ) Since lim f x = f 2 = −4 , we conclude that f is continuous at x = 2. x→ 2
SECTION 3.3 185
⎧39.99 63. The cost function is C(x) = ⎨ ⎩0.25 x − 47.51 (a) lim − C ( x ) = lim − ( 39.99 ) = 39.99 x → 350
(b)
if 0 < x ≤ 350 . if x > 350
x → 350
lim C ( x ) = lim + ( 0.25 x − 47.51) = 39.99
x → 350 +
x → 350
(c) The left limit equals the right limit, so lim C ( x ) exists. C is continuous at x = 350 x → 350
since lim C ( x ) = C(350) = 39.99. x → 350
(d) Answers will vary. ⎧10 ⎪ ⎪ 23 (10.45 + 10 v − v ) 65. (a) If t = 10º then W(v) = ⎨33 − 22.04 ⎪ ⎪⎩−3.7034 (b) lim+ W ( v ) = lim+ (10 ) = 10 v→ 0
(c) (d)
0 ≤ v < 1.79 1.79 ≤ v ≤ 20 . v > 20
v→ 0
lim W ( v ) = lim − (10 ) = 10
v →1.79 −
v →1.79
(
⎡ 23 10.45 + 10 1.79 − 1.79 lim + W ( v ) = lim + ⎢33 − v → 1.79 v → 1.79 ⎢ 22.04 ⎣
) ⎤⎥ = 10.00095 ⎥ ⎦
(e) W(1.79) = 10.00095 (f) W is not continuous at v = 1.79 since lim − W ( v ) ≠ lim + W ( v ) , and therefore, v →1.79
v → 1.79
lim W ( v ) does not exist. In order to be continuous at 1.79, lim W ( v ) must exist.
v → 1.79
v → 1.79
(g) Rounded to two decimal places, lim + W ( v ) = 10.00 lim − W ( v ) = 10.00 v →1.79
W(1.79) = 10.00
v → 1.79
Now the function W is continuous at 1.79. (h) Answers will vary.
(
⎡ 23 10.45 + 10 20 − 20 (i) lim− W ( v ) = lim− ⎢33 − v → 20 v → 20 ⎢ 22.04 ⎣
) ⎤⎥ = – 3.7033 ⎥ ⎦
(j) lim+ W ( v ) = lim+ ( −3.7034 ) = – 3.7034 v → 20
v → 20
(k) W(20) = – 3.7033 (l) W is not continuous at 20. The right limit is not equal to W(20).
186 SECTION 3.4
(m) Rounded to two decimal places, lim+ W ( v ) = – 3.70 lim − W ( v ) = – 3.70 v → 20
v → 20
W(20) = – 3.70
Now the function W is continuous at 20. (n) Answers will vary.
3.4 Limits at Infinity; Infinite Limits; End Behavior; Asymptotes 1. As x→∞, x3 + x 2 + 2 x − 1 = x3 , and x3 + x + 1 = x 3 , so
x3 + x 2 + 2 x − 1 x3 = lim =1 x→ ∞ x→ ∞ x3 + x + 1 x3 lim
3. As x→∞, 2x + 4 = 2x, and x – 1 = x, so 2x + 4 2x =2 lim = lim x→ ∞ x − 1 x→ ∞ x 5. As x→∞, 3x 2 – 1 = 3x 2, and x 2 + 4 = x 2, so
3x 2 − 1 3 x2 = lim =3 x→ ∞ x2 + 4 x→ ∞ x2 lim
7. As x→ – ∞, 5x 3 – 1 = 5x 3, and x 4 + 1 = x 4, so
5 x3 − 1 5 x3 5 1 = lim = lim = 5 lim =0 4 4 x→ − ∞ x + 1 x→ − ∞ x x→ − ∞ x x→ − ∞ x lim
9. As x→ ∞, 5x 3 + 3 = 5x 3, and x 2 + 1 = x 2, so 5 x3 + 3 5x 3 lim 2 = lim 2 = lim 5 x = ∞ x→ ∞ x + 1 x→ ∞ x→ ∞ x 11. As x→ – ∞, 4x 5 = 4x 5, and x 2 + 1 = x 2, so 4 x5 4x 5 3 = lim = lim 4 x 3 = 4 ⋅ lim x 3 = −∞ lim 2 x→ − ∞ x2 + 1 x→ − ∞ x→ − ∞ x→ − ∞ x
1 1 , we examine the values of f that are , x ≠ 2. To determine lim+ x→ 2 x − 2 x−2 close to 2, but remain greater than 2.
13. Here f (x) =
x f ( x) =
2.1 2.01 2.001 2.0001 2.00001 1 x−2
10
100
1000 10,000 100,000
186 SECTION 3.4
(m) Rounded to two decimal places, lim+ W ( v ) = – 3.70 lim − W ( v ) = – 3.70 v → 20
v → 20
W(20) = – 3.70
Now the function W is continuous at 20. (n) Answers will vary.
3.4 Limits at Infinity; Infinite Limits; End Behavior; Asymptotes 1. As x→∞, x3 + x 2 + 2 x − 1 = x3 , and x3 + x + 1 = x 3 , so
x3 + x 2 + 2 x − 1 x3 = lim =1 x→ ∞ x→ ∞ x3 + x + 1 x3 lim
3. As x→∞, 2x + 4 = 2x, and x – 1 = x, so 2x + 4 2x =2 lim = lim x→ ∞ x − 1 x→ ∞ x 5. As x→∞, 3x 2 – 1 = 3x 2, and x 2 + 4 = x 2, so
3x 2 − 1 3 x2 = lim =3 x→ ∞ x2 + 4 x→ ∞ x2 lim
7. As x→ – ∞, 5x 3 – 1 = 5x 3, and x 4 + 1 = x 4, so
5 x3 − 1 5 x3 5 1 = lim = lim = 5 lim =0 4 4 x→ − ∞ x + 1 x→ − ∞ x x→ − ∞ x x→ − ∞ x lim
9. As x→ ∞, 5x 3 + 3 = 5x 3, and x 2 + 1 = x 2, so 5 x3 + 3 5x 3 lim 2 = lim 2 = lim 5 x = ∞ x→ ∞ x + 1 x→ ∞ x→ ∞ x 11. As x→ – ∞, 4x 5 = 4x 5, and x 2 + 1 = x 2, so 4 x5 4x 5 3 = lim = lim 4 x 3 = 4 ⋅ lim x 3 = −∞ lim 2 x→ − ∞ x2 + 1 x→ − ∞ x→ − ∞ x→ − ∞ x
1 1 , we examine the values of f that are , x ≠ 2. To determine lim+ x→ 2 x − 2 x−2 close to 2, but remain greater than 2.
13. Here f (x) =
x f ( x) =
2.1 2.01 2.001 2.0001 2.00001 1 x−2
10
100
1000 10,000 100,000
SECTION 3.4 187
We see that as x gets closer to 2 from the right, the value of f (x) =
1 becomes x−2
unbounded in the positive direction, and we write 1 = ∞ lim x→ 2 + x − 2 15.
Here f (x) =
x
( x − 1)
2
, x ≠ 1. To determine lim− x →1
x
( x − 1)
2
, we examine the values of f that
are close to 1, but remain smaller than 1. x f ( x) =
x
( x − 1)
2
0.9
0.99
0.999
0.9999
90
9900 999,000 99,990,000
We see that as x gets closer to 1 from the left, the value of f (x) =
x
( x − 1)
2
becomes
unbounded in the positive direction, and we write x = ∞ lim 2 x →1 − x − 1 ( ) x2 + 1 x2 + 1 , we examine the values of f that , x ≠ 1 . To determine lim+ 3 x→ 1 x − 1 x3 − 1 are close to 1, but remain larger than 1.
17. Here f (x) =
x f ( x) =
1.1
1.01
1.001
1.0001
1.00001
x2 + 1 6.6767 66.66777 666.66678 6666.667 66,666.667 x3 − 1
We see that as x gets closer to 1 from the right, the value of f (x) =
x2 + 1 becomes x3 − 1
unbounded in the positive direction, and we write x2 + 1 lim 3 =∞ x→ 1+ x − 1 1− x 1− x , we examine the values of f that , x ≠ 2 . To determine lim− x → 2 3x − 6 3x − 6 are close to 2, but remain smaller than 2.
19. Here f (x) =
x f ( x) =
1.9 1.99 1.999 1.9999 1.99999 1− x 3x − 6
3
33
333
3333
33,333
We see that as x gets closer to 2 from the left, the value of f (x) =
1− x becomes 3x − 6
188 SECTION 3.4
unbounded in the positive direction, and we write 1− x lim− = ∞ x → 2 3x − 6 21.
To find horizontal asymptotes, we need to find two limits, lim f ( x ) and lim f ( x ) . x→ ∞
x→ − ∞
1 ⎞ 1 ⎛ lim f ( x ) = lim ⎜ 3 + 2 ⎟ = lim 3 + lim 2 = 3 + 0 = 3 x→ ∞ x→ ∞ x→ ∞ x x ⎠ x→ ∞ ⎝ We conclude that the line y = 3 is a horizontal asymptote of the graph when x becomes unbounded in the positive direction. 1 ⎞ 1 ⎛ lim f ( x ) = lim ⎜ 3 + 2 ⎟ = lim 3 + lim 2 = 3 + 0 = 3 x→ − ∞ x→ − ∞ x→ − ∞ x x ⎠ x→ − ∞ ⎝ We conclude that the line y = 3 is a horizontal asymptote of the graph when x becomes unbounded in the negative direction. To find vertical asymptotes, we need to examine the behavior of the graph of f when x is near 0, the point where f is not defined. This will require looking at the one-sided limits of f at 0. lim− f ( x ) : Since x → 0 − , we know x < 0, but x 2 > 0. It follows that the x→ 0
1 is positive and becomes unbounded as x → 0 − . 2 x 1 ⎞ ⎛ lim− f ( x ) = lim− ⎜ 3 + 2 ⎟ = ∞ x→ 0 x→ 0 ⎝ x ⎠ + lim+ f ( x ) : Since x → 0 , we know x > 0, and x 2 > 0. It follows that the
expression
x→ 0
1 is positive and becomes unbounded as x → 0 + . 2 x 1 ⎞ ⎛ lim+ f ( x ) = lim+ ⎜ 3 + 2 ⎟ = ∞ x→ 0 x→ 0 ⎝ x ⎠ We conclude that the graph of f has a vertical asymptote at x = 0.
expression
23.
To find horizontal asymptotes, we need to find two limits, lim f ( x ) and lim f ( x ) . x→ ∞
lim f ( x ) = lim
x→ ∞
x→ ∞
2 x2
( x − 1)
2
= lim
x→ ∞
x→ − ∞
2 x2 2 x2 = =2 lim x2 − 2 x + 1 x→ ∞ x2
We conclude that the line y = 2 is a horizontal asymptote of the graph when x becomes unbounded in the positive direction. lim f ( x ) = lim
x→ − ∞
x→ − ∞
2 x2
( x − 1)
2
= lim
x→ − ∞
2 x2 2 x2 =2 = lim x2 − 2 x + 1 x→ − ∞ x2
We conclude that the line y = 2 is a horizontal asymptote of the graph when x becomes unbounded in the negative direction.
SECTION 3.4 189
To find vertical asymptotes, we need to examine the behavior of the graph of f when x is near 1, the point where f is not defined. This will require looking at the one-sided limits of f at 1. lim− f ( x ) : Since x → 1 − , we know x < 1, so x – 1 < 0, but (x – 1) 2 > 0. It follows x →1
that the expression
2 x2
( x − 1)
2
is positive and becomes unbounded as x → 1 − .
lim− f ( x ) = lim−
x →1
x →1
2x2
( x − 1)
2
=∞
lim+ f ( x ) : Since x → 1 + , we know x > 1, so both x – 1 > 0 and (x – 1) 2 > 0. It
x →1
follows that the expression
2 x2
( x − 1)
2
is positive and becomes unbounded as x → 1 + .
lim+ f ( x ) = lim+
x →1
x →1
2x2
( x − 1)
2
=∞
We conclude that the graph of f has a vertical asymptote at x = 1. f ( x ) and lim f ( x ) . 25. To find horizontal asymptotes, we need to find two limits, xlim →∞ x→ − ∞ x2 x2 = lim = lim 1 = 1 x→ ∞ x→ ∞ x2 − 4 x→ ∞ x2 x→ ∞ We conclude that the line y = 1 is a horizontal asymptote of the graph when x becomes unbounded in the positive direction. x2 x2 lim f ( x ) = lim 2 = lim 2 = lim 1 = 1 x→ − ∞ x→ − ∞ x − 4 x→ − ∞ x x→ − ∞ We conclude that the line y = 1 is a horizontal asymptote of the graph when x becomes unbounded in the negative direction. lim f ( x ) = lim
To find vertical asymptotes, we need to examine the behavior of the graph of f when x is near –2 and 2, the points where f is not defined. This will require looking at the one-sided limits of f . lim − f ( x ) : Since x → − 2 − , we know x < – 2 and x2 > 4, so x2 – 4 > 0. It follows x→ − 2
x2 is positive and becomes unbounded as x → − 2 − . x2 − 4 x2 lim − f ( x ) = lim − 2 = ∞ x→ − 2 x→ − 2 x − 4 lim + f ( x ) : Since x → − 2 + , we know x > – 2 and x2 < 4, so x2 – 4 < 0. It
that the expression
x→ − 2
x2 is negative and becomes unbounded as x → − 2 + . follows that the expression 2 x −4 x2 lim + f ( x ) = lim + 2 = –∞ x→ − 2 x→ − 2 x − 4
190 SECTION 3.4
We now examine the limits as x → 2. lim− f ( x ) : Since x → 2 − , we know x < 2 and x2 < 4, so x2 – 4 < 0. It follows that x→ 2
x2 is negative and becomes unbounded as x → 2 − . the expression 2 x −4 x2 lim− f ( x ) = lim− 2 = –∞ x→ 2 x→ 2 x − 4 lim+ f ( x ) : Since x → 2 + , we know x > 2 and x2 > 4, so x2 – 4 > 0. It follows that
x→ 2
the expression
x2 is positive and becomes unbounded as x → 2 + . x2 − 4 x2 lim+ f ( x ) = lim+ 2 =∞ x→ 2 x→ 2 x − 4
We conclude that the graph of f has a vertical asymptotes at x = –2 and at x = 2. 27.
(a) We observe from the graph that the domain continues indefinitely toward the infinities, since there are arrows on both ends. We also observe a vertical asymptote at x = 6. We conclude that the domain of f is {x | x ≠ 6} or all real numbers except 6. (b) The arrows pointing upward as x approaches 6 (the vertical asymptote) indicate that the range of f is the set of positive numbers or {y | y ≥ 0} or the interval [0, ∞). (c) (-4, 0) and (0, 0) are the x-intercepts; (0, 0) is also the y-intercept. (d) Since f (x) = y, f (– 2) = 2. (e) If f (x) = 4, then x = 8 or x = 4. (f) f is discontinuous at x = 6; 6 is not in the domain of f . (g) The vertical asymptote is x = 6. (h) y = 4 is a horizontal asymptote of the graph when x becomes unbounded in the negative direction. (i) There is only one local maximum. It occurs at (– 2, 2) where the local maximum is y = 2. (j) There are 3 local minima. They occur at (– 4, 0) where the local minimum is y = 0; at (0, 0) where the local minimum is y = 0; and at (8, 4) where the local minimum is y = 4. (k) The function f is increasing on the intervals (– 4, – 2), (0, 6), and (8, ∞).
SECTION 3.4 191
(l) The function f is decreasing on the intervals (– ∞, – 4 ), (– 2, 0), and (6, 8). (m) As x approaches – ∞, y approaches 4, so lim f ( x ) = 4. x→ − ∞
(n) As x approaches ∞, y becomes unbounded in the positive direction, so lim f ( x ) = ∞. x→ ∞
(o) As x approaches 6 from the left, we see that y becomes unbounded in the positive direction, so lim− f ( x ) = ∞. x→ 6
(p) As x approaches 6 from the right, we see that y becomes unbounded in the positive direction, so lim+ f ( x ) = ∞. x→ 6
29.
x −1 x −1 . To determine the behavior of the graph near – 1 and 1, we = 2 x − 1 ( x − 1)( x + 1)
R ( x) =
look at lim R ( x ) and lim R ( x ) . x → −1
x→ 1
For lim R ( x ) , we have x → −1
lim R ( x ) = lim
x → −1
x→ − 1
x −1 x −1 1 = lim = lim 2 x − 1 x → − 1 ( x − 1) ( x + 1) x → − 1 x + 1
If x < – 1 and x is getting closer to – 1, the value of unbounded; that is, lim − R ( x ) = – ∞.
1 < 0 and is becoming x +1
x → −1
If x > – 1 and x is getting closer to – 1, the value of that is, lim + R ( x ) = ∞.
1 > 0 and is becoming unbounded; x +1
x → −1
The graph of R will have a vertical asymptote at x = – 1. For lim R ( x ) , we have x→ 1
lim R ( x ) = lim
x→ 1
x→ 1
x −1 x −1 1 1 = lim = = lim 2 x − 1 x → 1 ( x − 1) ( x + 1) x → 1 x + 1 2
As x gets closer to 1, the graph of R gets closer to
1 . Since R is not defined at 1, the 2
⎛ 1⎞ graph will have a hole at ⎜1, ⎟ . ⎝ 2⎠ 31.
x ( x + 1) x2 + x . To determine the behavior of the graph near – 1 and 1, R ( x) = 2 = x − 1 ( x − 1)( x + 1) look at lim R ( x ) and lim R ( x ) . x → −1
x→ 1
192 SECTION 3.4
For lim R ( x ) , we have x → −1
lim R ( x ) = lim
x → −1
x→ − 1
(
)
x x +1 x2 + x x −1 1 = lim = lim = = 2 x → − x → − 1 1 x −1 x − 1 −2 2 ( x − 1) ( x + 1)
As x gets closer to – 1, the graph of R gets closer to
1 . Since R is not defined at – 1, the 2
1⎞ ⎛ graph will have a hole at ⎜ −1, ⎟ . 2⎠ ⎝ For lim R ( x ) , we have x→ 1
(
x→ 1
)
x x +1 x2 + x x = lim = lim 2 x → x → 1 1 x −1 x −1 ( x − 1) ( x + 1)
lim R ( x ) = lim x→ 1
If x < 1 and x is getting closer to 1, the value of that is, lim− R ( x ) = – ∞.
x < 0 and is becoming unbounded; x −1
x→ 1
If x > 1 and x is getting closer to 1, the value of that is, lim+ R ( x ) = ∞.
x > 0 and is becoming unbounded; x −1
x→ 1
The graph of R will have a vertical asymptote at x = 1. 33. A rational function is undefined at every number that makes the denominator zero. So we solve x 4 − x3 + 8 x − 8 = 0 Set the denominator = 0. 3 x ( x − 1) + 8 ( x − 1) = 0 Factor by grouping.
(x
3
+ 8 ) ( x − 1) = 0
Apply the Zero-Product Property. x3 + 8 = 0 or x – 1 = 0 Solve for x. x = – 2 or x=1 To determine the behavior of the graph near – 2 and near 1, we look at lim R ( x ) and x→ − 2
lim R ( x ) .
x→ 1
For lim R ( x ) , we have x→ − 2
lim R ( x ) =
x→ − 2
lim
x→ − 2
= lim
x→ − 2
x 2 ( x − 1) + ( x − 1) x3 − x 2 + x − 1 = lim x 4 − x 3 + 8 x − 8 x → − 2 x 3 ( x − 1) + 8 ( x − 1)
(x (x
2
+ 1) ( x − 1)
x2 + 1 = lim 3 3 + 8 ) ( x − 1) x → − 2 x + 8
Since the limit of the denominator is 0, we use one-sided limits. If x < – 2 and x is
SECTION 3.4 193
getting closer to – 2, the value of x 3 + 8 < 0, so the quotient unbounded; that is, lim − R ( x ) = – ∞.
x2 + 1 < 0 and is becoming x3 + 8
x→ − 2
If x > – 2 and x is getting closer to – 2, the value of x 3 + 8 > 0, so the quotient x2 + 1 > 0 and is becoming unbounded; that is, lim + R ( x ) = ∞. x→ − 2 x3 + 8 The graph of R will have a vertical asymptote at x = – 2. For lim R ( x ) , we have x→ 1
x3 − x 2 + x − 1 x2 + 1 2 = = lim x→ 1 x → 1 x 4 − x3 + 8 x − 8 x → 1 x3 + 8 9 2 As x gets closer to 1, the graph of R gets closer to . Since R is not defined at 1, the 9 ⎛ 2⎞ graph will have a hole at ⎜1, ⎟ . ⎝ 9⎠ lim R ( x ) = lim
35. A rational function is undefined at every number that makes the denominator zero. So we solve x2 + x − 6 = 0 Set the denominator = 0. Factor. ( x + 3)( x − 2 ) = 0
x + 3 = 0 or x – 2 = 0 Apply the Zero-Product Property. Solve for x. x = – 3 or x=2 To determine the behavior of the graph near – 3 and near 2, we look at lim R ( x ) and x→ − 3
lim R ( x ) .
x→ 2
For lim R ( x ) , we have x→ − 3
lim R ( x ) =
x→ − 3
lim
x→ − 3
= lim
x→ − 3
x2 ( x − 2) + 4 ( x − 2) x3 − 2 x 2 + 4 x − 8 = lim x→ − 3 x2 + x − 6 ( x + 3)( x − 2 )
(x
2
+ 4) ( x − 2)
( x + 3) ( x − 2 )
x2 + 4 = lim x→ − 3 x + 3
Since the limit of the denominator is 0, we use one-sided limits. If x < – 3 and x is x2 + 4 < 0 and is becoming getting closer to – 3, the value of x + 3 < 0, so the quotient x+3 unbounded; that is, lim − R ( x ) = – ∞. x→ − 3
If x > – 3 and x is getting closer to – 3, the value of x + 3 > 0, so the quotient x +4 > 0 and is becoming unbounded; that is, lim + R ( x ) = ∞. x→ − 3 x+3 The graph of R will have a vertical asymptote at x = – 3. 2
194 SECTION 3.4
For lim R ( x ) , we have x→ 2
x3 − 2 x 2 + 4 x − 8 x2 + 4 8 = = lim x→ 2 x→ 2 x→ 2 x + 3 x2 + x − 6 5 8 As x gets closer to 2, the graph of R gets closer to . Since R is not defined at 2, the 5 ⎛ 8⎞ graph will have a hole at ⎜ 2, ⎟ . ⎝ 5⎠ lim R ( x ) = lim
37. A rational function is undefined at every number that makes the denominator zero. So we solve x 4 + x3 + x + 1 = 0 Set the denominator = 0. 3 x ( x + 1) + ( x + 1) = 0 Factor by grouping.
(x
3
+ 1) ( x + 1) = 0
x3 + 1 = 0 or x + 1 = 0 Apply the Zero-Product Property. x = – 1 or x=–1 Solve for x. To determine the behavior of the graph near – 1, we look at lim R ( x ) . x→ − 1
lim R ( x ) =
x→ − 1
x ( x + 2 x + 1) x3 + 2 x 2 + x = lim x 4 + x3 + x + 1 x → − 1 ( x3 + 1) ( x + 1) 2
lim
x → −1
= lim
x → −1
x ( x + 1) ( x + 1)
( x + 1) ( x
2
− x + 1) ( x + 1)
= lim
x → −1
As x gets closer to – 1, the graph of R gets closer to –
1 x =– x − x +1 3 2
1 . Since R is not defined at 1, 3
1⎞ ⎛ the graph will have a hole at ⎜ −1, − ⎟ . 3⎠ ⎝ 39.
(a) Production costs are the sum of fixed costs and variable costs. So the cost function C of producing x calculators is C = C(x) = 10x + 79,000 (b) The domain of C is {x | x ≥ 0}. (c) The average cost per calculator, when x calculators are produced is given by the C ( x ) 10 x + 79,000 79,000 function C ( x ) = = = 10 + . x x x (d) The domain of C is {x | x > 0}. 79,000 ⎞ ⎛ (e) lim+ C = lim+ ⎜ 10 + ⎟=∞ x→ 0 x→ 0 ⎝ x ⎠ The average cost of making nearly 0 calculators becomes unbounded. 10 x + 79,000 10 x = lim = 10 (f) lim C ( x ) = lim x→ ∞ x→ ∞ x→ ∞ x x
CHAPTER 3 REVIEW 195
The average cost of producing a calculator when a very large number of calculators are produced is $10. 41.
5x Since x is approaching 100, but is remaining less than x → 100 x → 100 100 − x 5x > 0 and is becoming unbounded; so 100, 100 – x > 0, and the quotient 100 − x lim − C ( x ) = ∞ .
(a)
lim − C ( x ) = lim −
x → 100
(b) It is not possible to remove 100% of the pollutant. Explanations will vary. 43. Graphs will vary.
Chapter 3 Review TRUE-FALSE ITEMS 1. True
3. True
5. True
7. True
FILL-IN-THE-BLANKS
lim f ( x ) = N
1.
x→ c
5. is not equal to
3. not exist 7. y = 2 … horizontal
REVIEW EXERCISES 1. Here f (x) =
x3 − 8 , and c = 2. We find the limit by evaluating the function f at values of x−2
x close to 2. x f ( x) =
x3 − 8 x−2
x f ( x) =
x3 − 8 x−2
1.9
1.99
1.999
1.9999
11.41
11.94
11.9940
11.9994
2.1
2.01
2.001
2.0001
12.61
12.0601
2.0060
12.0006
We infer from the table that lim f ( x ) = lim x→ 2
x→ 2
x3 − 8 = 12 . x−2
CHAPTER 3 REVIEW 195
The average cost of producing a calculator when a very large number of calculators are produced is $10. 41.
5x Since x is approaching 100, but is remaining less than x → 100 x → 100 100 − x 5x > 0 and is becoming unbounded; so 100, 100 – x > 0, and the quotient 100 − x lim − C ( x ) = ∞ .
(a)
lim − C ( x ) = lim −
x → 100
(b) It is not possible to remove 100% of the pollutant. Explanations will vary. 43. Graphs will vary.
Chapter 3 Review TRUE-FALSE ITEMS 1. True
3. True
5. True
7. True
FILL-IN-THE-BLANKS
lim f ( x ) = N
1.
x→ c
5. is not equal to
3. not exist 7. y = 2 … horizontal
REVIEW EXERCISES 1. Here f (x) =
x3 − 8 , and c = 2. We find the limit by evaluating the function f at values of x−2
x close to 2. x f ( x) =
x3 − 8 x−2
x f ( x) =
x3 − 8 x−2
1.9
1.99
1.999
1.9999
11.41
11.94
11.9940
11.9994
2.1
2.01
2.001
2.0001
12.61
12.0601
2.0060
12.0006
We infer from the table that lim f ( x ) = lim x→ 2
x→ 2
x3 − 8 = 12 . x−2
196 CHAPTER 3 REVIEW
3.
lim f ( x) = 0
x→ 0
5. f (x) = 3x 2 – 2x + 1 is a polynomial. We know that for polynomials, lim f ( x ) = f ( c ) . x→ c
lim ( 3 x − 2 x + 1) = 3 ⋅ 2 − 2 ⋅ 2 + 1 = 12 − 4 + 1 = 9 2
2
Limit of a polynomial.
x→ 2
7. f (x) = x 2 + 1 is a polynomial. We know that for polynomials, lim f ( x ) = f ( c ) .
(
lim ( x 2 + 1) = lim ( x 2 + 1) 2
x→ − 2
x→ − 2
↑ Limit of a Power
9.
2
2 2 = ⎡( −2 ) + 1⎤ = ( 5 ) = 25 ⎣ ⎦
↑ Limit of a polynomial
f ( x ) = x 2 + 7 ; its domain is all real numbers.
lim x→ 3
x 2 + 7 = lim ( x 2 + 7 ) = x→ 3
↑ Limit of a Root
11.
)
x→ c
2
(3
2
+ 7 ) = 16 = 4
↑ Limit of a polynomial
f ( x ) = 1 − x 2 . Its domain is the set of numbers that keeps 1 – x 2 ≥ 0.
1 – x 2 ≥ 0 or x 2 ≤ 1 or x ≥ – 1 and x ≤ 1 So the domain of f is {x | – 1 ≤ x ≤ 1} or x in the interval [– 1, 1] . As x → 1 − , x gets closer to 1, but remains less than 1; x is in the domain of f. So we need only to consider x as it approaches 1. lim− 1 − x 2 = lim 1 − x 2 = x→ 1
x→ 1
lim− (1 − x 2 ) = 1 − 1 = 0
x→ 1
↑ Limit of a Root
↑ Limit of a Polynomial
13. f (x) = 5x + 6 is a polynomial, so as x approaches c, f (x) approaches f (c). lim ( 5 x + 6 ) x→ 2
3/ 2
= ⎡lim ( 5 x + 6 ) ⎤ ⎣⎢ x → 2 ⎦⎥
↑ Limit of a Power
3/ 2
= (5 ⋅ 2 + 6)
↑ Limit of a Polynomial
3/ 2
= 16 3 / 2 = 64
CHAPTER 3 REVIEW 197 15. Here f (x) = x 2 + x + 2 and g(x) = x 2 – 9 are both polynomials. So, 2 2 lim ( x 2 + x + 2 )( x 2 − 9 ) = ⎡ lim ( x 2 + x + 2 ) ⎤ ⎡ lim ( x 2 − 9 ) ⎤ = ⎡( −1) + ( −1) + 2 ⎤ ⎡( −1) − 9 ⎤ = −16 ⎦⎣ ⎦ x→ −1 ⎣⎢ x → − 1 ⎦⎥ ⎣⎢ x → − 1 ⎦⎥ ⎣ ↑ Limit of a Product
↑ Limits of Polynomials
x −1 . As x approaches 1, the limit of the denominator equals zero, so x3 − 1 Formula (11) cannot be used directly. We factor the expression first. lim 1 1 1 x −1 x −1 x→ 1 lim lim 3 = lim = = = 2 2 x→ 1 x2 + x + 1 x→ 1 x − 1 x→ 1 lim x + x + 1 3 ( x − 1) ( x + x + 1)
17. Here f ( x ) =
x→ 1
↑ Factor.
↑ Limit of a Quotient
x2 − 9 . As x approaches – 3, the limit of the denominator equals zero, x 2 − x − 12 so Formula (11) cannot be used directly. We factor the expression first. lim ( x − 3) −6 6 ( x − 3) x + 3 x − 3 x→ x2 − 9 = lim = −3 = = lim 2 = lim x → − 3 x − x − 12 x→ − 3 ( x − 4 ) x + 3 x → − 3 x − 4 lim ( x − 4 ) −7 7
19. Here f ( x ) =
( (
) )
↑ Factor.
21.
↑ Limit of a Quotient
x2 − 1 . As x approaches – 1 from the left, the limit of the denominator x3 − 1 equals zero, so Formula (11) cannot be used directly. We factor the expression first.
Here f ( x ) =
lim−
x →−1
lim ( x + 1) ( x − 1) ( x + 1) x2 − 1 x +1 0 x →−1− = lim lim = = = =0 2 3 − − 2 2 x − 1 x →−1 ( x − 1) x + x + 1 x→−1 x + x + 1 lim− x + x + 1 1
(
)
↑ Factor.
23.
x→ − 3
x →−1
(
)
↑ Limit of a Quotient
x3 − 8 . As x approaches 2, the limit of the denominator equals x3 − 2 x 2 + 4 x − 8 zero, so Formula (11) cannot be used directly. We factor the expression first. x − 2 ( x2 + 2 x + 4) ( x − 2) ( x2 + 2 x + 4) x3 − 8 lim = lim 2 = lim x → 2 x3 − 2 x 2 + 4 x − 8 x→ 2 x ( x − 2) + 4 ( x − 2) x→ 2 ( x2 + 4) x − 2
Here f ( x ) =
(
lim x→ 2
x2 + 2x + 4 = x2 + 4 .
= lim ( x + 2 x + 4 ) 2
x→ 2
lim ( x 2 + 4 ) x→ 2
=
22 + 2 ⋅ 2 + 4 12 3 = = 22 + 4 8 2 ↑ Limit of a Quotient
)
(
)
198 CHAPTER 3 REVIEW
25.
x 4 − 3 x3 + x − 3 . As x approaches 3, the limit of the denominator equals x3 − 3x 2 + 2 x − 6 zero, so Formula (11) cannot be used directly. We factor the expression first.
Here f ( x ) =
3 x 3 ( x − 3) + 1( x − 3) ( x + 1) ( x − 3) x 4 − 3 x3 + x − 3 = = x 3 − 3 x 2 + 2 x − 6 x 2 ( x − 3) + 2 ( x − 3) ( x 2 + 2 ) ( x − 3)
Factor both the numerator and the denominator by grouping.
lim ( x3 + 1) 28 x3 + 1) x − 3 ( x 4 − 3x3 + x − 3 = x→ 3 2 = lim 3 = lim 2 x → 3 x − 3x 2 + 2 x − 6 x→ 3 ( x + 2 ) x − 3 lim ( x + 2 ) 11
( (
.
27.
29.
31.
33.
35.
5 x 4 − 8 x3 + x 5x4 lim = lim 4 x → ∞ 3x 4 + x 2 + 5 x → ∞ 3x 5 5 = lim = x→ ∞ 3 3
) )
x→ 3
↑ Limit of a Quotient
As x → ∞, 5x 4 – 8x 3 + x = 5x 4 and 3x 4 + x 2 + 5 = 3x 4.
x2 is not defined at x = 3. When x → 3 − , x – 3< 0. Since x 2 ≥ 0, it follows x−3 x2 is negative and becomes unbounded as x → 3 − . that the expression x−3 x2 lim =−∞ x→ 3 − x − 3 f ( x) =
8x 4 − x 2 + 2 8x 4 = lim x→ ∞ − 4x 3 + 1 x→ ∞ − 4x 3 8x = lim =−∞ x→ ∞ − 4 lim
As x → ∞, 8x 4 – x 2 + 2 = 8x 4 and – 4x 3 + 1 = – 4x 3.
1 − 9x 2 is not defined at x = – 3. When x → − 3 + , x > – 3 and x 2 – 9 < 0. Since 2 x −9 1 − 9x 2 is positive and as becomes unbounded x → − 3 + . 1 – 9x2 < 0, it follows that 2 x −9 1 − 9x 2 =∞ lim + 2 x→ − 3 x − 9 f ( x) =
f ( x ) = 3 x 4 − x 2 + 2 is a polynomial function, and polynomial functions are continuous
at all values of x. So f (x) is continuous at c = 5. 37.
f ( x) =
x4 −4 is a rational function which is continuous at all values of x in its domain. x+2
CHAPTER 3 REVIEW 199
Since x = – 2 is not in the domain of f, the function f is not continuous at c = – 2. 39. The function f is defined at c = – 2; f (– 2) = 4. x2 −4 The lim f ( x ) = lim = lim ( x − 2 ) = − 4 x→ − 2 x→ − 2 x + 2 x→ − 2 Since the limit as x approaches – 2 does not equal f (– 2), the function is not continuous at c = – 2. 41. The function f is defined at c = – 2; f (– 2) = – 4. x2 −4 The lim f ( x ) = lim = lim ( x − 2 ) = − 4 x→ − 2 x→ − 2 x + 2 x→ − 2 Since the limit as x approaches – 2 equals f (– 2), the function is continuous at c = – 2 43. To find any horizontal asymptotes we need to find lim f ( x ) and lim f ( x ) . x→ ∞
x→ − ∞
3x 3x 3 = lim 2 = lim = 0 2 x →∞ x →∞ x − 1 x →∞ x x →∞ x The line y = 0 is a horizontal asymptote of the graph when x is sufficiently positive. lim f ( x ) = lim
3x 3x 3 = lim 2 = lim = 0 2 x → − ∞ x → − ∞ x −1 x x The line y = 0 is a horizontal asymptote of the graph when x is sufficiently negative. lim f ( x ) = lim
x→ − ∞
x→ − ∞
The domain of f is {x | x ≠ – 1, x ≠ 1}. To locate any vertical asymptotes we look at lim f ( x ) and lim f ( x ) . x → −1
x→ 1
Looking at one-sided limits of f at – 1, we find lim − f ( x ) : When x → – 1 from the left, x < – 1 and x 2 > 1 or x 2 – 1 > 0. So, the x → −1
expression
3x is negative and becomes unbounded. x 2 −1 3x =–∞ lim − f ( x ) = lim − 2 x→ − 1 x → −1 x − 1
lim f ( x ) : When x → – 1 from the right, x > – 1 and x 2 < 1 or x 2 – 1 < 0. So, the
x → −1+
expression
3x is positive and becomes unbounded. x 2 −1 3x =∞ lim + f ( x ) = lim + 2 x → −1 x → −1 x − 1
We conclude f has a vertical asymptote at x = – 1. lim f ( x ) : When x → 1 from the left, x < 1 and x 2 < 1 or x 2 – 1 < 0. So, the
x→ 1−
200 CHAPTER 3 REVIEW
expression
3x is negative and becomes unbounded. x 2 −1 3x =–∞ lim− f ( x ) = lim− 2 x→ 1 x → 1 x −1
lim+ f ( x ) : When x → 1 from the right, x > 1 and x 2 – 1 > 0. So, the expression
x→ 1
3x is positive and becomes unbounded. x 2 −1 3x =∞ lim+ f ( x ) = lim+ 2 x→ 1 x → 1 x −1 We conclude f has a vertical asymptote at x = 1. 45.
To find any horizontal asymptotes we need to find lim f ( x ) and lim f ( x ) . x→ ∞
x→ − ∞
5x 5x 5 = lim = lim = 5 lim f ( x ) = lim x→ ∞ x→ ∞ x + 2 x→ ∞ x x→ ∞ 1 The line y = 5 is a horizontal asymptote of the graph when x is sufficiently positive. 5x 5x 5 = lim = lim = 5 x→ − ∞ x→ − ∞ x + 2 x→ − ∞ x x→ − ∞ 1 The line y = 5 is a horizontal asymptote of the graph when x is sufficiently negative. lim f ( x ) = lim
The domain of f is {x | x ≠ – 2}. To locate any vertical asymptotes we look at lim f ( x ) . x→ − 2
Looking at one-sided limits of f at – 2, we find lim − f ( x ) : When x → – 2 from the left, x < – 2 and x + 2 < 0. So, the expression x→ −2
5x is positive and becomes unbounded. x+2 5x = ∞ lim − f ( x ) = lim − x→ − 2 x→ − 2 x + 2 lim f ( x ) : When x → – 2 from the right, x > – 2 and x + 2 > 0. So, the expression
x→ −2+
5x is negative and becomes unbounded. x+2 5x = –∞ lim + f ( x ) = lim + x→ − 2 x→ − 2 x + 2 We conclude f has a vertical asymptote at x = – 2.
CHAPTER 3 REVIEW 201 47. (a) There is a vertical asymptote at x = 2 and f is not defined at 2, so the domain of f is the intervals (– ∞, 2) or (2, 5) or (5, ∞).
(b) The range of f is the set of all real numbers, that is all y in the interval (– ∞, ∞). (c) The x-intercepts are the points at which the graph crosses or touches the x-axis. The x-intercepts are (– 2, 0), (0, 0), (1, 0), and (6, 0). (d) The y-intercept is (0, 0). (e) f (–6) = 2 and f (– 4) =1 (f) f (– 2) = 0 and f (6) = 0 (g) (h)
lim f ( x ) = 4 ; lim + f ( x ) = − 2
x→ − 4 −
x→ − 4
lim f ( x ) = − 2 ; lim + f ( x ) = 2
x→ − 2 −
x→ − 2
(i) lim− f ( x ) = 2 ; lim+ f ( x ) = 2 x →5
x →5
(j) The lim f ( x ) does not exist since lim− f ( x ) = 4 and lim+ f ( x ) = 1 are not equal. x→ 0
x→ 0
x→ 0
(k) The lim f ( x ) does not exist since lim− f ( x ) = − ∞ and lim+ f ( x ) = ∞ . x→ 2
x→ 2
x→ 2
(l) f is not continuous at – 2 since lim f ( x ) does not exist. x→ − 2
(m) f is not continuous at – 4 since lim f ( x ) does not exist. x→ − 4
(n) f is not continuous at 0 since lim f ( x ) does not exist. x→ 0
(o) f is not continuous at 2; there is a vertical asymptote at 2. (p) f is continuous at 4. (q) f is not continuous at 5 since f is not defined at x = 5. (r) f is increasing on the open intervals (– 6, – 4), (– 2, 0), and (6, ∞). (s) f is decreasing on the open intervals (– ∞, – 6), (0, 2), (2, 5), and (5, 6). (t) lim f ( x ) = ∞ and lim f ( x ) = 2 x→ − ∞
x→ ∞
(u) There are no local maxima. There is a local minimum of 2 at x = – 6, a local
202 CHAPTER 3 REVIEW
minimum of 0 at x = 0, and a local minimum of 0 at x = 6. (v) There is a horizontal asymptote of y = 2 as x becomes unbounded in the positive direction, and a vertical asymptote at x =2. 49. The average rate of change of f (x) from – 2 to x is
(
2 ∆ y f ( x ) − f ( − 2 ) ( 2 x − 3 x ) − 2 ( −2 ) − 3 ( − 2 ) = = ∆x x − ( − 2) x − ( −2 ) 2
2 x 2 − 3x − 8 − 6 x+2 ( 2 x − 7 )( x + 2 ) = x+2
=
)
Remove parentheses. Factor.
The limit as x → − 2 is lim
( 2x − 7) ( x + 2) x+2
x →− 2
= lim ( 2 x − 7 ) = −11 x →− 2
51. The average rate of change of f (x) from 3 to x is x 3 x 3 − − ∆y f ( x ) − f ( 3) x − 1 3 − 1 x − 1 2 = = = ∆x x−3 x−3 x−3 2 x − 3 ( x − 1) Write as a single fraction. = 2 ( x − 1)( x − 3) =
−x+3 2 ( x − 1)( x − 3)
The limit as x → 3 is −1
lim
x→3
53.
−x+3
2 ( x − 1) ( x − 3)
= lim
x→3
−1 1 =− 2 ( x − 1) 4
x+4 . To determine the behavior of the graph near – 4 and 4, we look at x 2 − 16 lim R ( x ) and lim R ( x ) .
R ( x) = x→ − 4
x→ 4
For lim R ( x ) , we have x→ − 4
lim R ( x ) = lim
x→ − 4
x→ − 4
x+4 x+4 1 1 = lim =− = lim 2 4 x → − x → − 4 x−4 x − 16 8 ( x + 4) ( x − 4)
As x gets closer to – 4, the graph of R gets closer to − 1⎞ ⎛ the graph will have a hole at ⎜ − 4, − ⎟ . 8⎠ ⎝
1 . Since R is not defined at – 4, 8
CHAPTER 3 REVIEW 203
For lim R ( x ) , we have x→ 4
lim R ( x ) = lim
x→ 4
x→ 4
x+4 x+4 1 = lim = lim 2 x → x → 4 4 x−4 x − 16 ( x + 4) ( x − 4)
Since the limit of the denominator is 0, we use one-sided limits to investigate lim
x→ 4
If x < 4 and x is getting closer to 4, the value of that is, lim− x→ 4
1 . x−4
1 < 0 and is becoming unbounded; x−4
1 = – ∞. x−4
If x > 4 and x is getting closer to 4, the value of
1 > 0 and is becoming unbounded; x−4
1 = ∞. x→ 4 x − 4 The graph of R will have a vertical asymptote at x = 4. that is, lim+
55. Rational functions are undefined at values of x that would make the denominator of the function equal zero. Solving x 2 − 11x + 18 = 0 or ( x − 9 )( x − 2 ) = 0 we get x = 9 or x = 2.
So R is undefined at x = 2 and x = 9. To analyze the behavior of the graph near 2 and 9, we look at lim R ( x ) and lim R ( x ) . x→ 2
x→ 9
For lim R ( x ) , we have x→ 2
lim R ( x ) = lim x→ 2
x→ 2
x 2 ( x − 2) + 4( x − 2) x 3 − 2x 2 + 4x − 8 = lim x→ 2 x 2 − 11x + 18 ( x − 2 )( x − 9 )
(x = lim x→ 2
2
+ 4) ( x − 2)
( x − 2) ( x − 9)
= lim x→ 2
x2 + 4 8 =− 7 x −9
As x gets closer to 4, the graph of R gets closer to −
8 . Since R is not defined at 2, the 7
8⎞ ⎛ graph will have a hole at ⎜ 2, − ⎟ . 7⎠ ⎝ For lim R ( x ) , we have x→ 9
x 2 + 4) ( x − 2) ( x2 + 4 x 3 − 2x 2 + 4x − 8 = lim = lim lim R ( x ) = lim x→ 9 x→ 9 x→9 x 2 − 11x + 18 ( x − 2 ) ( x − 9 ) x→9 x − 9
Since the limit of the denominator is 0, we use one-sided limits to investigate x2 + 4 . lim x→9 x − 9 x2 + 4 If x < 9 and x is getting closer to 9, the value of < 0 and is becoming unbounded; x −9
204 CHAPTER 3 PROJECT
that is, lim− x→9
x2 + 4 = – ∞. x −9
If x > 9 and x is getting closer to 9, the value of that is, lim+ x→9
x2 + 4 > 0 and is becoming unbounded; x−9
x2 + 4 = ∞. x −9
The graph of R will have a vertical asymptote at x = 9. 57. Answers will vary. 59. (a) lim S ( x ) = lim x→ ∞
x→ ∞
2000 x 2 2000 x 2 2000 = = lim = 571.43 lim 2 2 x → ∞ x → ∞ 3.5 x + 1000 3.5 x 3.5
CHAPTER 3 PROJECT 1.
⎧0.10 ⎪0.15 ⎪ ⎪0.25 R ( x) = ⎨ ⎪0.28 ⎪0.33 ⎪ ⎩0.35
0 < x ≤ 7000 7000 < x ≤ 28, 400 28, 400 < x ≤ 68,800 68,800 < x ≤ 143,500 143,500 < x ≤ 311,950 x > 311,950
3. The function R is not continuous. It is discontinuous at the endpoints of each tax bracket. 5.
0.10 x ⎧ ⎪ 700 + 0.15 x − 7000 ( ) ⎪ ⎪⎪ 3910 + 0.25 ( x − 28, 400 ) A( x) = ⎨ ⎪ 14, 010 + 0.28 ( x − 68,800 ) ⎪34,926 + 0.33 ( x − 143,500 ) ⎪ ⎪⎩90,514 + 0.35 ( x − 311,950 )
1 < x ≤ 7000 7000 < x ≤ 28, 400 28, 400 < x ≤ 68,800 68,800 < x ≤ 143,500 143,500 < x ≤ 311,950 x > 311,950
7. The function A is not continuous if your income is $311,950. 9. To compute column 3, we find the amount of tax paid if a person earns the highest dollar amount allowable in the previous row. That is Row 2, Column 3: A couple earning $14,000 pays 0.10(14, 000) = $1400 So the entry will be $1400. Row 3, Column 3: We calculate the taxes paid by a couple earning $56,800.
MATHEMATICAL QUESTIONS 205
$1400 + 0.15($56,800 − $14, 000) = $9710 So the entry will be $9710. Row 4, Column 3: We calculate the taxes paid by a couple earning $114,650. $9710 + 0.25($114,650 – $56,800) = $24,172.50 So the entry will be $24,172.50. Row 5, Column 3: We calculate the taxes paid by a couple earning $174,700. $24,172.50 + 0.28($174,700 – $114,650) = $40,986.50 So the entry will be $40,986.50. Row 6, Column 3: We calculate the taxes paid by a couple earning $311,950. $40,986.50 + 0.33($311,950 – $174,700) = $86,279.00 So the entry will be $86,279.00. MATHEMATICAL QUESTIONS FROM PROFESSIONAL EXAMS 1. (b) 5 6
3. (d)
2 4
lim
x →3
lim
h→0
x2 − x−6 = lim x →3 x2 −9
( x − 3) ( x + 2 ) ( x + 2) = 5 = lim ( x − 3) ( x + 3) x → 3 ( x + 3) 6
⎛ 2+h − 2 2+h − 2 = lim ⎜ ⋅ h→0 h h ⎝ ⎛ 2+h 2 − = lim ⎜ h→0 ⎜ ⎜ h 2+h + ⎝ h = lim h→0 h 2+h + 2
(
) ( ) ( )
(
=
1 2 2
=
2+h + 2 ⎞ ⎟ 2+h + 2 ⎠ 2 ⎛ ⎞ 2 ⎟⎞ 2+h−2 ⎟ = lim ⎜ 2 ⎟⎟ h → 0 ⎜ h 2 + h + 2 ⎟ ⎝ ⎠ ⎠ 1 = lim h→0 2+h + 2
2 4
)
(
(
)
)
Chapter 3 – The Limit of a Function Section 3.1 Finding Limits Using Tables and Graphs In Problems 41–46, use a graphing utility to find the indicated limit rounded to two decimal places. 41.
lim x →1
x3 − x 2 + x − 1 x 4 − x3 + 2 x − 2
We will determine the limit using a table of values, which we can create using our graphing calculator. There are two options when creating a table using the calculator. We can have the calculator automatically generate a table given a starting value and an increment for x, or we can create a table by picking values for x ourselves. We will use the second option. We must first set the TABLE feature in the correct mode so we can enter values for x.
yp††~Í
Enter the expression into Y1 in the function editor.
73
Go to TABLE and enter values for x that are close to 1, but less than 1. . ysÊÍË·ÍË®ÍË® ® Í Ë ® ® ® Í Ë ® ® ® ® ÍË ® ®®®®Í Use the } key to move the cursor to the top of the x column and now enter values for x that are close to 1 but greater than 1.
Note that the last entry in the x column appears to be 1. However, since that entry is highlighted, its value is displayed at the bottom of the screen. The entry is really 1.00001. The reason that the calculator displays a 1 in the table is that it only shows entries to five significant digits and 1.00001 rounded to five significant digits is 1. Thus, rounding to two decimal places, lim x →1
43.
lim x→2
x3 − x 2 + x − 1 ≈ 0.67 x 4 − x3 + 2 x − 2
x3 − 2 x 2 + 4 x − 8 x2 + x − 6
Enter the expression into Y1 in the function editor.
74
Go to TABLE and enter values for x that are close to 2, but less than 2.
Note that the last entry in the x column appears to be 2. However, since that entry is highlighted, its value is displayed at the bottom of the screen. The entry is really 1.99999. The reason that the calculator displays a 2 in the table is that it only shows entries to five significant digits and 1.99999 rounded to five significant digits is 2. Use the } key to move the cursor to the top of the x column and now enter values for x that are close to 2 but greater than 2.
Thus, lim x→2
45.
x3 − 2 x 2 + 4 x − 8 = 1.6 x2 + x − 6
x3 + 2 x 2 + x x →−1 x 4 + x 3 + 2 x + 2 lim
75
Enter the expression into Y1 in the function editor.
Go to TABLE and enter values for x that are close to −1 , but less than −1 .
Note that the last two entries in the x column appears to be −1 . However, since the last entry is highlighted, its value is displayed at the bottom. The entry is really −1.00001 . The reason that the calculator displays a −1 in the table is that it only shows entries to five significant digits and −1.00001 rounded to five significant digits is −1 . The same idea is true for the previous entry. The outputs 1E-4 and 1E-5 represent numbers in scientific notation. The output 1E-4 represents the number 1×10−4 = 0.0001 and the 1E-5 represents the number −5 1×10 = 0.00001 . Use the } key to move the cursor to the top of the x column and now enter values for x that are close to −1 but greater than −1 .
The output -1E-4 represents the number −1× 10−4 = −0.0001 and the -1E-5 represents the number −1×10−5 = −0.00001 .
76
Thus, x3 + 2 x 2 + x lim =0 x →−1 x 4 + x 3 + 2 x + 2
________________________________________________________________________
77
Summary The command introduced in this chapter was: TABLE
78
Chapter 4 The Derivative of a Function 4.1 The Definition of a Derivative 1.
The slope of the tangent line to the graph of f (x) = 3x + 5 at the point (1, 8) is ( 3x + 5) − 8 3 ( x − 1) f ( x ) − f (1) 3x − 3 = lim = lim = lim mtan = lim = lim 3 = 3 x →1 x →1 x →1 x − 1 x →1 x→1 x −1 x −1 x −1 An equation of the tangent line is y – f(c) = mtan (x – c) y – 8 = 3(x – 1) y = 3x + 5 Simplify.
3.
The slope of the tangent line to the graph of f (x) = x 2 + 2 at the point (– 1, 3) is ( x 2 + 2 ) − ( 3) ( x − 1) ( x + 1) f ( x ) − f ( −1) x2 − 1 mtan = lim = lim = lim = lim x→ − 1 x → −1 x→ − 1 x + 1 x→ − 1 x − ( −1) x − ( −1) ( x + 1)
= lim ( x − 1) = −2 x → −1
An equation of the tangent line is y – 3 = (– 2)[x – (– 1)] y – 3 = – 2x – 2 y = – 2x + 1
5.
y – f(c) = mtan (x – c) Simplify. Add 3 to both sides.
The slope of the tangent line to the graph of f (x) = 3x 2 at the point (2, 12) is 3 ( x − 2) ( x + 2) f ( x ) − f ( 2) 3 x 2 − 12 3( x2 − 4 ) = lim = lim = lim mtan = lim x→ 2 x→ 2 x→ 2 x→ 2 x−2 x−2 x−2 x−2 = lim [3 ( x + 2 )] = 12 x→ 2
An equation of the tangent line is y – 12 = 12(x – 2)
y – f(c) = mtan (x – c)
SECTION 4.1 207
y – 12 = 12x – 24 y = 12x – 12
Simplify. Add 12 to both sides.
7.
The slope of the tangent line to the graph of f (x) = 2x 2 + x at the point (1, 3) is ( 2x2 + x ) − 3 ( 2 x + 3) ( x − 1) f ( x ) − f (1) mtan = lim = lim = lim = lim ( 2 x + 3) = 5 x→ 1 x→ 1 x→ 1 x→ 1 x −1 x −1 x −1 An equation of the tangent line is y – 3 = 5(x – 1) y – f(c) = mtan (x – c) y = 5x – 2 Simplify.
9.
The slope of the tangent line to the graph of f (x) = x 2 – 2x + 3 at the point (– 1, 6) is ( x 2 − 2 x + 3) − 6 f ( x ) − f ( −1) x2 − 2 x − 3 mtan = lim = lim = lim x→ − 1 x→ − 1 x→ − 1 x − ( −1) x +1 x +1 ( x − 3) ( x + 1) = lim = lim ( x − 3) = −4 x→ − 1 x→ − 1 x +1 An equation of the tangent line is y – 6 = (– 4)[x – (– 1)] y – 6 = – 4x – 4 y = – 4x + 2
11.
y – f (c) = mtan (x – c) Simplify. Add 6 to both sides.
The slope of the tangent line to the graph of f (x) = x 3 + x 2 at the point (–1, 0) is ( x3 + x 2 ) − 0 x 2 ( x + 1) = lim = lim x 2 = 1 mtan = lim x → −1 x → −1 ( x + 1) x → −1 x − ( −1)
208 SECTION 4.1
An equation of the tangent line is y – 0 = 1[x – (– 1)] y=x+1
y – f(c) = mtan (x – c) Simplify.
13.
To find f ′(3), we follow the three steps outlined in the text. Step 1: f (3) = – 4(3) + 5 = – 12 + 5 = – 7 f ( x ) − f ( 3) ( −4 x + 5 ) − ( −7 ) −4 x + 12 ( −4 ) ( x − 3) Step 2: = = = x−3 x−3 x−3 x−3 Step 3: The derivative of f at 3 is ( −4 ) ( x − 3) f ( x ) − f ( 3) = −4 f ' ( 3) = lim = lim x→ 3 x→ 3 x−3 x−3
15.
To find f ′(0), we follow the three steps outlined in the text. Step 1: f (0) = (0) 2 – 3 = – 3 f ( x ) − f ( 0 ) ( x 2 − 3) − ( −3) x 2 = = Step 2: x−0 x x Step 3: The derivative of f at 0 is f ( x ) − f ( 0) x2 ( ) = lim x = 0 f ' 0 = lim = lim x→ 0 x→ 0 x x→ 0 x−0
17.
To find f ′(1), we follow the three steps outlined in the text. Step 1: f (1) = 2 · 12 + 3 · 1 = 5 f ( x ) − f (1) ( 2 x 2 + 3 x ) − ( 5 ) 2 x 2 + 3 x − 5 ( 2 x + 5 )( x − 1) = = = Step 2: x −1 x −1 x −1 x −1 Step 3: The derivative of f at 1 is ( 2 x + 5 ) ( x − 1) f ( x ) − f (1) = lim ( 2 x + 5 ) = 7 f ' (1) = lim = lim x→ 1 x→ 1 x→ 1 x −1 x −1
19.
To find f ′(0), we follow the three steps outlined in the text. Step 1: f (0) = 03 + 4 · 0 = 0 f ( x ) − f ( 0 ) ( x3 + 4 x ) − ( 0 ) x ( x 2 + 4 ) = = Step 2: x−0 x x Step 3: The derivative of f at 0 is f ( x ) − f (0) x ( x2 + 4) f ' ( 0 ) = lim = lim = lim ( x 2 + 4 ) = 4 x→ 0 x→ 0 x→ 0 x−0 x
SECTION 4.1 209 21.
To find f ′(1), we follow the three steps outlined in the text. Step 1: f (1) = 13 + 12 – 2 · 1 = 0 f ( x ) − f (1) ( x3 + x 2 − 2 x ) − 0 x ( x 2 + x − 2 ) x ( x + 2 )( x − 1) Step 2: = = = x −1 x −1 x −1 x −1 Step 3: The derivative of f at 1 is x ( x + 2 ) ( x − 1) f ( x ) − f (1) f ' (1) = lim = lim x→ 1 x→ 1 x −1 x −1 = lim [ x ( x + 2 )] = lim x ⋅ lim ( x + 2 ) = 1 ⋅ 3 = 3 x→ 1
x→ 1
x→ 1
23.
To find f ′(1), we follow the three steps outlined in the text. 1 Step 1: f (1) = = 1 1 ⎛ 1 ⎞ ( ) 1− x − 1 ( ) ( −1) ( x − 1) f ( x ) − f 1 ⎜⎝ x ⎟⎠ Step 2: = = x = x −1 x −1 x −1 x ( x − 1) Step 3: The derivative of f at 1 is ( −1) ( x − 1) −1 f ( x ) − f (1) = lim = −1 f ' (1) = lim = lim x→ 1 x→ 1 x→ 1 x x −1 x ( x − 1)
25.
First we find the difference quotient of f (x) = 2x. f ( x + h) − f ( x) 2 ( x + h ) − 2 x 2 x + 2h − 2 x 2h = = =2 = h h h h ↑ Simplify
↑ Cancel the h’s.
The derivative of f is the limit of the difference quotient as h → 0, that is, f ( x + h) − f ( x) f ' ( x ) = lim = lim 2 = 2 h→ 0 h→ 0 h 27.
First we find the difference quotient of f (x) = 1 – 2x. f ( x + h ) − f ( x ) [1 − 2 ( x + h )] − [1 − 2 x ] 1 − 2 x − 2h − 1 + 2 x −2h = = −2 = = h h h h ↑ Simplify
↑ Cancel the h’s.
The derivative of f is the limit of the difference quotient as h → 0, that is, f ( x + h) − f ( x) f ' ( x ) = lim = lim ( −2 ) = −2 h→ 0 h→ 0 h 29.
First we find the difference quotient of f (x) = x 2 + 2. ⎡⎣( x + h )2 + 2 ⎤⎦ − [ x 2 + 2] f ( x + h) − f ( x) = h h 2 x + 2 xh + h 2 + 2 − x 2 − 2 = h
210 SECTION 4.1
2 xh + h 2 h ( h 2x + h) = h = 2x + h =
Simplify. Factor out h. Cancel the h’s.
The derivative of f is the limit of the difference quotient as h → 0, that is, f ( x + h) − f ( x) f ' ( x ) = lim = lim ( 2 x + h ) = 2 x h→ 0 h→ 0 h 31.
First we find the difference quotient of f (x) = 3x 2 – 2x + 1. ⎡⎣3 ( x + h )2 − 2 ( x + h ) + 1⎤⎦ − [3 x 2 − 2 x + 1] f ( x + h) − f ( x) = h h 2 2 3x + 6 xh + 3h − 2 x − 2h + 1 − 3x 2 + 2 x − 1 = h 2 6 xh + 3h − 2h = Simplify. h h ( 6 x + 3h − 2 ) h = 6x + 3h – 2
=
Factor out h. Cancel the h’s.
The derivative of f is the limit of the difference quotient as h → 0, that is, f ( x + h) − f ( x) f ' ( x ) = lim = lim ( 6 x + 3h − 2 ) = 6 x − 2 h→ 0 h→ 0 h 33.
First we find the difference quotient of f (x) = x 3. ( x + h )3 − x 3 f ( x + h) − f ( x) = h h 3 x + 3 x 2 h + 3xh 2 + h3 − x3 = h 2 2 3x h + 3 xh + h3 = h 2 ( h 3x + 3 xh + h 2 ) = h 2 = 3x + 3xh + h 2
Simplify. Factor out h. Cancel the h’s.
The derivative of f is the limit of the difference quotient as h → 0, that is, f ( x + h) − f ( x) = lim ( 3 x 2 + 3 xh + h 2 ) = 3 x 2 f ' ( x ) = lim h→ 0 h→ 0 h 35.
First we find the difference quotient of f (x) = mx + b. f ( x + h ) − f ( x ) [ m ( x + h ) + b ] − [ mx + b ] = h h
SECTION 4.1 211
mx + mh + b − mx − b h mh = h =m
=
Simplify Cancel the h’s.
The derivative of f is the limit of the difference quotient as h → 0, that is, f ( x + h) − f ( x) = lim m = m f ' ( x ) = lim h→ 0 h→ 0 h 37.
(a) The average rate of change of f (x) = 3x + 4 as x changes from 1 to 3 is ( 3 i 3 + 4 ) − ( 3 i 1 + 4 ) 13 − 7 6 ∆f f ( 3) − f (1) = = = = =3 ∆x 3 −1 3 −1 2 2 (b) The instantaneous rate of change at x = 1 is the derivative of f at 1. ( 3x + 4 ) − 7 3 ( x − 1) f ( x ) − f (1) 3x − 3 f ' (1) = lim = lim = lim = lim =3 x→ 1 x→ 1 x→ 1 x − 1 x→ 1 x −1 x −1 x −1 The instantaneous rate of change of f at 1 is 3.
39.
(a) The average rate of change of f (x) = 3x 2 + 1 as x changes from 1 to 3 is ( 3 i 32 + 1) − ( 3 i 12 + 1) = 28 − 4 = 24 = 12 ∆f f ( 3) − f (1) = = ∆x 3 −1 2 2 2 (b) The instantaneous rate of change at x = 1 is the derivative of f at 1. ( 3x 2 + 1) − ( 4 ) f ( x ) − f (1) 3x2 − 3 f ' (1) = lim = lim = lim x→ 1 x→ 1 x→ 1 x − 1 x −1 x −1 3 ( x − 1) ( x + 1) 3 ( x 2 − 1) = lim 3 ( x + 1) = 6 = lim = lim x→ 1 x→ 1 x→ 1 x −1 x −1 The instantaneous rate of change of f at 1 is 6.
41.
(a) The average rate of change of f (x) = x 2 + 2x as x changes from 1 to 3 is ( 32 + 2 ⋅ 3) − (12 + 2 ⋅ 1) = 15 − 3 = 12 = 6 ∆f f ( 3) − f (1) = = 2 2 2 ∆x 3 −1 (b) The instantaneous rate of change at x = 1 is the derivative of f at 1. ( x 2 + 2 x ) − ( 3) f ( x ) − f (1) x2 + 2x − 3 f ' (1) = lim = lim = lim x→ 1 x→ 1 x→ 1 x −1 x −1 x −1 ( x + 3) ( x − 1) = lim = lim ( x + 3) = 4 x→ 1 x→ 1 x −1 The instantaneous rate of change of f at 1 is 4.
212 SECTION 4.1 43.
(a) The average rate of change of f (x) = 2x 2 – x + 1 as x changes from 1 to 3 is ( 2 ⋅ 32 − 3 + 1) − ( 2 ⋅ 12 − 1 + 1) 16 − 2 14 ∆f f ( 3) − f (1) = = = = =7 2 2 2 ∆x 3 −1 (b) The instantaneous rate of change at x = 1 is the derivative of f at 1. ( 2 x 2 − x + 1) − ( 2 ) f ( x ) − f (1) 2 x2 − x − 1 f ' (1) = lim = lim = lim x→ 1 x→ 1 x→ 1 x −1 x −1 x −1 ( 2 x + 1) ( x − 1) = lim = lim ( 2 x + 1) = 3 x→ 1 x→ 1 x −1 The instantaneous rate of change of f at 1 is 3.
45.
The display below is from a TI-83 Plus graphing calculator.
47.
So f ′(– 2) = 60. 49.
The display below is from a TI-83 Plus graphing calculator.
So f ′(0) = 1. 53.
The display below is from a TI-83 Plus graphing calculator.
So f ′(8) = – 0.85878. 51.
The display below is from a TI-83 Plus graphing calculator.
So f ′(1) = 8.15485.
The display below is from a TI-83 Plus graphing calculator.
So f ′(1) = 0. 55.
We first find an equation of the tangent line to the graph of y = x 2 at (1, 1). The slope of the tangent line at (1, 1) is
SECTION 4.1 213
mtan = lim
x→ 1
( x + 1) ( x − 1) f ( x ) − f (1) x2 −1 = lim = lim = lim ( x + 1) = 2 x→ 1 x − 1 x→ 1 x→ 1 x −1 ( x − 1)
An equation of the tangent line is y – 1 = 2(x – 1) y – 1 = 2x – 2 y = 2x – 1
y – f(c) = mtan (x – c) Simplify. Add 1 to both sides.
Now we see if the point (2, 5) satisfies the equation of the tangent line. 2·2–1=3 y = 2x – 1; x = 2, y = 5. 5≠3 So the graph of the tangent line does not pass through the point (2, 5). 57.
For the rocket bomb to hit its target, the point (1, 0) must be on the graph of the tangent line to the graph of y = x 2 at some point (c, c 2). The slope of the tangent line at (c, c 2) is ( x − c) ( x + c) x2 − c2 = lim lim = lim ( x + c ) = 2c x→ c x − c x→ c x→ c x−c An equation of the tangent line is y – c 2 = 2c(x – c) y – f(c) = mtan (x – c) 2 2 y – c = 2cx – 2c Simplify. y = 2cx – c 2 Add c 2 to both sides. The point (1, 0) satisfies the equation of the tangent line, so 0 = 2c(1) – c 2 y = 2cx – c 2; x = 1; y = 0 2 c – 2c = 0 Simplify. c(c – 2) = 0 Factor. c=0 c=2 Apply the Zero-Product Property. Since the dive bomber is flying from right to left, the bomber reaches c = 2 first and should release the bomb at point (2, 4).
59.
(a) The average rate of change in sales S from day x = 1 to day x = 5 is 2 2 ∆S S ( 5 ) − S (1) ( 4 ( 5 ) + 50 ( 5 ) + 5000 ) − ( 4 (1) + 50 (1) + 5000 ) = = ∆x 5 −1 4 5350 − 5054 296 = = = 74 tickets per day. 4 4 (b) The average rate of change in sales S from day x = 1 to day x = 10 is 2 2 ∆S S (10 ) − S (1) ( 4 (10 ) + 50 (10 ) + 5000 ) − ( 4 (1) + 50 (1) + 5000 ) = = ∆x 10 − 1 9 5900 − 5054 846 = = = 94 tickets per day. 9 9 (c) The average rate of change in sales S from day x = 5 to day x = 10 is 2 2 ∆S S (10 ) − S ( 5 ) ( 4 (10 ) + 50 (10 ) + 5000 ) − ( 4 ( 5 ) + 50 ( 5 ) + 5000 ) = = ∆x 10 − 5 5
214 SECTION 4.1
=
5900 − 5350 550 = = 110 tickets per day. 5 5
(d) The instantaneous rate of change in sales on day 5 is the derivative of S at x = 5. S ( x ) − S ( 5) (4 x 2 + 50 x + 5000) − ( 5350 ) 4 x 2 + 50 x − 350 S ′(5) = lim = lim = lim x→ 5 x→ 5 x→ 5 x −5 x−5 x−5 2 ( 2 x + 35 ) ( x − 5 ) = lim = lim [ 2 ( 2 x + 35 )] = 2 lim ( 2 x + 35 ) = 2 ⋅ (10 + 35 ) = 90 x→ 5 x→ 5 x→ 5 x−5 The instantaneous rate of change of S on day 5 is 90 ticket sales per day. (e) The instantaneous rate of change in sales on day 10 is the derivative of S at x = 10. S ( x ) − S (10 ) (4 x 2 + 50 x + 5000) − ( 5900 ) 4 x 2 + 50 x − 900 = lim = lim S ′(10) = lim x → 10 x → 10 x → 10 x − 10 x − 10 x − 10 2 ( 2 x + 45 ) ( x − 10 ) = lim [ 2 ( 2 x + 45 )] = 2 lim ( 2 x + 45 ) = 2 ⋅ 65 = 130 x → 10 x → 10 x → 10 x − 10 The instantaneous rate of change of S on day 10 is 130 ticket sales per day.
= lim
61.
(a) At x = $10 per crate, the farmer is willing to supply S(10) = 50 · 10 2 – 50 · 10 = 4500 crates of grapefruits. ↑ S(x) = 50x 2 – 50x
(b) At x = $13 per crate, the farmer is willing to supply S(13) = 50 · 13 2 – 50 · 13 = 7800 crates of grapefruits. (c) The average rate of change in supply from $10 to $13 is ∆S S (13) − S (10 ) 7800 − 4500 3300 = = = = 1100 ∆x 13 − 10 3 3 The average rate of change in crates of grapefruit supplied is 1100 crates per dollar increase in price. (d) The instantaneous rate of change in supply at x = 10 is the derivative S′(10). [50 x 2 − 50 x ] − [ 4500] S ( x ) − S (10 ) 50 ( x 2 − x − 90 ) S ′(10) = lim = lim = lim x →10 x→10 x →10 x − 10 x − 10 x − 10 50 ( x − 10 ) ( x + 9 ) = lim = lim [50 ( x + 9 )] = 50 lim ( x + 9 ) = 50 ⋅ 19 = 950 x →10 x →10 x →10 x − 10 The instantaneous rate of change in supply at x = $10 is 950 crates. (e) The average rate of change in supply over the price interval from $10 to $13 is 1100 crates of grapefruit per $1.00 change in price. The instantaneous rate of change in supply of 950 crates is the increase in supply of grapefruit as the price changes from $10 to $11.
SECTION 4.1 215 63.
(a) The marginal revenue is the derivative R′(x). ⎡⎣8 ( x + h ) − ( x + h )2 ⎤⎦ − [8 x − x 2 ] R ( x + h) − R ( x) R′(x) = lim = lim h→0 h→0 h h 2 2 2 8 x + 8h − x − 2 xh − h − 8 x + x = lim Simplify. h→0 h 8h − 2 xh − h 2 = lim Simplify. h→0 h h ( 8 − 2x − h ) = lim Factor out the h. h→0 h = lim ( 8 − 2 x − h ) = 8 − 2 x Cancel the h’s. Go to the limit. h→0
The marginal revenue is R′(x) = 8 – 2x. (b) The marginal cost is the derivative C′(x). [ 2 ( x + h ) + 5] − [ 2 x + 5] C ( x + h) − C ( x) = lim C′(x) = lim h→0 h→0 h h 2 x + 2h + 5 − 2 x − 5 2h = lim = lim = lim 2 = 2 h→0 h → 0 h→0 h h (c) To find the break-even point we solve the equation R(x) = C(x). 8x – x 2 = 2x + 5 x 2 – 6x + 5 = 0 Put the quadratic equation in standard form. (x – 5)(x – 1) = 0 Factor. x–5=0 x–1=0 Apply the Zero-Product Property. x=5 x=1 Solve. There are two break-even points. One is when 1000 units are produced, and the other is when 5000 units are produced. (d) To find the number x for which marginal revenue equals marginal cost, we solve the equation R′(x) = C′(x). 8 – 2x = 2 2x = 6 x=3 Marginal revenue equals marginal cost when 3000 units are produced and sold. (e)
216 SECTION 4.1 65.
(a) The revenue function R(x) = xp = x(–10x + 2000) = –10x 2 + 2000x. (b) The marginal revenue is the derivative R′(x). ⎡⎣ −10 ( x + h )2 + 2000 ( x + h ) ⎤⎦ − [ −10 x 2 + 2000 x ] R ( x + h) − R ( x) R′(x) = lim = lim h→ 0 h→ 0 h h 2 2 2 −10 x − 20 xh − 10h + 2000 x + 2000h + 10 x − 2000 x = lim Simplify. h→ 0 h −20 xh − 10h 2 + 2000h = lim Simplify. h→ 0 h h ( −20 x − 10h + 2000 ) = lim Factor out an h. h→ 0 h = lim ( −20 x − 10h + 2000 ) Cancel out the h’s. h→ 0
= −20 x + 2000
Go to the limit.
(c) The marginal revenue at x = 100 tons is R′(100) = (–20) · 100 + 2000 = 0 dollars. (d) The average rate in change in revenue from x = 100 to x = 101 tons is ∆R R (101) − R (100 ) ⎡⎣( −10 ) (1012 ) + 2000 (101) ⎤⎦ − ⎡⎣( −10 ) (1002 ) + 2000 (1000 ) ⎤⎦ = = 101 − 100 1 ∆x = 99,990 − 100,000 = −10 (e) R′(100) = 0 indicates that there is no additional revenue gained by selling the 101st ton of cement. The average rate of change in revenue from selling the 101st ton of cement represents a decrease in revenue of $10. 67.
(a) The revenue function R(x) = xp, where p is the unit price and x is the number of units sold. R(x) = xp = x(90 – 0.02x) = 90x – 0.02x 2 (b) The marginal revenue is the derivative R′(x). ⎡⎣90 ( x + h ) − 0.02 ( x + h )2 ⎤⎦ − [90 x − 0.02 x 2 ] R ( x + h) − R ( x) R′(x) = lim = lim h→0 h→0 h h 2 2 90 x + 90h − 0.02 x − 0.04 xh − 0.02h − 90 x + 0.02 x 2 = lim Simplify. h→0 h 90h − 0.04 xh − 0.02h 2 = lim Simplify. h→0 h h ( 90 − 0.04 x − 0.02h ) = lim Factor out an h. h→0 h = lim ( 90 − 0.04 x − 0.02h ) = 90 − 0.04 x Cancel the h’s; take the limit. h→0
(c) It costs $10 per unit to produce the product, so the cost function C = C(x) = 10x. The marginal cost is the derivative C′(x). [10 ( x + h )] − [10 x ] C ( x + h) − C ( x) = lim C′(x) = lim h→0 h→0 h h
SECTION 4.2 217
10 x + 10h − 10 x 10 h = lim = lim 10 = 10 h→0 h→0 h→0 h h
= lim
(d) A break-even point is a number x for which R(x) = C(x). We solve the equation R(x) = C(x) 90x – 0.02x 2 = 10x Put the quadratic equation in standard form. 0.02x 2 – 80x = 0 2 x – 4000x = 0 Multiply both sides by 50. x(x – 4000) = 0 Factor. x = 0 x – 4000 = 0 Apply the Zero-Product Property. x=0 x = 4000 Solve for x. There are two break-even points. One is when no units are produced and sold; the other is when x = 4000 units are produced and sold. (e) The marginal revenue equals marginal cost when R′(x) = C′(x). R′(x) = C′(x) 90 – 0.04x = 10 80 = 0.04x x = 2000 The marginal revenue equals marginal cost when 2000 units are produced and sold. 69.
The instantaneous rate of change of the volume of the cylinder with respect to the radius r at r = 3 is the derivative V′(3). V ( r ) − V ( 3) 3π r 2 − 3π ( 32 ) 3π r 2 − 27π 3π ( r 2 − 9 ) = lim = lim = lim V′(3) = lim r →3 r→3 r→3 r →3 r −3 r −3 r −3 r −3 3π ( r − 3) ( r + 3) = lim = lim [3π ( r + 3)] = 18π ≈ 56.55 r →3 r →3 r −3
4.2 The Derivative of a Power Function; Sum and Difference Formulas 1.
The function f (x) = 4 is a constant. f ′(x) = 0
3.
The function f (x) = x 5 is a power function. f ′(x) = 5x 5 – 1 = 5x 4
5.
f ′(x) =
d ( 2) d 6 x = 6 x 2 = 6 ⋅ 2 x = 12 x dx dx
7.
f ′(t) =
d ⎛ t4 ⎞ 1 d 4 1 t = ⋅ 4t 4−1 = t 3 ⎜ ⎟= dt ⎝ 4 ⎠ 4 dt 4
9.
f ′(x) =
d ( 2 d 2 d x + x) = x + x dx dx dx = 2x +1
Use the derivative of a sum formula (Formula (5)).
d dx
x =1
SECTION 4.2 217
10 x + 10h − 10 x 10 h = lim = lim 10 = 10 h→0 h→0 h→0 h h
= lim
(d) A break-even point is a number x for which R(x) = C(x). We solve the equation R(x) = C(x) 90x – 0.02x 2 = 10x Put the quadratic equation in standard form. 0.02x 2 – 80x = 0 2 x – 4000x = 0 Multiply both sides by 50. x(x – 4000) = 0 Factor. x = 0 x – 4000 = 0 Apply the Zero-Product Property. x=0 x = 4000 Solve for x. There are two break-even points. One is when no units are produced and sold; the other is when x = 4000 units are produced and sold. (e) The marginal revenue equals marginal cost when R′(x) = C′(x). R′(x) = C′(x) 90 – 0.04x = 10 80 = 0.04x x = 2000 The marginal revenue equals marginal cost when 2000 units are produced and sold. 69.
The instantaneous rate of change of the volume of the cylinder with respect to the radius r at r = 3 is the derivative V′(3). V ( r ) − V ( 3) 3π r 2 − 3π ( 32 ) 3π r 2 − 27π 3π ( r 2 − 9 ) = lim = lim = lim V′(3) = lim r →3 r→3 r→3 r →3 r −3 r −3 r −3 r −3 3π ( r − 3) ( r + 3) = lim = lim [3π ( r + 3)] = 18π ≈ 56.55 r →3 r →3 r −3
4.2 The Derivative of a Power Function; Sum and Difference Formulas 1.
The function f (x) = 4 is a constant. f ′(x) = 0
3.
The function f (x) = x 5 is a power function. f ′(x) = 5x 5 – 1 = 5x 4
5.
f ′(x) =
d ( 2) d 6 x = 6 x 2 = 6 ⋅ 2 x = 12 x dx dx
7.
f ′(t) =
d ⎛ t4 ⎞ 1 d 4 1 t = ⋅ 4t 4−1 = t 3 ⎜ ⎟= dt ⎝ 4 ⎠ 4 dt 4
9.
f ′(x) =
d ( 2 d 2 d x + x) = x + x dx dx dx = 2x +1
Use the derivative of a sum formula (Formula (5)).
d dx
x =1
218 SECTION 4.2
11.
f ′(x) =
d ( 3 d 3 d 2 d x − x 2 + 1) = x − x + 1 dx dx dx dx
Use Formulas (6) and (5). d
= 3x 2 − 2 x
13.
15.
17.
f ′(t) =
f ′(x) =
f ′(x) =
d( 2 d d d 2t − t + 4 ) = ( 2t 2 ) − t + 4 dt dt dt dt d 2 d = 2 t − t +0 dt dt = 4t − 1
f ′(x) =
1= 0
Use Formulas (5) and (6). Use Formulas(4) and (2). Differentiate.
d ⎛1 8 2⎞ d ⎛1 8⎞ d d 2 ⎜ x + 3x + ⎟ = ⎜ x ⎟ + ( 3x ) + 3 ⎠ dx ⎝ 2 ⎠ dx dx ⎝ 2 dx 3 1 d 8 d = x +3 x+0 2 dx dx 1 = ⋅ 8x7 + 3 = 4 x7 + 3 2
Use Formula (5). Use Formula (4);
d
2
dx
3
= 0.
Differentiate and simplify.
d ⎡ 1 ( 5 )⎤ 1 d ( 5 ) 1 ⎡ d 5 d ⎤ 1 [ 4 5 x −8 ⎥ = x − 8 = ⎢ x − 8⎥ = 5 x − 0] = x 4 ⎢ 3 ⎣ dx 3 dx ⎣ 3 dx ⎦ 3 ⎦ 3 dx ↑ ↑ ↑ ↑ Use Formula (4).
19.
dx
Use Formula (6).
Differentiate.
Simplify.
d [ 2 d d d d d ax + bx + c ] = ( ax 2 ) + ( bx ) + c = a x 2 + b x + 0 dx dx dx dx dx dx ↑ ↑ Use Formula (5).
Use Formula (4);
= a ⋅ 2 x + b ⋅ 1 = 2ax + b ↑ ↑ Differentiate.
d
c=0
dx
Simplify.
21.
d ( d d d d −6 x 2 + x + 4 ) = ( −6 x 2 ) + x + 4 = ( −6 ) x 2 + 1 + 0 = ( −6 ) ⋅ 2 x + 1 = −12 x + 1 dx dx dx dx dx
23.
d( d d d d −16t 2 + 80t ) = ( −16t 2 ) + ( 80t ) = ( −16 ) t 2 + 80 t = ( −16 ) ⋅ 2t + 80 ⋅ 1 = −32t + 80 dt dt dt dt dt
25.
dA d ( 2 ) d = π r = π r 2 = π ⋅ 2r = 2π r dr dr dr
27.
dV d ⎛ 4 3 ⎞ 4 d 3 4 = ⎜ π r ⎟ = π r = π ⋅ 3r 2 = 4π r 2 3 dr dr ⎝ 3 ⎠ 3 dr
SECTION 4.2 219 29.
To find f ′(– 3), first we find the derivative of the function f. d ( 2) f ′(x) = 4 x = 8x dx Then we substitute – 3 for x. f ′(– 3) = 8 · (– 3) = – 24
31.
To find f ′(4), first we find the derivative of the function f. d ( 2 d d d f ′(x) = 2 x − x ) = ( 2 x2 ) − x = 2 x2 − 1 = 4 x − 1 dx dx dx dx Then we substitute 4 for x. f ′(4) = 4 · 4 – 1 = 15
33.
To find f ′(3), first we find the derivative of the function f. d⎛ 1 1d 3 d 1 ⎞ d⎛ 1 ⎞ d t + 5 t = − ⋅ 3t 2 + 5 = −t 2 + 5 f ′(x) = ⎜ − t 3 + 5t ⎟ = ⎜ − t 3 ⎟ + ( 5t ) = − 3 dt 3 dt ⎝ 3 dt ⎠ dt ⎝ 3 ⎠ dt Then we substitute 3 for t. f ′(3) = – 32 + 5 = – 4
35.
To find f ′(1), first we find the derivative of the function f. d ⎡1 1⎡ d d ⎤ ⎤ 1 d ( 6 f ′(x) = ⎢ ( x 6 − x 4 ) ⎥ = x − x4 ) = ⎢ x6 − x4 ⎥ 2 ⎣ dx dx ⎣ 2 dx ⎦ ⎦ 2 dx 1 = ( 6 x5 − 4 x3 ) = 3x5 − 2 x3 2 Then we substitute 1 for x. f ′(1) = 3 · 15 – 2 · 13 = 1
37.
First we find the derivative of the function f. In Problem 19 we found f ′(x) = 2ax + b. b So we now substitute − for x. 2a ⎛ b ⎞ ⎛ b ⎞ f ′ ⎜ − ⎟ = 2 a ⎜ − ⎟ + b = −b + b = 0 ⎝ 2a ⎠ ⎝ 2a ⎠
39.
41.
dy = 4x 3 dx Then we evaluate the derivative at the point (1, 1) by substituting 1 for x. dy = 4 · 13 = 4 dx
First we find the derivative
dy = 2x – 0 = 2x dx Then we evaluate the derivative at the point (4, 2) by substituting 4 for x. dy =2·4=8 dx
First we find the derivative
220 SECTION 4.2
43.
45.
dy = 6x – 1 dx Then we evaluate the derivative at the point (–1, 4) by substituting –1 for x. dy = 6 · (–1) – 1 = – 7 dx
First we find the derivative
First we find the derivative
dy 1 = ⋅ 2x = x dx 2
⎛ 1⎞ Then we evaluate the derivative at the point ⎜1, ⎟ by substituting 1 for x. ⎝ 2⎠ dy =1 dx dy = 0 – 2 + 3x 2 = – 2 + 3x 2 dx Then we evaluate the derivative at the point (2, 6) by substituting 2 for x. dy = – 2 + 3 · 22 = 10 dx
47.
First we find the derivative
49.
The slope of the tangent line to the graph of f (x) = x 3 + 3x – 1 at the point (0, – 1) is the derivative of the function f evaluated at the point (0, – 1). The derivative of f is f ′(x) = 3x 2 + 3 mtan = f ′(0) = 3 · (0) 2 + 3 = 3 An equation of the tangent line is y – (–1)= 3(x – 0) y + 1 = 3x y = 3x – 1
51.
y – f(c) = mtan (x – c) Simplify. Subtract 1 from both sides.
We first find the derivative f ′(x). f ′(x) = 6x – 12 + 0 = 6x – 12 We then solve the equation f ′(x) = 0. 6x – 12 = 0 6x = 12 x=2
53.
We first find the derivative f ′(x). f ′(x) = 3x 2 – 3 + 0 = 3x 2 – 3 We then solve the equation f ′(x) = 0. 3x 2 – 3 = 0 3( x 2 – 1) = 0 x2 – 1 = 0 x=1 x=–1
f ′(x) = 3x 2 – 3 Factor out the 3. Divide both sides by 3. Solve using the Square Root Method.
SECTION 4.2 221 55.
We first find the derivative f ′(x). f ′(x) = 3x2 + 1 We then solve the equation f ′(x) = 0. 3x2 + 1 = 0 has no solutions.
f ′(x) = 3x2 + 1
57.
The slope of the tangent line to the function f (x) = 9x3 is the derivative of f . f ′(x) = 27x 2 To find the slope of the line 3x – y + 2 = 0, we put the equation in slope-intercept form. y = 3x + 2 The slope of the line is m = 3. For the tangent line to the graph of f to be parallel to the line y, mtan = 3. We solve the equation f ′(x) = 3. 27x 2 = 3 f ′(x) = 27x 2 1 x2 = Divide both sides by 27. 9 1 1 x= x=– Solve using the Square Root Method. 3 3 There are two points on the graph of the function y = 9x3 for which the slope of the tangent line is parallel to the graph of the line 3x – y + 2 = 0. They are ⎛ 1 ⎛ 1 ⎞⎞ ⎛ 1 1 ⎞ ⎛ 1 ⎛ 1 ⎞⎞ ⎛ 1 1 ⎞ ⎜ , f ⎜ ⎟ ⎟ = ⎜ , ⎟ and ⎜ − , f ⎜ − ⎟ ⎟ = ⎜ − , − ⎟ ⎝ 3 ⎝ 3 ⎠⎠ ⎝ 3 3 ⎠ ⎝ 3 ⎝ 3 ⎠⎠ ⎝ 3 3 ⎠
59.
If (x, y) = (x, 2x 2 – 4x + 1) is a point on the graph of the function y = 2x 2 – 4x + 1 for which the tangent line to the graph of y passes through the point (1, – 3), then the slope of the tangent line is given by ∆y y2 − y1 ( 2 x 2 − 4 x + 1) − ( −3) 2 x 2 − 4 x + 4 mtan = = = = x −1 x −1 ∆x x2 − x1 The slope of the tangent line to the graph of the function y is also given by the derivative of y. d ( 2 2 x − 4 x + 1) = 4 x − 4 mtan = dx Since both these expressions for the slope must be same, we will set them equal to each other and solve for x. 2x2 − 4x + 4 Set the slope equal to the derivative. = 4x – 4 x −1 2x 2 – 4x + 4 = (4x – 4)(x – 1) Multiply both sides by x – 1. 2x 2 – 4x + 4 = 4x 2 – 8x + 4 Multiply out the right side. 2x 2 – 4x = 0 Put the quadratic equation in standard form. 2x(x – 2) = 0 Factor. 2x = 0 x – 2 = 0 Apply the Zero-Product Property. x=0 x=2 Solve for x.
We evaluate the tangent lines to the graph of y using each of these two values of x. If x = 0 then y = f (0) = 1. The point on the graph of y is (0, 1). The slope of the tangent line to the graph of y at (0, 1) is f ′(0) = 4 · 0 – 4 = – 4. The equation of the tangent line through (0, 1) and (1, –3) is
222 SECTION 4.2
y – 1 = – 4(x – 0) y = – 4x + 1 If x = 2 then y = f (2) = 2 · 22 – 4 · 2 + 1 = 1. The point on the graph of y is (2, 1). The slope of the tangent line to the graph of y at (2, 1) is f ′(2) = 4 · 2 – 4 = 4. The equation of the tangent line through (2, 1) and (1, –3) is y – 1 = 4(x – 2) y – 1 = 4x – 8 y = 4x – 7 61. (a) The average cost of producing 10 additional pairs of eyeglasses is 2 2 ∆y C (110) − C (100 ) ⎡⎣0.2 ⋅ 110 + 3 ⋅ 110 + 1000 ⎤⎦ − ⎡⎣ 0.2 ⋅ 100 + 3 ⋅ 100 + 1000 ⎤⎦ = = ∆x 110 − 100 10 3750 − 3300 450 = = = 45 dollars per pair. 10 10
(b) The marginal cost of producing an additional pair of eyeglasses is C′(x) = 0.4x + 3 (c) The marginal cost at x = 100 is C′(100) = 0.4(100) + 3 = 43 dollars. (d) The marginal cost at x = 100 is the cost of producing one additional pair of eyeglasses when 100 pairs have already been produced.
63.
(a) The derivative V′(R) = 4kR 3. (b) V′(0.3) = 4k · 0.33 = 0.108k centimeters cubed (c) V′(0.4) = 4k · 0.43 = 0.256k centimeters cubed
65.
(a) It costs C(40) = 2000 + 50 · 40 – 0.05 · 402 = 3920 dollars to produce 40 microwave ovens. (b) The marginal cost function is the derivative of C(x). C′(x) = 50 – 0.10x (c) C′(40) = 50 – 0.10 · 40 = 46. The marginal cost at x = 40 indicates that the cost of producing the 41st microwave oven is $46.00. (d) An estimate of the cost of producing 41 microwave ovens can be obtained by adding C(40) and C′(40). C(40) + C′(40) = 3920 + 46 = 3966 dollars to produce 41 microwaves. (e) The actual cost of producing 41 microwave ovens is C(41) = 2000 + 50 · 41 – 0.05 · 412 = 3965.95 dollars. The actual cost of producing 41 microwave ovens is $0.05 less than the estimated cost.
SECTION 4.2 223
(f) The actual cost of producing the 41st microwave oven is $45.95. C(41) – C(40) = 3965.95 – 3920 = $45.95 (g) The average cost function for producing x microwave ovens is 2000 + 50 x − 0.05 x 2 2000 C ( x) = = + 50 − 0.05 x x x (h) The average cost of producing 41 microwave ovens is C ( 41) = $96.73 2000 + 50 ⋅ 41 − 0.05 ⋅ 412 C ( 41) = = 96.73 41 67.
The marginal price of beans in year t is the derivative of p(t). p′(t) = 0.021t 2 – 1.26t + 0.005 (a) The marginal price of beans in 1995 is p′(2), since t = 0 represents 1993, t = 2 represents 1995. p′(2) = 0.021 · 2 2 – 1.26 · 2 + 0.005 = – 2.431 (b) The marginal price of beans in 2002 is p′(9), since t = 0 represents 1993, t = 9 represents 2002. p′(9) = 0.021 · 9 2 – 1.26 · 9 + 0.005 = – 9.634 (c) Answers will vary.
69.
The instantaneous rate of change of the volume V of a sphere with respect to its radius r when r = 2 feet is V′(2). 4 V (r) = π r 3 3 V′(r) = 4 πr 2 V′(2) = 4 π · 2 2 = 16 π feet cubed
71.
The instantaneous rate of change of work output at time t is the derivative of A(t). A′(t) = 3a3 t 2 + 2a2 t + a1
73.
d n x = nx n−1 dx To prove this formula we begin with the difference quotient. d n f ( x + h) − f ( x) x = f ′ ( x ) = lim h→0 dx h n n ( x + h) − x = lim Use the difference quotient. h→0 h Formula (3) from Section 2 of the text is
⎡ n ⎤ n ( n − 1) n−2 2 n −1 x h + … + hn ⎥ − [ xn ] ⎢⎣ x + nx h + 2 ⎦ = lim h→0 h
Use the hint provided.
224 SECTION 4.3
n ( n − 1) n−2 2 x h + … + hn 2 = lim h→0 h ⎛ ⎞ n ( n − 1) n−2 h ⎜ nx n−1 + x h + … + h n−1 ⎟ 2 ⎠ = lim ⎝ h→0 h ⎛ ⎞ n ( n − 1) n−2 = lim ⎜ nx n−1 + x h + … + h n−1 ⎟ h→0 ⎝ 2 ⎠ = nx n – 1 nx n−1h +
Simplify.
Factor out an h. Cancel the h’s. Find the limit.
4.3 Product and Quotient Formulas 1.
The function f is the product of two functions g(x) = 2x + 1 and h(x) = 4x – 3 so that using the formula for the derivative of a product, we have ⎡d ⎤ ⎡d ⎤ f ′ ( x ) = ( 2 x + 1) ⎢ ( 4 x − 3) ⎥ + ( 4 x − 3) ⎢ ( 2 x + 1) ⎥ Derivative of a product formula. ⎣ dx ⎦ ⎣ dx ⎦ = (2x + 1)(4) + (4x – 3)(2) Differentiate. = 8x + 4 + 8x – 6 Simplify. = 16x – 2 Simplify.
3.
The function f is the product of two functions g(t) = t 2 + 1 and h(t) = t 2 – 4 so that using the formula for the derivative of a product, we have ⎡d ⎤ ⎡d ⎤ f ′ ( t ) = ( t 2 + 1) ⎢ ( t 2 − 4 ) ⎥ + ( t 2 − 4 ) ⎢ ( t 2 + 1) ⎥ Derivative of a product formula. ⎣ dt ⎦ ⎣ dt ⎦ = (t 2 + 1)(2t) + (t 2 – 4)(2t) Differentiate. = 2t 3 + 2t + 2t 3 – 8t Simplify. 3 = 4t – 6t Simplify.
5.
The function f is the product of two functions g(x) = 3x – 5 and h(x) = 2x 2 + 1 so that using the formula for the derivative of a product, we have ⎡d ⎤ ⎡d ⎤ f ′ ( x ) = ( 3 x − 5 ) ⎢ ( 2 x 2 + 1) ⎥ + ( 2 x 2 + 1) ⎢ ( 3 x − 5 ) ⎥ Derivative of a product formula. ⎣ dx ⎦ ⎣ dx ⎦ 2 = (3x – 5)(4x) + (2x + 1)(3) Differentiate. = 12x 2 – 20x + 6x 2 + 3 Simplify. 2 = 18x – 20x + 3 Simplify.
7.
The function f is the product of two functions g(x) = x 5 + 1 and h(x) = 3x 3 + 8 so that using the formula for the derivative of a product, we have ⎡d ⎤ ⎡d ⎤ f ′ ( x ) = ( x5 + 1) ⎢ ( 3 x 3 + 8 ) ⎥ + ( 3 x3 + 8 ) ⎢ ( x5 + 1) ⎥ Derivative of a product formula. ⎣ dx ⎦ ⎣ dx ⎦ 5 2 3 4 = (x + 1)(9x ) + (3x + 8)(5x ) Differentiate. = 9x 7 + 9x 2 + 15x 7 + 40x 4 Simplify. 7 4 2 = 24x + 40x + 9x Simplify.
224 SECTION 4.3
n ( n − 1) n−2 2 x h + … + hn 2 = lim h→0 h ⎛ ⎞ n ( n − 1) n−2 h ⎜ nx n−1 + x h + … + h n−1 ⎟ 2 ⎠ = lim ⎝ h→0 h ⎛ ⎞ n ( n − 1) n−2 = lim ⎜ nx n−1 + x h + … + h n−1 ⎟ h→0 ⎝ 2 ⎠ = nx n – 1 nx n−1h +
Simplify.
Factor out an h. Cancel the h’s. Find the limit.
4.3 Product and Quotient Formulas 1.
The function f is the product of two functions g(x) = 2x + 1 and h(x) = 4x – 3 so that using the formula for the derivative of a product, we have ⎡d ⎤ ⎡d ⎤ f ′ ( x ) = ( 2 x + 1) ⎢ ( 4 x − 3) ⎥ + ( 4 x − 3) ⎢ ( 2 x + 1) ⎥ Derivative of a product formula. ⎣ dx ⎦ ⎣ dx ⎦ = (2x + 1)(4) + (4x – 3)(2) Differentiate. = 8x + 4 + 8x – 6 Simplify. = 16x – 2 Simplify.
3.
The function f is the product of two functions g(t) = t 2 + 1 and h(t) = t 2 – 4 so that using the formula for the derivative of a product, we have ⎡d ⎤ ⎡d ⎤ f ′ ( t ) = ( t 2 + 1) ⎢ ( t 2 − 4 ) ⎥ + ( t 2 − 4 ) ⎢ ( t 2 + 1) ⎥ Derivative of a product formula. ⎣ dt ⎦ ⎣ dt ⎦ = (t 2 + 1)(2t) + (t 2 – 4)(2t) Differentiate. = 2t 3 + 2t + 2t 3 – 8t Simplify. 3 = 4t – 6t Simplify.
5.
The function f is the product of two functions g(x) = 3x – 5 and h(x) = 2x 2 + 1 so that using the formula for the derivative of a product, we have ⎡d ⎤ ⎡d ⎤ f ′ ( x ) = ( 3 x − 5 ) ⎢ ( 2 x 2 + 1) ⎥ + ( 2 x 2 + 1) ⎢ ( 3 x − 5 ) ⎥ Derivative of a product formula. ⎣ dx ⎦ ⎣ dx ⎦ 2 = (3x – 5)(4x) + (2x + 1)(3) Differentiate. = 12x 2 – 20x + 6x 2 + 3 Simplify. 2 = 18x – 20x + 3 Simplify.
7.
The function f is the product of two functions g(x) = x 5 + 1 and h(x) = 3x 3 + 8 so that using the formula for the derivative of a product, we have ⎡d ⎤ ⎡d ⎤ f ′ ( x ) = ( x5 + 1) ⎢ ( 3 x 3 + 8 ) ⎥ + ( 3 x3 + 8 ) ⎢ ( x5 + 1) ⎥ Derivative of a product formula. ⎣ dx ⎦ ⎣ dx ⎦ 5 2 3 4 = (x + 1)(9x ) + (3x + 8)(5x ) Differentiate. = 9x 7 + 9x 2 + 15x 7 + 40x 4 Simplify. 7 4 2 = 24x + 40x + 9x Simplify.
SECTION 4.3 225 9.
The function f is the quotient of two functions g(x) = x and h(x) = x + 1. We use the formula for the derivative of a quotient to get ( x + 1) d x − x d ( x + 1) d ⎛ x ⎞ dx dx Derivative of a quotient formula. ⎜ ⎟= 2 dx ⎝ x + 1 ⎠ ( x + 1) ( x + 1) (1) − x (1) = Differentiate. ( x + 1)2 x +1− x = Simplify. ( x + 1)2 1 = Simplify. ( x + 1)2
11.
The function f is the quotient of two functions g(x) = 3x + 4 and h(x) = 2x – 1. We use the formula for the derivative of a quotient to get ( 2 x − 1) d ( 3x + 4 ) − ( 3x + 4 ) d ( 2 x − 1) d ⎛ 3x + 4 ⎞ dx dx Derivative of a quotient formula. ⎜ ⎟= dx ⎝ 2 x − 1 ⎠ ( 2 x − 1)2 ( 2 x − 1)( 3) − ( 3x + 4 ) ( 2 ) = Differentiate. ( 2 x − 1)2 6x − 3 − 6x − 8 = Simplify. ( 2 x − 1)2 11 = − Simplify. ( 2 x − 1)2
13.
The function f is the quotient of two functions g(x) = x 2 and h(x) = x – 4. We use the formula for the derivative of a quotient to get ( x − 4) d x2 − x2 d ( x − 4) 2 d ⎛ x ⎞ dx dx Derivative of a quotient formula. ⎜ ⎟= dx ⎝ x − 4 ⎠ ( x − 4 )2 ( x − 4 )( 2 x ) − x 2 (1) = Differentiate. ( x − 4 )2 2x2 − 8x − x2 = Simplify. ( x − 4 )2 =
15.
x2 − 8x ( x − 4 )2
Simplify.
The function f is the quotient of two functions g(x) = 2x + 1 and h(x) = 3x 2 + 4. We use the formula for the derivative of a quotient to get ( 3x 2 + 4 ) d ( 2 x + 1) − ( 2 x + 1) d ( 3x 2 + 4 ) d ⎛ 2x +1 ⎞ dx dx Derivative of a quotient formula. ⎜ 2 ⎟= 2 2 dx ⎝ 3 x + 4 ⎠ ( 3x + 4 )
226 SECTION 4.3
= =
( 3x 2 + 4 ) ( 2 ) − ( 2 x + 1)( 6 x ) ( 3x 2 + 4 )2
Differentiate.
6 x 2 + 8 − 12 x 2 − 6 x
Simplify.
( 3 x 2 + 4 )2
= −
6x2 + 6x − 8
Simplify.
( 3 x 2 + 4 )2
17.
d −2 d ( −2 ) d 4 = −2t = −2 t −2 = −2 ( −2t −3 ) = 3 2 dx t dx dx t
19.
d ⎛ 1 1 ⎞ d ( d d d 1 2 1 + x −1 + x −2 ) = 1 + x −1 + x −2 = 0 − 1x −2 − 2 x −3 = − 2 − 3 ⎜1 + + 2 ⎟ = dx ⎝ x x ⎠ dx dx dx dx x x
21.
The slope of the tangent line to the function f at (1, 2) is the derivative of f at x = 1. d f ′ ( x ) = ( x 3 − 2 x + 2 ) ( x + 1) dx d d = ( x3 − 2 x + 2 ) ( x + 1) + ( x + 1) ( x3 − 2 x + 2 ) Derivative of a product. dx dx 3 2 = ( x − 2 x + 2 ) (1) + ( x + 1) ( 3x − 2 ) Differentiate. 3 3 2 = x − 2 x + 2 + 3x + 3x − 2 x − 2 Simplify. 3 2 = 4 x + 3x − 4 x Simplify. mtan = f ′(1) = 3
An equation of the tangent line is y – 2 = 3(x – 1) y – 2 = 3x – 3 y = 3x – 1 23.
y – y1 = m(x – x1) Simplify. Add 2 to both sides.
⎛ 1⎞ The slope of the tangent line to the function f at ⎜1, ⎟ is the derivative of f at x = 1. ⎝ 2⎠ ( x + 1) d x3 − x3 d ( x + 1) 3 d x dx dx Derivative of a quotient. = f ′( x ) = 2 dx x + 1 ( x + 1) ( x + 1) ( 3x 2 ) − x3 (1) = Differentiate. ( x + 1)2 3x3 + 3x 2 − x3 = Simplify. ( x + 1)2 2 x3 + 3x 2 = ( x + 1)2 mtan = f ′(1) =
2 (13 ) + 3 (12 ) 5 = 4 (1 + 1)2
Simplify.
SECTION 4.3 227
An equation of the tangent line is 1 5 y − = ( x − 1) 2 4 1 5 5 y− = x− 2 4 4 5 3 y = x− 4 4
y – y1 = m(x – x1) Simplify. Add
1 2
to both sides.
25.
We first find the derivative f ′(x). d ⎡( 2 d d 2 2 f ′(x) = ⎣ x − 2 ) ( 2 x − 1) ⎤⎦ = ( x − 2 ) ( 2 x − 1) + ( 2 x − 1) ( x − 2 ) dx dx dx 2 = ( x − 2 ) ( 2 ) + ( 2 x − 1)( 2 x ) = 2 x2 − 4 + 4 x2 − 2 x = 6x2 − 2 x − 4 We then solve the equation f ′(x) = 0. 6x 2 – 2x – 4 = 0 f ′(x) = 6x 2 – 2x – 4 3x 2 – 1x – 2 = 0 (3x +2)(x – 1) = 0 Factor. 2 x= Apply the Zero-Product Property. x=1 3
27.
We first find the derivative f ′(x). ( x + 1) d x 2 − x 2 d ( x + 1) d x2 dx dx f ′(x) = = 2 dx x + 1 ( x + 1) ( x + 1)( 2 x ) − x 2 (1) = ( x + 1)2 2 x2 + 2 x − x2 x2 + 2 x = ( x + 1)2 ( x + 1)2 We then solve the equation f ′(x) = 0. 2 x2 + 2 x x + 2x f ′(x) = = 0 ( x + 1) 2 ( x + 1)2 x 2 + 2x = 0 Multiply both sides by (x + 2)2; x ≠ – 2. x(x + 2) = 0 Factor. Apply the Zero-Product Property. x=0 x+2=0 Solve for x. x=–2 =
29.
y is the product of two functions, we will use the formula for the derivative of a product. d 2( [ x 3x − 2 )] = x 2 d ( 3x − 2 ) + ( 3x − 2 ) d x 2 y′ = dx dx dx 2 = x ( 3) + ( 3x − 2 )( 2 x ) Differentiate. 2 2 = 3x + 6 x − 4 x Simplify. 2 = 9x – 4x Simplify.
228 SECTION 4.3 31.
y is the product of two functions, we will use the formula for the derivative of a product. d ⎡( 2 d d 2 2 2 2 2 y′ = ⎣ x + 4 )( 4 x + 3) ⎦⎤ = ( x + 4 ) ( 4 x + 3) + ( 4 x + 3) ( x + 4 ) dx dx dx = ( x 2 + 4 ) ( 8 x ) + ( 4 x 2 + 3) ( 2 x ) Differentiate. = 8x 3 + 32x + 8x 3 + 6x Simplify. = 16x 3 + 38x Simplify.
33.
y is the quotient of two functions, we will use the formula for the derivative of a quotient. ( 3x + 5 ) d ( 2 x + 3) − ( 2 x + 3) d ( 3x + 5 ) d 2x + 3 dx dx y′ = = 2 dx 3x + 5 ( 3x + 5) ( 3 x + 5 ) ( 2 ) − ( 2 x + 3)( 3) Differentiate. = ( 3x + 5 )2 6 x + 10 − 6 x − 9 = Simplify. ( 3x + 5)2 1 = Simplify. ( 3x + 5)2
35.
y is the quotient of two functions, so we will use the formula for the derivative of a quotient. ( x2 − 4) d x2 − x2 d ( x2 − 4) 2 d x dx dx = y′ = 2 2 dx x − 4 ( x2 − 4) = =
( x2 − 4) ( 2x ) − x2 ( 2x ) ( x 2 − 4 )2 2 x3 − 8 x − 2 x3
=− 37.
( x 2 − 4 )2 8x
( x 2 − 4 )2
Differentiate.
Simplify. Simplify.
y is the quotient of two functions, but its numerator is the product of two more functions. So we will use the formula for the derivative of a quotient and when differentiating the numerator the formula for the derivative of a product. ( 2 x + 1) d [( 3x + 4 )( 2 x − 3)] − [( 3x + 4 )( 2 x − 3)] d ( 2 x + 1) d ( 3 x + 4 )( 2 x − 3) dx dx y′ = = 2 2x +1 dx ( 2 x + 1) ( 2 x + 1) ⎡( 3x + 4 ) d ( 2 x − 3) + ( 2 x − 3) d ( 3x + 4 ) ⎤ − ( 3x + 4 )( 2 x − 3) d ( 2 x + 1) ⎢⎣ ⎥⎦ dx dx dx = 2 ( 2 x + 1)
SECTION 4.3 229
=
( 2 x + 1) [( 3x + 4 ) ( 2 ) + ( 2 x − 3)( 3)] − ( 3x + 4 )( 2 x − 3) ( 2 )
( 2 x + 1)2 ( 2 x + 1) [ 6 x + 8 + 6 x − 9] − ( 6 x 2 + 8 x − 9 x − 12 ) ( 2 ) = ( 2 x + 1)2 ( 2 x + 1)(12 x − 1) − ( 6 x 2 − x − 12 ) ( 2 ) = ( 2 x + 1)2 24 x 2 − 2 x + 12 x − 1 − 12 x 2 + 2 x + 24 = ( 2 x + 1)2 12 x 2 + 12 x + 23 = ( 2 x + 1)2
Differentiate.
Simplify.
Add like terms. Multiply. Simplify.
39.
y is the quotient of two functions, so we will use the formula for the derivative of a quotient. ( x 2 + 4 ) d ( 4 x3 ) − 4 x3 d ( x 2 + 4 ) 3 d 4x dx dx y′ = = 2 2 2 dx x + 4 ( x + 4) ( x 2 + 4 )(12 x 2 ) − 4 x3 ( 2 x ) = Differentiate. 2 ( x2 + 4) 12 x 4 + 48 x 2 − 8 x 4 Simplify. = 2 ( x2 + 4) 4 x 4 + 48 x 2 = Simplify. 2 ( x2 + 4)
41.
(a) The average change in value from t = 2 to t = 5 is ⎡10,000 ⎤ ⎡10,000 ⎤ + 6000 ⎥ − ⎢ + 6000 ⎥ ∆V V ( 5 ) − V ( 2 ) ⎢⎣ 5 ⎦ ⎦ ⎣ 2 = = ∆t 5−2 3 2000 − 5000 −3000 == = = −1000 3 3 (b) The instantaneous rate of change in value is the derivative of function V. 10,000 + 6000 = 10,000t – 1 + 6000 V(t) = t 10,000 V′(t) = – 10,000t – 2 = − t2 (c) The instantaneous rate of change after 2 years is V′(2). 10,000 10,000 =− = −2500 V′(2) = − 2 2 4
230 SECTION 4.3
(d) The instantaneous rate of change after 5 years is V′(5). 10,000 10,000 V′(5) = − =− = −400 2 5 25 (e) Answers will vary. 43.
(a) The revenue function R is the product of the unit price and the number of units sold. 40 ⎞ ⎛ R = R(x) = px = ⎜10 + ⎟ x = 10 x + 40 x ⎠ ⎝ (b) The marginal revenue is the derivative of the revenue function R. R′(x) = 10 (c) The marginal revenue when x = 4, is R′(4) = 10. (d) The marginal revenue when x = 6, is R′(6) = 10
d 100,000 = 2 dp p + 10 p + 50
45. (a) D′ ( p ) =
= =
(b) D′(5) =
dp
D′(15) =
(5
2
dp
( p 2 + 10 p + 50 )
2
( p 2 + 10 p + 50 ) ( 0 ) − 100,000 ( 2 p + 10 ) 2 ( p 2 + 10 p + 50 )
d dp
100, 000 = 0
−200,000 p − 1,000,000
( p 2 + 10 p + 50 )
−200,000(5) − 1,000,000
D′(10) =
47.
( p 2 + 10 p + 50 ) d (100,000 ) − 100,000 d ( p 2 + 10 p + 50 )
+ 10(5) + 50 )
2
= – 128
−200,000(10) − 1,000,000
(102 + 10(10) + 50 )
2
−200, 000(15) − 1, 000, 000
(152 + 10(15) + 50 )
2
2
= – 48
= – 22.145
The rate at which the population is growing is given by the derivative of the population function P. d
P′(t) =
d 4t ⎞ d 4t ⎤ ⎛ ⎡d 1000 ⎜1 + = 1000 ⎢ 1 + 2 ⎟ dt dt 100 + t 2 ⎦⎥ ⎝ 100 + t ⎠ ⎣ dt
d( ⎡( 2 d 2 ⎤ ) ( ) ) ⎢ 100 + t dt 4t − 4t dt 100 + t ⎥ = 1000 ⎢ ⎥ 2 (100 + t 2 ) ⎢⎣ ⎥⎦ ⎡ (100 + t 2 ) ( 4 ) − 4t ( 2t ) ⎤ = 1000 ⎢ ⎥ 2 (100 + t 2 ) ⎢⎣ ⎥⎦
( cf ( x ) ) = c
d
f ( x ); dx dx The derivative of a sum is the sum of the derivatives
Derivative of a quotient formula.
Differentiate.
SECTION 4.3 231
⎡ 400 + 4t 2 − 8t 2 ⎤ 4000 (100 − t 2 ) = 1000 ⎢ ⎥= 2 2 ( ) (100 + t 2 )2 + 100 t ⎣ ⎦
Simplify.
In Parts (a) – (d), we evaluate the derivative P′(t) at the indicated time. 4000 (100 − 12 ) = 38.820 (a) t = 1 hour, P′(1) = 2 (100 + 12 ) (b) t = 2 hours, P′(2) =
(c) t = 3 hours, P′(3) =
(d) t = 4 hours, P′(4) = 49.
4000 (100 − 22 )
(100 + 22 )
2
4000 (100 − 32 )
(100 + 32 )
2
4000 (100 − 42 )
(100 + 42 )
2
= 35.503
= 30.637
= 24.970
First we must find the function I that describes the intensity of light with respect to the distance r of the object from the source of the light. Since I is inversely related to the square of distance, and we are told that I = 1000 units when r = 1 meter, we solve for the constant of proportionality k. I (1) =
k = 1000 12 k = 1000
I (r) =
k 2
r Solve for k.
So, we have 1000 r2 The rate of change of intensity with respect to distance r is the derivative I ′(r) of the function I. dI d 1000 d ( d −2 2000 I ′(r) = = = 1000r −2 ) = 1000 r = 1000(−2r −3 ) = − 3 2 dr dr r dr dr r When r = 10 meters the rate of change of the intensity of the light is 2000 2000 I ′(10) = − 3 = − = −2 units per meter. 10 1000 I = I (r) =
51.
(a) The marginal cost is the derivative of the cost function C. d ⎛ x 36,000 ⎞ d ⎛ x −1 ⎞ C′(x) = ⎜ 100 + + ⎟ = ⎜100 + + 36,000 x ⎟ 10 10 dx ⎝ x ⎠ dx ⎝ ⎠ =
d d −1 1 d 100 + ⋅ x + 36,000 x 10 dx dx dx
The derivative of a sum is the sum of the derivatives; d d ( cf ( x ) ) = c f ( x ) dx dx
232 SECTION 4.4
1 + 36,000 ( −1x −2 ) 10 1 36,000 = − x2 10
= 0+
Differentiate.
In Parts (b) – (d), the derivative C′(x) is evaluated at the indicated ground speeds. 1 36,000 (b) ground speed x = 500 mph C′(500) = − = – 0.044 10 5002 1 36,000 − = – 0.019 (c) ground speed x = 550 mph C′(550) = 10 5502 1 36,000 = – 0.078 (d) ground speed x = 450 mph C′(450) = − 10 4502 53.
(a) First we find the derivative of the function S with respect to reward r. d d ( g − r ) ( ar ) − ar ( g − r ) d ar dr dr = Use the derivative of a quotient formula. S′(r) = 2 dr g − r (g − r) ( g − r ) ( a ) − ar ( −1) Differentiate. = 2 (g − r) ag − ar + ar ag = = Simplify. 2 2 (g − r) (g − r) Since both a and g are constants for a given individual k = ag, and S′(r) is of the form, k S′(r) = which is inversely proportional to the square of the difference between 2 (g − r) the personal goal of the individual and the amount of reward received. (b) Answers may vary.
4.4 The Power Rule 1.
f ′(x) =
d ( ) d ( 4 3 d ( 2 x − 3 ) = 4 ( 2 x − 3 )3 ( 2 ) = 8 ( 2 x − 3 ) 3 f x = 2 x − 3) = 4 ( 2 x − 3) dx dx dx ↑ ↑ ↑ Use the Power Rule.
3.
f ′(x) =
f ′(x) =
Simplify.
3 2 d 2 2 d ( ) d ( 2 ( x 2 + 4 ) = 3( x 2 + 4 ) ( 2 x ) = 6 x ( x2 + 4 ) f x = x + 4) = 3( x2 + 4) dx dx dx ↑ ↑ ↑
Use the Power Rule.
5.
Differentiate.
Differentiate.
Simplify.
2 d d ( 2 d f ( x) = 3x + 4 ) = 2 ( 3x 2 + 4 ) ( 3x 2 + 4 ) = 2 ( 3x 2 + 4 ) ( 6 x ) dx dx dx ↑ ↑
Use the Power Rule.
Differentiate.
232 SECTION 4.4
1 + 36,000 ( −1x −2 ) 10 1 36,000 = − x2 10
= 0+
Differentiate.
In Parts (b) – (d), the derivative C′(x) is evaluated at the indicated ground speeds. 1 36,000 (b) ground speed x = 500 mph C′(500) = − = – 0.044 10 5002 1 36,000 − = – 0.019 (c) ground speed x = 550 mph C′(550) = 10 5502 1 36,000 = – 0.078 (d) ground speed x = 450 mph C′(450) = − 10 4502 53.
(a) First we find the derivative of the function S with respect to reward r. d d ( g − r ) ( ar ) − ar ( g − r ) d ar dr dr = Use the derivative of a quotient formula. S′(r) = 2 dr g − r (g − r) ( g − r ) ( a ) − ar ( −1) Differentiate. = 2 (g − r) ag − ar + ar ag = = Simplify. 2 2 (g − r) (g − r) Since both a and g are constants for a given individual k = ag, and S′(r) is of the form, k S′(r) = which is inversely proportional to the square of the difference between 2 (g − r) the personal goal of the individual and the amount of reward received. (b) Answers may vary.
4.4 The Power Rule 1.
f ′(x) =
d ( ) d ( 4 3 d ( 2 x − 3 ) = 4 ( 2 x − 3 )3 ( 2 ) = 8 ( 2 x − 3 ) 3 f x = 2 x − 3) = 4 ( 2 x − 3) dx dx dx ↑ ↑ ↑ Use the Power Rule.
3.
f ′(x) =
f ′(x) =
Simplify.
3 2 d 2 2 d ( ) d ( 2 ( x 2 + 4 ) = 3( x 2 + 4 ) ( 2 x ) = 6 x ( x2 + 4 ) f x = x + 4) = 3( x2 + 4) dx dx dx ↑ ↑ ↑
Use the Power Rule.
5.
Differentiate.
Differentiate.
Simplify.
2 d d ( 2 d f ( x) = 3x + 4 ) = 2 ( 3x 2 + 4 ) ( 3x 2 + 4 ) = 2 ( 3x 2 + 4 ) ( 6 x ) dx dx dx ↑ ↑
Use the Power Rule.
Differentiate.
SECTION 4.4 233
= 12 x ( 3 x 2 + 4 ) = 36 x 3 + 48 x ↑ ↑ Simplify.
7.
Simplify.
The function f is the product of x and (x + 1)3. We begin by using the formula for the derivative of a product. That is, d d 3 3 d f ′(x) = f ( x ) = x ( x + 1) + ( x + 1) x dx dx dx We continue by using the Power Rule: 2 d ⎡ ( x + 1) ⎤ + ( x + 1)3 d x f ′(x) = x ⎢3 ( x + 1) ⎥⎦ dx dx ⎣ 2 3 = x ⎡⎣3 ( x + 1) ⋅ 1⎤⎦ + ( x + 1) ⋅ 1 Differentiate. 2 3 = 3x ( x + 1) + ( x + 1) 2 = ( x + 1) [3 x + ( x + 1)] = (x + 1) 2 (4x + 1)
9.
Simplify. Factor. Simplify.
The function f is the product of 4x 2 and (2x + 1) 4. We begin by using the formula for the derivative of a product. That is, d 4 4 d ( 4 x 2 ) = 4 x 2 d ( 2 x + 1)4 + ( 2 x + 1)4 ⋅ 4 ⋅ d x 2 f ′(x) = 4 x 2 ( 2 x + 1) + ( 2 x + 1) dx dx dx dx We continue by using the Power Rule: 3 d ⎡ ( 2 x + 1) ⎤ + ( 2 x + 1)4 ⋅ 4 ⋅ d x 2 f ′(x) = 4 x 2 ⎢ 4 ( 2 x + 1) ⎥⎦ dx dx ⎣ 3 4 = 4 x 2 ⎡⎣ 4 ( 2 x + 1) ( 2 ) ⎤⎦ + ( 2 x + 1) ⋅ 4 ⋅ 2 x Differentiate. 3 4 = 32 x 2 ( 2 x + 1) + 8 x ( 2 x + 1) 3 = 8 x ( 2 x + 1) [ 4 x + ( 2 x + 1)]
Simplify. Factor.
= 8 x ( 2 x + 1) ( 6 x + 1) 3
Simplify.
11. Before differentiating the function f , we simplify it, then we use the Power Rule. f (x) = [x(x – 1)] 3 = (x 2 – x)3 3 2 d d ( 2 ( x2 − x ) x − x ) = 3( x2 − x ) Use the Power Rule. f ′(x) = dx dx
= 3 ( x 2 − x ) ( 2 x − 1) 2 = 3x 2 ( x − 1) ( 2 x − 1) 2
13. f ′(x) =
d ( 3x − 1)−2 = −2 ( 3x − 1)−3 d ( 3 x − 1) dx dx −3 = −2 ( 3x − 1) ( 3) 6 =− ( 3x − 1)3
Differentiate. Factor.
Use the Power Rule. Differentiate. Simplify.
234 SECTION 4.4 15. We rewrite f (x) as f (x) = 4(x 2 + 4) – 1. Then we use the Power Rule. −1 −1 −2 d d ⎡ ( 2 d 2 2 ( x2 + 4) f ′(x) = ⎣ 4 x + 4 ) ⎤⎦ = 4 ( x + 4 ) = 4 ⋅ ( −1) ( x + 4 ) dx dx dx ↑ ↑ f ′(cx) = cf ′(x)
Use the Power Rule.
= −4 ( x 2 + 4 ) ( 2 x ) = −
8x
−2
↑
2
↑
Differentiate.
17.
( x2 + 4)
Simplify.
We rewrite f (x) as f (x) = – 4(x 2 – 9) – 3. Then we use the Power Rule. −3 −3 −4 d d ⎡ ( 2 d 2 2 ⎤ ( x2 − 9) f ′(x) = ⎣ − 4 x − 9 ) ⎦ = − 4 ⋅ ( x − 9 ) = − 4 ⋅ ( −3 ) ( x − 9 ) dx dx dx ↑ ↑ f ′(cx) = cf ′(x)
Use the Power Rule.
= 12 ( x 2 − 9 ) ( 2 x ) = −4
24 x
( x2 − 9)
↑
↑
Differentiate.
Simplify.
4
19. In this problem the function f is a quotient raised to the power 3. We begin with the Power Rule and then use the formula for the derivative of a quotient. 3
2
d ⎛ x ⎞ ⎛ x ⎞ ⎡ d ⎛ x ⎞⎤ f ′(x) = ⎜ ⎟ = 3⎜ ⎟ ⎜ ⎟ dx ⎝ x + 1 ⎠ ⎝ x + 1 ⎠ ⎢⎣ dx ⎝ x + 1 ⎠ ⎥⎦ d ⎡ ( x ) − x dy ( x + 1) ⎤ 2 ( x + 1) ⎢ ⎥ ⎛ x ⎞ dx dx = 3⎜ ⎟ ⎢ ⎥ ⎝ x +1⎠ ⎣ ( x + 1)2 ⎦ 2 ⎛ x ⎞ ⎡ ( x + 1) (1) − x (1) ⎤ = 3⎜ ⎟ ⎥ ⎝ x + 1 ⎠ ⎣⎢ ( x + 1)2 ⎦
The Power Rule.
The derivative of a quotient.
Differentiate.
2
3x 2 ⎛ x ⎞ ⎡ 1 ⎤ = = 3⎜ ⎟ ⎝ x + 1 ⎠ ⎣⎢ ( x + 1)2 ⎦⎥ ( x + 1)4
Simplify.
21. Here the function f is the quotient of two functions, the numerator of which is raised to a power. We will first use the formula for the derivative of a quotient and then use the power rule when differentiating the numerator. d 4 4 d ( 3x 2 ) 4 3 x 2 ( 2 x + 1) − ( 2 x + 1) d ⎡ ( 2 x + 1) ⎤ dx dx f ′(x) = ⎢ ⎥= 2 2 2 dx ⎣ 3 x ⎦ ( 3x ) 3 d ⎡ ( 2 x + 1) ⎤ − ( 2 x + 1)4 d ( 3x 2 ) 3x 2 ⎢ 4 ( 2 x + 1) ⎥⎦ dx dx ⎣ = The Power Rule.
( 3x 2 )2
SECTION 4.4 235
=
3 4 3 x 2 ⎡⎣ 4 ( 2 x + 1) ( 2 ) ⎤⎦ − ( 2 x + 1) ( 6 x )
( 3x 2 )
24 x 2 ( 2 x + 1) − 6 x ( 2 x + 1) 9x4 3 6 x ( 2 x + 1) [ 4 x − ( 2 x + 1)] 3
= =
=
Differentiate.
2 4
Simplify. Factor.
9x4 2 3 6 x ( 2 x + 1) ( 2 x − 1)
Simplify.
3
9 x43
2 ( 2 x + 1) ( 2 x − 1) = 3 x3 3
23.
Cancel.
Here the function f is the quotient of two functions, the numerator of which is raised to a power. We will first use the formula for the derivative of a quotient and then use the power rule when differentiating the numerator. 3 3 d d 3 x ( x 2 + 1) − ( x 2 + 1) x d ⎡ ( x 2 + 1) ⎤ dx ⎢ ⎥ = dx f ′(x) = dx ⎣ x x2 ⎦ 2 d 3 ⎡ ( x 2 + 1) ⎤⎥ − ( x 2 + 1) d x x ⎢3 ( x 2 + 1) dx dx ⎦ = ⎣ 2 x 2 3 2 x ⎡⎣3 ( x + 1) ( 2 x ) ⎤⎦ − ( x 2 + 1) (1) = x2
The Power Rule.
Differentiate.
6 x 2 ( x 2 + 1) − ( x 2 + 1) = x2 2 ( x 2 + 1) ⎡⎣6 x 2 − ( x 2 + 1)⎤⎦ = x2 2
3
Simplify. Factor.
( x 2 + 1) ( 5 x 2 − 1) 2
= 25.
Simplify.
x2
We rewrite f as f (x) = (x + x – 1) 3 and use the Power Rule. 3 2 d 2 d ( ) d ( ( x + x − 1 ) = 3 ( x + x − 1 ) (1 − x − 2 ) f ′(x) = f x = x + x − 1 ) = 3( x + x − 1 ) dx dx dx ↑ ↑ Use the Power Rule.
Differentiate.
1⎞ ⎛ 1 ⎞ ⎛ x 2 + 1 ⎞ ⎛ x 2 − 1 ⎞ 3 ( x 2 + 1) ( x 2 − 1) ⎛ = 3⎜ x + ⎟ ⎜1 − 2 ⎟ = 3⎜ ⎟ ⎜ ⎟= x ⎠ ⎝ x ⎠ ⎝ x ⎠ ⎝ x2 ⎠ x4 ⎝ ↑ ↑ ↑ Rewrite with positive exponents.
2
2
2
Write with a single denominator.
Simplify.
236 SECTION 4.4 27. Here the function f is the quotient of two functions, the denominator of which is raised to a power. We will first use the formula for the derivative of a quotient and then use the Power Rule when differentiating the denominator. 2
f ′(x) =
d 3x = dx ( x 2 + 1)2
2 2 ( x 2 + 1) d ( 3x 2 ) − 3x 2 d ( x 2 + 1)
dx
⎡⎣( x 2 + 1) ⎤⎦ 2
2
dx
The derivative of a quotient.
2 ( x 2 + 1) d ( 3 x 2 ) − 3 x 2 ⋅ 2 ( x 2 + 1) d ( x 2 + 1)
=
dx
dx
( x + 1) 2 ( x 2 + 1) ( 6 x ) − 3 x 2 ⋅ 2 ( x 2 + 1) ( 2 x ) = 4 ( x 2 + 1) 2 ( x 2 + 1) ( 6 x ) − 12 x3 ( x 2 + 1) = 4 ( x 2 + 1) 6 x ( x 2 + 1) ⎡⎣( x 2 + 1) − 2 x 2 ⎤⎦ =
( x 2 + 1) 6 x (1 − x 2 ) = 3 ( x 2 + 1) 29.
2
4
43
Use the Power Rule.
Differentiate.
Simplify.
Factor.
Simplify.
The rate at which the car is depreciating is the derivative V′(t). (1 + 0.4t + 0.1t 2 ) d 29,000 − 29,000 d (1 + 0.4t + 0.1t 2 ) 29,000 d dt dt V′(t) = = 2 2 dt 1 + 0.4t + 0.1t 2 (1 + 0.4t + 0.1t ) −29,000 ( 0.4 + 0.2t ) = 2 (1 + 0.4t + 0.1t 2 )
(a) The rate of depreciation 1 year after purchase is V′(1) −29,000 ( 0.4 + 0.2 ) −29,000 ( 0.6 ) = = − 7733.33 V′(1) = 1.5 2 (1 + 0.4 + 0.1)2 The car is depreciating at a rate of $7733.33 per year when it is one year old. (b) The rate of depreciation 2 years after purchase is V′(2) −29,000 ( 0.4 + 0.2 ⋅ 2 ) −29,000 ( 0.8 ) = = − 4793.39 V′(2) = 2 2.22 (1 + 0.4 ⋅ 2 + 0.1 ⋅ 22 ) The car is depreciating at a rate of $4793.39 per year when it is two years old. (c) The rate of depreciation 3 years after purchase is V′(3)
SECTION 4.4 237
V′(3) =
−29,000 ( 0.4 + 0.2 ⋅ 3)
(1 + 0.4 ⋅ 3 + 0.1 ⋅ 3 )
2 2
=
−29,000 (1.0 ) = − 3017.69 3.12
The car is depreciating at a rate of $3017.69 per year when it is three years old. (d) The rate of depreciation 4 years after purchase is V′(4) −29,000 ( 0.4 + 0.2 ⋅ 4 ) −29,000 (1.2 ) = = − 1972.79 V′(4) = 2 4.22 (1 + 0.4 ⋅ 4 + 0.1 ⋅ 42 ) The car is depreciating at a rate of $1972.79 per year when it is four years old. 31.
(a) The rate of change is given by the derivative. dp d ⎛ 10,000 d −1 ⎞ d 10,000 d = ⎜ − 5⎟ = − 5 = ⎡⎣10,000 ( 5 x + 100 ) ⎤⎦ dx dx ⎝ 5 x + 100 ⎠ dx 5 x + 100 dx dx ↑
↑
Derivative of a difference.
= 10,000 ⋅
d dx
5=0
d ( 50,000 2000 −1 −2 =− 5 x + 100 ) = 10,000 ⋅ ( −1) ( 5 x + 100 ) ( 5 ) = − 2 2 dx ( 5 x + 100 ) ( x + 20 )
↑ f ′(cx) = cf ′(x)
↑ Differentiate.
↑ Simplify.
(b) The revenue function R = px. 10,000 x ⎛ 10,000 ⎞ − 5⎟ x = − 5x R = R(x) = ⎜ 5 x + 100 ⎝ 5 x + 100 ⎠ (c) The marginal revenue is the derivative R′(x). d ⎛ 10,000 x ⎞ d 10,000 x d ( ) − 5x ⎟ = − 5x R′(x) = ⎜ dx ⎝ 5 x + 100 ⎠ dx 5 x + 100 dx ( 5 x + 100 ) (10,000 ) − 10,000 x ( 5 ) 1,000,000 40,000 = −5 −5 = −5 = 2 2 2 ( 5 x + 100 ) ( 5 x + 100 ) ( x + 20 ) (d) R′(10) =
R′(40) =
33.
1,000,000 2
−5 =
1,000,000 − 5 = 39.44 dollars 1502
2
−5 =
1,000,000 − 5 = 6.11 dollars 3002
( 5 ⋅ 10 + 100 ) 1,000,000
( 5 ⋅ 40 + 100 )
(a) The average rate of change in the mass of the protein is 28 28 − 7 ∆M M ( 2 ) − M ( 0 ) 2 + 2 0 + 2 7 − 14 = = = =− 2−0 2 2 2 ∆t grams per hour.
238 SECTION 4.5
(b) M ′ ( t ) =
d 28 d 28 −1 −2 = 28 ( t + 2 ) = 28 ⋅ ( −1)( t + 2 ) ⋅ 1 = − 2 dt t + 2 dt (t + 2)
M ′( 0) = −
28
( 0 + 2)
2
=−
28 = −7 4
4.5 The Derivatives of the Exponential and Logarithmic Functions; The Chain Rule d ( 3 x) d 3 d x x −e = x − e = 3x 2 − e x dx dx dx
1.
f ′(x) =
3.
Using the formula for the derivative of a product, d ( 2 x) d d 2 x e = x2 ex + ex x = x 2 e x + 2 xe x = x ( x + 2 ) e x f ′(x) = dx dx dx
5.
Using the formula for the derivative of a quotient, d d 2 x2 ex − ex x x d e x2ex − ex ⋅ 2 x x ( x − 2) ex ( x − 2) ex dx dx f ′(x) = = = = = 2 dx x 2 x4 x4 x3 ( x2 ) ↑ Differentiate.
7.
↑ Simplify.
Using the formula for the derivative of a quotient, d d ex ( 4 x2 ) − 4 x2 ex 2 x x x d 4 x2 dx dx = e ⋅ 8 x − 4 x e = 4 x ( 2 − x ) e = 4 x ( 2 − x ) f ′(x) = = 2 dx e x e2 x e2 x ex (ex ) ↑ Differentiate.
= 9.
↑ Factor.
↑ Factor.
↑ Simplify.
8x − 4 x 2 ex
y = f (u) = u 5 = (x 3 + 1) 5 = f (x) 4 d 4 4 dy d ( 3 )5 ( x3 + 1) = 5 ( x3 + 1) ( 3x 2 ) = 15 x 2 ( x3 + 1) = x + 1 = 5 ( x 3 + 1) dx dx dx ↑ Chain Rule.
11. y = f (u) =
↑ Differentiate.
u x2 + 1 x2 + 1 = 2 = 2 = f (x) u + 1 ( x + 1) + 1 x + 2
↑ Simplify.
238 SECTION 4.5
(b) M ′ ( t ) =
d 28 d 28 −1 −2 = 28 ( t + 2 ) = 28 ⋅ ( −1)( t + 2 ) ⋅ 1 = − 2 dt t + 2 dt (t + 2)
M ′( 0) = −
28
( 0 + 2)
2
=−
28 = −7 4
4.5 The Derivatives of the Exponential and Logarithmic Functions; The Chain Rule d ( 3 x) d 3 d x x −e = x − e = 3x 2 − e x dx dx dx
1.
f ′(x) =
3.
Using the formula for the derivative of a product, d ( 2 x) d d 2 x e = x2 ex + ex x = x 2 e x + 2 xe x = x ( x + 2 ) e x f ′(x) = dx dx dx
5.
Using the formula for the derivative of a quotient, d d 2 x2 ex − ex x x d e x2ex − ex ⋅ 2 x x ( x − 2) ex ( x − 2) ex dx dx f ′(x) = = = = = 2 dx x 2 x4 x4 x3 ( x2 ) ↑ Differentiate.
7.
↑ Simplify.
Using the formula for the derivative of a quotient, d d ex ( 4 x2 ) − 4 x2 ex 2 x x x d 4 x2 dx dx = e ⋅ 8 x − 4 x e = 4 x ( 2 − x ) e = 4 x ( 2 − x ) f ′(x) = = 2 dx e x e2 x e2 x ex (ex ) ↑ Differentiate.
= 9.
↑ Factor.
↑ Factor.
↑ Simplify.
8x − 4 x 2 ex
y = f (u) = u 5 = (x 3 + 1) 5 = f (x) 4 d 4 4 dy d ( 3 )5 ( x3 + 1) = 5 ( x3 + 1) ( 3x 2 ) = 15 x 2 ( x3 + 1) = x + 1 = 5 ( x 3 + 1) dx dx dx ↑ Chain Rule.
11. y = f (u) =
↑ Differentiate.
u x2 + 1 x2 + 1 = 2 = 2 = f (x) u + 1 ( x + 1) + 1 x + 2
↑ Simplify.
SECTION 4.5 239
dy d x + 1 = = dx dx x 2 + 2 2
( x 2 + 2 ) d ( x 2 + 1) − ( x 2 + 1) d ( x 2 + 2 ) dx
( x2 + 2)
dx
2
↑ Derivative of a quotient
=
( x 2 + 2 ) ( 2 x ) − ( x 2 + 1) ( 2 x ) 2 x ⎡⎣( x 2 + 2 ) − ( x 2 + 1)⎤⎦ 2x = = 2 2 ( x2 + 2) ( x2 + 2) ( x 2 + 2 )2
↑ Differentiate.
↑ Factor.
↑ Simplify.
2
13.
2 2 ⎛1 ⎞ y = f (u) = ( u + 1) = ⎜ + 1⎟ = ( x −1 + 1) = f (x) ⎝x ⎠
dy d ( −1 ) d x + 1 = 2 ( x −1 + 1) ( x −1 + 1) = 2 ( x −1 + 1) ⋅ ( −1) x − 2 = dx dx dx 2
↑ Use the Chain Rule.
↑ Differentiate.
⎛1 ⎞ −2 ⎜ + 1⎟ 2 (1 + x ) = ⎝ x2 ⎠ = − x x3
↑ Write with positive exponents.
↑ Simplify.
5
5 3 5 15. y = f (u) = ( u 3 − 1) = ⎡⎣( x − 2 ) − 1⎤⎦ = ( x − 6 − 1) = f (x) 4 d dy d ( − 6 )5 ( x − 6 − 1) = 5 ( x − 6 − 1)4 ⋅ ( − 6 ) x − 7 = x − 1 = 5 ( x − 6 − 1) dx dx dx
↑ Use the Chain Rule.
↑ Differentiate. 4
4
⎛ 1 − x6 ⎞ ⎛ 1 ⎞ 4 4 4 − 30 1 30 ⎜ 6 ⎟ ⎜ 6 ⎟ 30 (1 − x 6 ) 30 (1 − x 6 ) −30 ( x − 6 − 1) x x ⎝ ⎠ ⎝ ⎠ = =− =− =− =− 4 x7 x7 x7 x 31 ( x6 ) x7 ↑ Simplify.
↑ Write with positive exponent.
↑ Write the numerator as a single quotient.
3 3x x 17. y = f (u) = u = ( e ) = e = f (x) dy d 3 x d = e = e3 x ⋅ ( 3 x ) = e3 x ⋅ 3 = 3e3 x dx dx dx 3
↑ ↑ Chain Rule Differentiate.
↑ Simplify.
3
u x 19. y = f (u) = e = e = f (x) 3 3 3 dy d x3 d 3 = e = ex ⋅ x = e x ⋅ 3x 2 = 3x 2 e x dx dx dx
↑ Chain Rule
↑ Differentiate.
↑ Simplify.
240 SECTION 4.5 21. (a) Using the Chain Rule, y = (x 3 + 1)2 is thought of as y = u 2 and u = x 3 + 1. dy du = 2u and = 3x 2 du dx dy dy du = ⋅ = 2u ⋅ 3x 2 = 2 ( x 3 + 1) ⋅ 3x 2 = 6 x 2 ( x 3 + 1) = 6 x 5 + 6 x 2 dx du dx ↑ Substitute u = x 3 + 1
↑ Simplify.
(b) Using the Power Rule, dy d ( 3 )2 d ( 3 ) = x + 1 = 2 ( x 3 + 1) ⋅ x + 1 = 2 ( x 3 + 1) ⋅ ( 3 x 2 ) = 6 x 2 ( x 3 + 1) = 6 x5 + 6 x 2 dx dx dx ↑ Power Rule
↑ Differentiate.
↑ Simplify.
(c) Expanding, y = (x 3 + 1)2 = x 6 + 2x3 + 1, and dy d ( 6 d 6 d d = x + 2 x 3 + 1) = x + 2 x 3 + 1 = 6 x 5 + 6 x 2 = 6 x 2 ( x3 + 1) dx dx dx dx dx ↑ Differentiate.
↑ Factor.
23. f ′(x) =
d 5x d e = e5 x ⋅ ( 5 x ) = e5 x ⋅ 5 = 5e5 x dx dx
25. f ′(x) =
2 d 2 2 d − x2 d − x2 8e = 8 ⋅ − x 2 ) = 8e − x ⋅ ( −2 x ) = − 16 xe − x e = 8e− x ( dx dx dx
27. f ′(x) =
2 d 2 d 2 x2 d 2 ⎛ 2 d 2⎞ x e = x2 e x + e x x2 = x2 ⎜ e x x ⎟ + e x ⋅ 2x dx dx dx dx ⎠ ⎝
↑ The derivative of a product
↑ The Chain Rule
= x 2 e x ⋅ 2 x + e x ⋅ 2 x = e x ( 2 x 3 + 2 x ) = 2 x e x ( x 2 + 1) 2
↑ Differentiate.
29. f ′(x) =
2
2
2
↑ Factor.
2 d ⎡ ( x )3 ⎤ d x 3 x 2 d x e = 15 ( e x ) e x = 15 e3 x ⎣5 e ⎦ = 5 ( e ) = 5 ⋅ 3 ( e ) ⋅ dx dx dx
↑ Use the Power Rule.
31. f (x) =
f ′(x) =
↑ Differentiate.
↑ Simplify.
x2 2 −x x = x e e d 2 −x ( x e ) = x2 d e− x + e− x d x 2 = x 2 ⎛⎜ e− x ⋅ d ( − x ) ⎞⎟ + e− x ⋅ 2 x dx dx dx dx ⎝ ⎠ ↑ Derivative of a product
↑ Chain Rule
SECTION 4.5 241
2x − x 2 = x 2 ⋅ e − x ⋅ ( −1) + 2 x e− x = − x 2 e− x + 2 x e− x = x e− x ( 2 − x ) = ex
33. f (x) =
( e x )2
↑
↑
Differentiate.
Simplify.
↑
Factor.
= x −1 e 2 x
x d −1 2 x f ′(x) = ( x e ) = x −1 d e2 x + e2 x d x −1 = x −1 ⎛⎜ e2 x d ( 2 x ) ⎞⎟ + e2 x ⋅ ( −1) x− 2 dx dx dx dx ⎝ ⎠ ↑ The derivative of a product
↑ Apply the Chain Rule.
= 2 x −1 e2 x − x −2 e 2 x = e 2 x ( 2 x −1 − x − 2 ) = ↑ Differentiate.
35. f ′(x) =
↑ Factor.
f ′(x) =
↑ Simplify.
d ( 2 d 2 d 1 3 x − 3ln x ) = x − 3 ln x = 2 x − 3 ⋅ = 2 x − dx dx dx x x ↑ The derivative of a difference
37.
e 2 x ( 2 x − 1) x2
↑ Differentiate.
↑ Simplify.
d ( 2 d d 2 1 x ln x ) = x 2 ln x + ln x ⋅ x = x 2 ⋅ + ln x ⋅ 2 x dx dx dx x ↑ The derivative of a product
↑ Differentiate.
= x + 2 x ln x = x (1 + 2 ln x ) = x (1 + ln x 2 ) ↑ Simplify.
↑ Factor.
↑ Alternate form of the answer
d ( ) 5x 5 3 d d dx [3ln ( 5 x )] = 3 ln ( 5 x ) = 3 ⋅ 39. f ′(x) = =3⋅ = 5x 5x x dx dx ↑ Derivative of ln g(x)
41. Here the f is the product of two functions, we use the product rule first and then the Chain Rule when differentiating ln (x 2 + 1).
f ′(x) =
d ⎡ d ( 2 ) d 2 ln x + 1 + ln ( x 2 + 1) ⋅ x ⎣ x ln ( x + 1) ⎤⎦ = x ⋅ dx dx dx d ( 2 ) x +1 dx =x⋅ + ln ( x 2 + 1) ⋅ 1 2 x +1 2x =x⋅ 2 + ln ( x 2 + 1) x +1
The derivative of a product.
Use the Chain Rule; Differentiate.
d dx
x = 1.
242 SECTION 4.5
=
2 x2 + ln ( x 2 + 1) 2 x +1
Simplify.
d ( ) 3x d d d ( ) d dx ( ) 43. f ′(x) = x + x = x + x = x + ⋅ 8ln 3 8 ln 3 8 [ ] dx dx dx dx 3x ↑ The derivative of a sum
=1+ 8 ⋅ ↑ Differentiate.
↑ The Chain Rule
3 8 x+8 =1+ = x x 3x ↑ ↑ Simplify. Alternate form of the answer.
45. f ′(x) = d ⎡8 ( ln x )3 ⎤ = 8 d ( ln x )3 = 8 ⋅ 3 ( ln x ) 2 ⋅ d ln x = 24 ( ln x ) 2 ⋅ 1 = 24 ( ln x ) ⎦ dx ⎣ dx dx x x ↑ Use the Power Rule.
47.
f ′(x) =
49. f ′(x) =
51.
f ′(x) =
53. f ′(x) =
d 1 log 3 x = dx x ln 3
d dx
log a x =
↑ Differentiate.
d dx
↑ Simplify.
1 x ln a
d 2 ( x log 2 x ) = x 2 ⋅ d log 2 x + log 2 x ⋅ d x 2 dx dx dx 1 = x2 ⋅ + log 2 x ⋅ 2 x x ln 2 x x ln x = + 2 x log 2 x = + 2x ln 2 ln 2 ln 2 x + 2 x ln x = ln 2 d x 3 = 3x ln 3 dx
2
x
Use the Change of Base Formula.
x
a = a ln a
d 2 x ( x ⋅ 2 ) = x 2 ⋅ d 2 x + 2 x ⋅ d x 2 = x 2 ⋅ 2 x ln 2 + 2 x ⋅ 2 x = 2 x ( x 2 ln 2 + 2 x ) dx dx dx ↑ The derivative of a product
↑ Differentiate;
↑ d dx
x
x
a = a ln a .
Factor.
55. The slope of the tangent line to the graph of f (x) = e 3x at the point (0, 1) is the derivative of the function f evaluated at the point (0, 1). The derivative of f is f ′(x) = 3e 3x The slope of the tangent line is mtan = f ′(0) = 3 · e 0 = 3.
SECTION 4.5 243
An equation of the tangent line is y − 1 = 3 ( x − 0 ) or y = 3x + 1 . 57. The slope of the tangent line to the graph of f (x) = ln x at the point (1, 0) is the derivative of the function f evaluated at the point (1, 0). The derivative of f is 1 f ′(x) = x The slope of the tangent line is mtan = f ′(1) = 1.
An equation of the tangent line is y – 0 = 1(x – 1) or y = x – 1. ⎛2 ⎞ 59. The slope of the tangent line to the graph of f (x) = e 3x – 2 at the point ⎜ , 1⎟ is the ⎝3 ⎠ ⎛2 ⎞ derivative of the function f evaluated at the point ⎜ , 1⎟ . The derivative of f is ⎝3 ⎠ 3x – 2 f ′(x) = 3e The slope of the tangent line is mtan = 3e3 ⋅ 2 3 − 2 = 3e0 = 3 2⎞ ⎛ An equation of the tangent line is y – 1 = 3 ⎜ x − ⎟ or y = 3x – 1. 3⎠ ⎝ 61. The slope of the tangent line to the graph of f (x) = x ln x at the point (1, 0) is the derivative of the function f evaluated at the point (1, 0). The derivative of f is d d 1 f ′( x) = x ⋅ x = x ⋅ + ln x ⋅ 1 = 1 + ln x ln x + ln x ⋅ dx dx x ↑ The derivative of a product
↑ Differentiate.
↑ Simplify.
The slope of the tangent line is mtan = f ′(1) = 1 + ln 1 = 1 + 0 = 1 An equation of the tangent line is y – 0 = 1(x – 1) or y = x – 1. 63. Parallel lines have the same slope. Since the slope of the line y = x is 1, the slope of the tangent line we seek is also 1. Moreover, the slope of a tangent line is given by the derivative of the function. So we need mtan = f ′(x) = 1. f ′(x) = e x = 1 when x = 0, and f (0) = e 0 = 1 which means an equation of a tangent line to the function is y – 1 = 1(x – 0) or y = x + 1. 65. (a) The reaction rate for a dose of 5 units is given by the derivative of R, R′(x) evaluated at x = 5 units. d 5.5 R′ ( x ) = ( 5.5ln x + 10 ) = dx x 5.5 R′ ( 5 ) = = 1.1 5
244 SECTION 4.5
(b) The reaction rate for a dose of 10 units is given by R′(10). 5.5 R′ (10 ) = = 0.55 10 67. The rate of change in atmospheric pressure is given by the derivative of the atmospheric pressure P. d d P′ ( x ) = (104 e − 0.00012 x ) = 104 e− 0.00012 x = 104 ⋅ ( −0.00012 ) e− 0.00012 x = −1.2e− 0.00012 x dx dx The rate of change of atmospheric pressure at x = 500 meters is ( ) P′ ( 500 ) = −1.2e( − 0.00012) 500 = −1.2e − 0.060 = −1.130 kilograms per square meter.
The rate of change of atmospheric pressure at x = 700 meters is ( ) P′ ( 700 ) = −1.2e( − 0.00012) 700 = −1.2e − 0.060 = −1.103 kilograms per square meter. 69. (a) The rate of change of A with respect to time is the derivative A′( t). d d d A′( t) = 102 − 90e − 0.21t ) = 102 − 90 ⋅ e − 0.21t = − 90 ⋅ ( − 0.21) e − 0.21t = 18.9e − 0.21t ( dx dx dx
(b) At t = 5, A′( t ) = A′(5) = 18.9e – 0.21(5) = 18.9e – 1.05 = 6.614 (c) At t = 10, A′( t) = A′(10) = 18.9e – 0.21(10) = 18.9e – 2.1 = 2.314 (d) At t = 30, A′( t) = A′(30) = 18.9e – 0.21(30) = 18.9e – 6.3 = 0.035 71. (a) The rate of change of S with respect to x is the derivative S ′(x). d d d S ′(x) = (100,000 + 400,000ln x ) = 100,000 + 400,000 ln x dx dx dx 1 400,000 = 400,000 ⋅ = x x 400,000 (b) S ′(10) = = 40,000 10
(c) S ′(20) =
400,000 = 20,000 20
73. (a) For x = 1000, the price p is ⎛ 1000 ⎞ p = 50 − 4ln ⎜ + 1⎟ = 50 − 4ln11 = $40.41 ⎝ 100 ⎠
(b) For x = 5000, the price p is ⎛ 5000 ⎞ p = 50 − 4ln ⎜ + 1⎟ = 50 − 4ln 51 = $34.27 ⎝ 100 ⎠ (c) The marginal demand is the derivative of the function p.
SECTION 4.5 245
d d ⎡ d ⎛ x ⎛ x ⎞⎤ d ⎞ p = ⎢50 − 4ln ⎜ + 1⎟ ⎥ = 50 − 4 ln ⎜ + 1⎟ dx dx ⎣ dx ⎝ 100 ⎠ ⎝ 100 ⎠ ⎦ dx ↑ The derivative of a difference.
=
d d ⎛ x + 100 ⎞ d 50 − 4 ln ⎜ ⎟ = − 4 ⎡⎣ ln ( x + 100 ) − ln100 ⎤⎦ dx dx ⎝ 100 ⎠ dx
↑
Write
x 100
↑
+ 1 as a single fraction.
The logarithm of a quotient is the difference of the logarithms.
d 1 4 ⎡d ⎤ =− = − 4 ⎢ ln ( x + 100 ) − ln100 ⎥ = − 4 ⋅ x + 100 x + 100 dx ⎣ dx ⎦ ↑ The derivative of a difference.
↑ Differentiate.
↑ Simplify.
The marginal demand for 1000 t-shirts is 4 4 p′(1000) = − =− = – 0.0036 1000 + 100 1100 (d) The marginal demand for 5000 t-shirts is 4 4 p′(5000) = − =− = – 0.00078 5000 + 100 5100 ⎡ ⎛ x ⎞⎤ ⎛ x ⎞ (e) The revenue function R(x) = p · x = ⎢50 − 4ln ⎜ + 1⎟ ⎥ ⋅ x = 50 x − 4 x ln ⎜ + 1⎟ . ⎝ 100 ⎠ ⎦ ⎝ 100 ⎠ ⎣ (f) The marginal revenue is the derivative of the function R. d ⎡ d ⎡ ⎛ x ⎞⎤ d ⎛ x ⎞⎤ R′ ( x ) = ⎢50 x − 4 x ln ⎜ + 1⎟ ⎥ = ( 50 x ) − ⎢ 4 x ln ⎜ + 1⎟ ⎥ dx ⎣ dx ⎣ ⎝ 100 ⎠ ⎦ dx ⎝ 100 ⎠ ⎦ = 50
⎡ d ⎛ x d ⎞ ⎛ x ⎞d ⎤ x − 4 ⎢ x ln ⎜ + 1⎟ + ln ⎜ + 1⎟ x ⎥ dx ⎝ 100 ⎠ dx ⎦ ⎣ dx ⎝ 100 ⎠
⎡ ⎡ x 1 ⎛ x ⎞⎤ ⎛ x ⎞⎤ = 50 − 4 ⎢ x ⋅ + ln ⎜ + 1⎟ ⎥ = 50 − 4 ⎢ + ln ⎜ + 1⎟ ⎥ ⎝ 100 ⎠ ⎦ ⎝ 100 ⎠ ⎦ ⎣ x + 100 ⎣ x + 100 When x = 1000 t-shirts are sold the marginal revenue is ⎡ 1000 ⎛ 1000 ⎞ ⎤ ⎡10 ⎤ R′ (1000 ) = 50 − 4 ⎢ + ln ⎜ + 1⎟ ⎥ = 50 − 4 ⎢ + ln11⎥ = $36.77 ⎝ 100 ⎠⎦ ⎣ 11 ⎦ ⎣1000 + 100 (g) When x = 5000 t-shirts are sold the marginal revenue is ⎡ 5000 ⎛ 5000 ⎞ ⎤ ⎡ 50 ⎤ R′ ( 5000 ) = 50 − 4 ⎢ + ln ⎜ + 1⎟ ⎥ = 50 − 4 ⎢ + ln 51⎥ = $30.35 ⎝ 100 ⎠⎦ ⎣ 51 ⎦ ⎣ 5000 + 100 (h) Profit is the difference between revenue and cost. P(x) = R(x) – C(x)
246 SECTION 4.5
⎡ ⎛ x ⎞⎤ ⎛ x ⎞ = ⎢50 x − 4 x ln ⎜ + 1⎟ ⎥ − [ 4 x ] = 46 x − 4 x ln ⎜ + 1⎟ ⎝ 100 ⎠ ⎦ ⎝ 100 ⎠ ⎣ (i) If 1000 t-shirts are sold the profit is ⎛ 1000 ⎞ P(1000) = 46 (1000 ) − 4 (1000 ) ln ⎜ + 1⎟ = 46,000 − 4000ln11 = $36,408.42 ⎝ 100 ⎠ (j) If 5000 t-shirts are sold the profit is ⎛ 5000 ⎞ P(5000) = 46 ( 5000 ) − 4 ( 5000 ) ln ⎜ + 1⎟ = 230,000 − 20,000 ln 51 = $151,363.49 ⎝ 100 ⎠ (k) To use TABLE to find the quantity x that maximizes profit, we enter the profit function into Y1, then in TBLSET we select a large value for x, choose ∆Tbl, and select the Auto option.
We chose x = 3,000,000 and ∆Tbl = 1000. Then using TABLE, we increased x until the profit function stopped increasing and began decreasing in magnitude. The quantity (to the nearest thousand) that maximizes profit is 3,632,000 t-shirts. The maximum profit is $14,525,251. ⎛ 3,632,000 ⎞ + 1⎟ = 50 − 4ln ( 36,321) = 8.00 (l) p(3,632,000) = p = 50 − 4ln ⎜ ⎝ 100 ⎠ To maximize profit, the t-shirts should be sold for $8.00 each. 75. (a) The rate of change of p with respect to t is the derivative of p. d d d d 1 0.026 p = ( 0.470 + 0.026 ln t ) = 0.470 + 0.026 ln t = 0.026 ⋅ = dt dt dt dt t t
(b) In 2002 the rate of change of p was 0.026 p′(5) = = 0.052 5 (c) In 2007 the rate of change of p was 0.026 = 0.026 p′(10) = 10
d g ( x ) g′ x ( ) d 77. Prove: ln g ( x ) = dx = dx g ( x) g ( x) Proof: Let y = f (u) = ln u and u = g(x).
SECTION 4.6 247
dy d 1 du = = g′(x) ln u = and u′ = du u du dx According to the Chain Rule if both f and g are differentiable functions, then g′( x ) dy dy du 1 1 = ⋅ = g′(x) = ⋅ g′( x ) = u dx du dx g ( x) g ( x) Then y′ = f ′(u) =
4.6
Higher-Order Derivatives
1.
f (x) = 2x + 5 f ′(x) = 2 f ′′(x) = 0
3. f (x) = 3x 2 + x – 2 f ′(x) = 6x + 1 f ′′(x) = 6
5.
f (x) = – 3x 4 + 2x 2 f ′(x) = – 12x 3 + 4x f ′′(x) = – 36x 2 + 4
7.
f ( x) =
1 = x–1 x
f ′′(x) = 2x – 3 9.
11.
1 = x + x −1 x 1 f ′(x) = 1 − x − 2 = 1 − 2 x 2 f ′′(x) = 2 x − 3 = 3 x
f ( x) = x +
f ( x) =
x x +1
d ⎛ x ⎞ ( f ′(x) = ⎜ ⎟= dx ⎝ x + 1 ⎠
x + 1)
d d x − x ( x + 1) ( x + 1) ⋅ 1 − x ⋅ 1 = 1 = x + 1 − 2 dx dx = ( ) 2 2 2 ( x + 1) ( x + 1) ( x + 1)
↑ The derivative of a quotient.
f ′′(x) = − 2 ( x + 1)
−3
↑ Use the Power Rule.
13.
1 x2 2 = 3 x
f ′(x) = – x – 2 = −
f ( x) = e x f ′(x) = e x f ′′(x) = e x
↑ Differentiate.
d 2 −3 ( x + 1) = − 2 ( x + 1) ⋅ 1 = − 3 dx ( x + 1) ↑ Simplify.
↑ Simplify.
↑ Write with a negative exponent.
SECTION 4.6 247
dy d 1 du = = g′(x) ln u = and u′ = du u du dx According to the Chain Rule if both f and g are differentiable functions, then g′( x ) dy dy du 1 1 = ⋅ = g′(x) = ⋅ g′( x ) = u dx du dx g ( x) g ( x) Then y′ = f ′(u) =
4.6
Higher-Order Derivatives
1.
f (x) = 2x + 5 f ′(x) = 2 f ′′(x) = 0
3. f (x) = 3x 2 + x – 2 f ′(x) = 6x + 1 f ′′(x) = 6
5.
f (x) = – 3x 4 + 2x 2 f ′(x) = – 12x 3 + 4x f ′′(x) = – 36x 2 + 4
7.
f ( x) =
1 = x–1 x
f ′′(x) = 2x – 3 9.
11.
1 = x + x −1 x 1 f ′(x) = 1 − x − 2 = 1 − 2 x 2 f ′′(x) = 2 x − 3 = 3 x
f ( x) = x +
f ( x) =
x x +1
d ⎛ x ⎞ ( f ′(x) = ⎜ ⎟= dx ⎝ x + 1 ⎠
x + 1)
d d x − x ( x + 1) ( x + 1) ⋅ 1 − x ⋅ 1 = 1 = x + 1 − 2 dx dx = ( ) 2 2 2 ( x + 1) ( x + 1) ( x + 1)
↑ The derivative of a quotient.
f ′′(x) = − 2 ( x + 1)
−3
↑ Use the Power Rule.
13.
1 x2 2 = 3 x
f ′(x) = – x – 2 = −
f ( x) = e x f ′(x) = e x f ′′(x) = e x
↑ Differentiate.
d 2 −3 ( x + 1) = − 2 ( x + 1) ⋅ 1 = − 3 dx ( x + 1) ↑ Simplify.
↑ Simplify.
↑ Write with a negative exponent.
248 SECTION 4.6 15.
f ( x) = ( x 2 + 4 )
3
f ′(x) = 3 ( x 2 + 4 )
2
2 2 d x 2 + 4) = 3( x 2 + 4) ⋅ 2 x = 6 x ( x 2 + 4) ( dx
↑ Use the Power Rule.
f ′′(x) =
↑ Simplify.
2 2 2 d d ⎡ d 6x ( x 2 + 4) ⎤ = 6 x ( x 2 + 4) + ( x 2 + 4) ( 6 x ) Derivative of a product. ⎦⎥ dx ⎣⎢ dx dx 2 d = 6 x ⋅ 2 ( x 2 + 4 ) ( x 2 + 4 ) + ( x 2 + 4 ) ⋅ 6 Use the Power Rule. dx
= 6x ⋅ 2 ( x 2 + 4) ⋅ 2x + ( x 2 + 4) ⋅ 6 2
= 24 x 2 ( x 2 + 4 ) + 6 ( x 2 + 4 )
2
Simplify.
= 6 ( x 2 + 4 )( 4 x 2 + x 2 + 4 )
Factor.
= 6 ( x 2 + 4 )( 5 x 2 + 4 )
17.
Simplify.
f (x) = ln x 1 f ′(x) = = x − 1 x f ′′(x) = − x − 2 = −
19.
Differentiate.
1 x2
f (x) = xe x d d f ′(x) = x e x + e x x = xe x + e x = e x ( x + 1) dx dx ↑ ↑ ↑ Derivative of a product. Differentiate. Factor.
f ′′(x) =
d d d xe x + e x ) = xe x + e x = xe x + e x + e x = e x ( x + 2 ) ( dx dx dx ↑ Derivative of a sum.
21.
↑ Differentiate; use f ′(x).
f (x) = (e x) 2 2 d x e = 2 ( e x ) = 2e 2 x dx 2 d f ′′(x) = 4 ( e x ) e x = 4 ( e x ) = 4e 2 x dx
f ′(x) = 2 ( e x )
23.
↑ Factor.
f (x) =
1 = (ln x) – 1 ln x
Use the Chain Rule. Use the Chain Rule.
SECTION 4.6 249
f ′(x) = − ( ln x )
d 1 −2 1 ln x = − ( ln x ) ⋅ = − 2 dx x x ( ln x )
−2
↑ Use the Chain Rule.
f ′′(x) = −
x ( ln x )
2
↑ Simplify.
d d d 2 2 2 d 1 − ⎡ x ( ln x ) ⎤ x ( ln x ) + ( ln x ) x ⎣ ⎦ dx dx dx dx = 2 4 x 2 ( ln x ) ⎡ x ( ln x ) 2 ⎤ ⎣ ⎦
↑
↑
The derivative of a quotient.
d dx
1 = 0 ; the derivative of a product.
1 2 + ( ln x ) 2 + ln x x = 2 4 3 2 x ( ln x ) x ( ln x )
2 x ln x ⋅ =
↑ Use the Power Rule.
25.
↑ Simplify.
(a) The function f is a polynomial, so the domain of f is all real numbers. (b) f ′(x) = 2x (c) f ′(x) is a monomial, so the domain of f ′ is all real numbers. (d) f ′(x) = 0 when 2x = 0, or when x = 0. (e) There are no numbers in the domain of f for which f ′(x) does not exist. (f) f ′′(x) = 2 (g) The domain of f ′′ is all real numbers.
27. (a) The function f is a polynomial, so the domain of f is all real numbers.
(b) f ′(x) = 3x 2 – 18x + 27 (c) f ′(x) is a polynomial, so the domain of f ′ is all real numbers. (d) f ′(x) = 0 when 3x 2 – 18x + 27 = 0, or when x = 3. 3x 2 – 18x + 27 = 0 x 2 – 6x + 9 = 0 (x – 3) 2 = 0 x =3
f ′(x) = 0 Divide both sides by 3. Factor. Use the square root method.
(e) There are no numbers in the domain of f for which f ′(x) does not exist.
250 SECTION 4.6
(f) f ′′(x) = 6x – 18 (g) The domain of f ′′ is all real numbers. 29.
(a) The function f is a polynomial, so the domain of f is all real numbers. (b) f ′(x) = 12x 3 – 36x 2 (c) f ′(x) is a polynomial, so the domain of f ′ is all real numbers. (d) f ′(x) = 0 when 12x 3 – 36x 2 = 0, or when x = 0 or x = 3. 12x 3 – 36x 2 = 0 x 3 – 3x 2 = 0 x 2 (x – 3) 2 = 0 x = 0 or x = 3
f ′(x) = 0 Divide both sides by 12. Factor. Use the square root method.
(e) There are no numbers in the domain of f for which f ′(x) does not exist. (f) f ′′(x) = 36x 2 – 72x (g) f ′′ is a polynomial, so the domain of f ′′ is all real numbers. 31. (a) The domain of the function f is all real numbers except x = 2 and x = – 2.
(b) f ′(x) =
(x
2
− 4)
d d x − x ( x 2 − 4) ( x 2 − 4) − x ⋅ 2 x x2 +4 dx dx = = − 2 2 2 ( x 2 − 4) ( x 2 − 4) ( x 2 − 4)
(c) The domain of the function f ′(x) is all real numbers except x = 2 and x = – 2. (d) f ′(x) is never equal to zero. (e) There are no numbers in the domain of f for which f ′(x) does not exist. 2 d 2⎤ d ⎡ 2 2 2 2 ⎢ ( x − 4 ) dx ( x + 4 ) − ( x + 4 ) dx ( x − 4 ) ⎥ (f) f ′′(x) = − ⎢ ⎥ 4 ( x2 − 4) ⎢ ⎥ ⎣ ⎦ 2 ⎡ ( x 2 − 4) ⋅ ( 2 x ) − ( x 2 + 4) ⋅ 2 ( x 2 − 4) ( 2 x ) ⎤ ⎥ = −⎢ 4 2 ⎢ ⎥ ( x − 4) ⎣ ⎦
⎡ 2x x 2 − 4 2 − 4x x 2 + 4 x 2 − 4 ⎤ ( ) ( )( )⎥ = − ⎢⎢ 43 ⎥ ( x2 − 4) ⎢⎣ ⎥⎦
SECTION 4.6 251
⎡ 3 ⎤ 2 x − 8 x − 4 x 3 − 16 x ⎥ = −⎢ 3 2 ⎢ ⎥ x − 4 ( ) ⎣ ⎦ ⎡ ⎤ − 2 x 3 − 24 x ⎥ 2 x 3 + 24 x = = −⎢ ⎢ ( x2 − 4) 3 ⎥ ( x2 − 4) 3 ⎣ ⎦ (g) The domain of the function f ′′(x) is all real numbers except x = 2 and x = – 2. 33. The function f is a polynomial of degree 3, so the fourth derivative is zero. 35. The function f is a polynomial of degree 19, so the twentieth derivative is zero. 37. The function f is a polynomial of degree 8, so the eighth derivative is equal to the constant 1 8! · = 7! = 5040 8 39. v = s′( t ) = 32t + 20 a = v′( t ) = s′′( t ) = 32
41. v = s′( t ) = 9.8 t + 4 a = v′( t ) = s′′( t ) = 9.8
43. To find a formula for the nth derivative, we take successive derivatives until we see a pattern. f ( x) = e x f ′( x ) = e x f ′′ ( x ) = e x
We see that each order derivative is e x, so we conclude that a formula for f (n) is n f ( ) ( x) = e x 45. To find a formula for the nth derivative, we take successive derivatives until we see a pattern. f ( x ) = ln x
f ′( x ) =
1 = x −1 x
f ′′ ( x ) = − x − 2 = − f ′′′ ( x ) = 2 x − 3 =
1 x2
2 x3 3! x4
f
( 4)
( x) = − 3 ⋅ 2x − 4 = −
f
( 5)
( x ) = ( − 4 ) ⋅ ( − 3! x − 5 ) =
4! x5
252 SECTION 4.6
f
( 6)
( x ) = ( − 5 ) ⋅ ( 4! x − 5 ) = −
5! x6
We see a pattern, noticing that the sign of the derivative alternates from positive to negative and conclude the formula for f (n) is ( n − 1)! n n −1 f ( ) ( x ) = ( −1) ( ) ⋅ xn 47. To find a formula for the nth derivative, we take successive derivatives until we see a pattern. f ( x ) = x ln x
d d 1 x = x ⋅ + ln x ⋅ 1 ln x + ln x ⋅ dx dx x = 1 + ln x 1 1 f ′′ ( x ) = 0 + = = x − 1 x x 1 f ′′′ ( x ) = − x − 2 = − 2 x 2 4 f ( ) ( x) = 2x − 3 = 3 x 3! 5 f ( ) ( x) = − 3 ⋅ 2x − 4 = − 4 x 4! 6 f ( ) ( x ) = ( − 4 ) ⋅ ( − 3! x − 5 ) = 5 x 5! 7 f ( ) ( x ) = ( − 5 ) ⋅ ( 4! x − 5 ) = − 6 x f ′( x ) = x ⋅
We see a pattern, noticing that the sign of the derivative alternates from positive to negative and conclude the formula for f (n) is ( n − 2 )! provided n > 1. n n f ( ) ( x ) = ( −1) ( ) ⋅ x n −1 49. To find a formula for the nth derivative, we take successive derivatives until we see a pattern. n f ( x ) = ( 2 x + 3)
d n −1 Use the Power Rule. ( 2 x + 3 ) = 2n ( 2 x + 3 ) dx n −2 d n −2 Use the Power Rule. f ′′ ( x ) = 2n ⋅ ( n − 1)( 2 x + 3) ( 2 x + 3) = 4n ( n − 1)( 2 x + 3) dx n −2 = 22 n ( n − 1)( 2 x + 3) f ′ ( x ) = n ( 2 x + 3)
n −1
f ′′′ ( x ) = 22 n ( n − 1) ⋅ ( n − 2 )( 2 x + 3)
n −3
d ( 2 x + 3) dx
SECTION 4.6 253
= 23 n ( n − 1)( n − 2 )( 2 x + 3) f
( 4)
n −3
( x ) = 24 n ( n − 1)( n − 2 )( n − 3)( 2 x + 3)
n −4
We see a pattern and conclude the formula for f (n) is n f ( ) ( x ) = 2 n ⋅ n! 51. To find a formula for the nth derivative, we take successive derivatives until we see a pattern. f ( x ) = e ax
d ( ax ) = ae ax dx d f ′′ ( x ) = ae ax ( ax ) = a 2 e ax dx d f ′′′ ( x ) = a 2 e ax ( ax ) = a 3 e ax dx We see a pattern and conclude the formula for f (n) is n f ( ) ( x ) = a n ⋅ e ax f ′ ( x ) = e ax
53.
To find a formula for the nth derivative, we take successive derivatives until we see a pattern. f ( x ) = ln(ax) a 1 0! = = ax x x1 1 1! f ′′ ( x ) = − 2 = − 2 x x 2 2! f ′′′ ( x ) = 3 = 3 x x 6 3! f (4) ( x ) = − 4 = − 4 x x 24 4! f (5) ( x ) = 5 = 5 x x (n) We see a pattern and conclude the formula for f is ( n − 1)! n f ( ) ( x ) = (−1) n −1 ⋅ xn f ′( x) =
55.
y = e 2x d d y′ = e 2 x = e 2 x ( 2 x ) = e 2 x ⋅ 2 = 2e 2 x dx dx d d d y′′ = y′ = ( 2e 2 x ) = 2e 2 x ( 2 x ) = 2e 2 x ⋅ 2 = 4e 2 x dx dx dx 2x 2x So, y′′ − 4 y = 4e − 4e = 0
254 SECTION 4.6 57.
f ( x) = x 2g ( x)
f ′( x ) = x 2
d d g ( x ) + g ( x ) x 2 = x 2 g′( x ) + 2x g ( x ) dx dx
↑ Derivative of a product
↑ Differentiate.
d 2 d x g′( x ) + 2x g ( x ) dx dx d d ⎡ d ′ ⎤ ⎡ d ⎤ = ⎢x 2 g ( x ) + g ′ ( x ) x 2 ⎥ + ⎢ 2 x g ( x ) + g ( x ) ( 2 x )⎥ dx ⎦ ⎣ dx dx ⎣ dx ⎦ 2 = ⎡⎣ x g ′′ ( x ) + 2 x g ′ ( x ) ⎤⎦ + ⎡⎣ 2 x g ′ ( x ) + 2 g ( x ) ⎤⎦
f ′′ ( x ) =
= x 2 g ′′ ( x ) + 4 x g ′ ( x ) + 2 g ( x )
59. (a) The velocity is
d 6 + 80t − 16t 2 ) = 80 − 32t ( dx At t = 2 seconds the velocity is v(2) = 80 – 32(2) = 16 feet per second. v = s′ ( t ) =
(b) The ball reaches its maximum height when v = 0. 80 v = 80 – 32t = 0 when t = = 2.5 seconds 32 The ball reaches its maximum height 2.5 seconds after it is thrown. (c) At t = 2.5, s(2.5) = 6 + 80(2.5) – 16(2.5.2) = 106 feet. The ball reaches a maximum height of 106 feet. (d) The acceleration is a = v′( t ) = – 32 feet per second per second. (e) The ball strikes the ground when s(t ) = 0. That is when 6 + 80t − 16t 2 = 0 Using the quadratic formula to solve for t, we find t=
− 80 ±
(80 ) − 4 ( −16 )( 6 ) 2 ( −16 ) 2
80 ± 6784 32 We need only the positive answer, since t represents time, and we get t = 5.0739. So the ball is in the air for 5.0739 seconds. t=
(f) The ball hits the ground at t = 5.0739 seconds, the velocity is v(5.0739) = 80 – 32(5.0739) = – 82.365 feet per second. The ball is moving at a speed of 82.365 feet per second in a downward direction. (g) The total distance traveled by the ball is the distance up plus the distance down or (106 – 6) + 106 = 206 feet.
SECTION 4.7 255 61. The velocity of the bullet is d ⎡ d d 3 3 2 d 2 v = s′ ( t ) = 8 − ( 2 − t ) ⎤ = 8 − ( 2 − t ) = −3 ( 2 − t ) ( 2 − t ) = 3( 2 − t ) ⎣ ⎦ dx dx dx dx meters per second. After 1 second, the bullet is traveling at a velocity v(1) = 3(2 – 1) 2 = 3 · 1 2 = 3 meters per second.
The acceleration is a = v′ ( t ) = – 6(2 – t ) = 6t – 12 meters per second per second. 63.
(a) The rock hits the ground when its height is zero. Since the rock started from a height of 88.2 meters, when it hits the ground the rock has traveled 88.2 meters. 4.9t 2 = 88.2 88.2 t2 = = 18 4.9 t = 3 2 ≈ 4.24 It takes the rock approximately 4.24 seconds to hit the ground. (b) The average velocity is
(
)
− ( 88.2 − 4.9 ⋅ 0 2 ) − 20.8 ds = = = = 4.9 dt 3 2 −0 3 2 3 2 meters per second. That is, the rock is moving at an average speed of 4.9 meters per second in the downward direction.
(
)
s 3 2 − s ( 0)
88.2 − 4.9 ⋅ 3 2
2
(c) The average velocity in the first 3 seconds is 4.9 meters per second in a downward direction. (See part (b).) (d) The velocity of the rock is v = s′ ( t ) = – 9.8t meters per second. The rock hits the ground at t = 4.24 seconds. The velocity is v(4.24) = – 9.8(4.24) = – 41.6 meters per second.
4.7 1.
Implicit Differentiation d d x2 + y2) = 4 ( dx dx dy 2x + 2 y =0 dx This is a linear equation in 2y
dy = − 2x dx dy −2 x x = =− dx 2 y y
dy dy . Solving for , we have dx dx
provided y ≠ 0.
SECTION 4.7 255 61. The velocity of the bullet is d ⎡ d d 3 3 2 d 2 v = s′ ( t ) = 8 − ( 2 − t ) ⎤ = 8 − ( 2 − t ) = −3 ( 2 − t ) ( 2 − t ) = 3( 2 − t ) ⎣ ⎦ dx dx dx dx meters per second. After 1 second, the bullet is traveling at a velocity v(1) = 3(2 – 1) 2 = 3 · 1 2 = 3 meters per second.
The acceleration is a = v′ ( t ) = – 6(2 – t ) = 6t – 12 meters per second per second. 63.
(a) The rock hits the ground when its height is zero. Since the rock started from a height of 88.2 meters, when it hits the ground the rock has traveled 88.2 meters. 4.9t 2 = 88.2 88.2 t2 = = 18 4.9 t = 3 2 ≈ 4.24 It takes the rock approximately 4.24 seconds to hit the ground. (b) The average velocity is
(
)
− ( 88.2 − 4.9 ⋅ 0 2 ) − 20.8 ds = = = = 4.9 dt 3 2 −0 3 2 3 2 meters per second. That is, the rock is moving at an average speed of 4.9 meters per second in the downward direction.
(
)
s 3 2 − s ( 0)
88.2 − 4.9 ⋅ 3 2
2
(c) The average velocity in the first 3 seconds is 4.9 meters per second in a downward direction. (See part (b).) (d) The velocity of the rock is v = s′ ( t ) = – 9.8t meters per second. The rock hits the ground at t = 4.24 seconds. The velocity is v(4.24) = – 9.8(4.24) = – 41.6 meters per second.
4.7 1.
Implicit Differentiation d d x2 + y2) = 4 ( dx dx dy 2x + 2 y =0 dx This is a linear equation in 2y
dy = − 2x dx dy −2 x x = =− dx 2 y y
dy dy . Solving for , we have dx dx
provided y ≠ 0.
256 SECTION 4.7 3.
d d x 2 y) = 8 ( dx dx d d x2 y+ y x2 = 0 dx dx dy x2 + y ⋅ 2x = 0 dx This is a linear equation in x2
5.
dy dy . Solving for , we have dx dx
dy = − 2 xy dx dy − 2 xy 2y = =− provided x ≠ 0. 2 dx x x
d d x 2 + y 2 − xy ) = 2 ( dx dx dy ⎛ d d ⎞ 2x + 2 y − ⎜ x y + y x ⎟ = 0 dx ⎝ dx dx ⎠ dy dy 2x + 2 y − x − y = 0 dx dx dy dy . Solving for , we have dx dx dy 2x + (2 y − x) − y = 0 dx dy ( 2 y − x) = y − 2x dx dy y − 2 x = provided 2 y − x ≠ 0. dx 2 y − x
This is a linear equation in
7.
d d x 2 + 4 xy + y 2 ) = y ( dx dx d 2 d d d 2 d x + 4x y + y ( 4x) + y = y dx dx dx dx dx dy dy dy 2x + 4x + y ⋅ 4 + 2 y = dx dx dx dy dy . Solving for , we have dx dx dy dy 2x + 4 y + ( 4x + 2 y ) = dx dx dy dy (4x + 2 y ) − = − 2x − 4 y dx dx
This is a linear equation in
SECTION 4.7 257
( 4 x + 2 y − 1)
9.
d d 3x 2 + y 3 ) = 1 ( dx dx dy 6x + 3y 2 =0 dx This is a linear equation in 3y 2
11.
dy dy . Solving for , we have dx dx
dy = − 6x dx dy − 6 x 2x = =− 2 2 dx 3 y y
provided y ≠ 0.
d d 2 4x 3 + 2 y 3 ) = x ( dx dx dy 12 x 2 + 6 y 2 = 2x dx This is a linear equation in 6y 2
13.
dy = − 2x − 4 y dx dy − 2x − 4 y provided 4 x + 2 y − 1 ≠ 0. = dx 4 x + 2 y − 1
dy dy . Solving for , we have dx dx
dy = 2 x − 12 x 2 dx dy 2 x − 12 x 2 x − 6 x 2 provided y ≠ 0. = = dx 6y 2 3y 2
d ⎛ 1 1 ⎞ d d −2 −2 ⎜ 2 − 2 ⎟ = (x − y ) = 4 dx ⎝ x y ⎠ dx dx dy − 2x − 3 + 2 y − 3 =0 dx dy dy . Solving for , we have dx dx dy 2 y −3 = 2x − 3 dx dy 2 x − 3 y 3 = = provided x ≠ 0. dx 2 y − 3 x 3
This is a linear equation in
258 SECTION 4.7
15.
d ⎛1 1⎞ d d −1 −1 ⎜ + ⎟ = (x + y ) = 2 dx ⎝ x y ⎠ dx dx dy − x −2 − y −2 =0 dx dy dy . Solving for , we have dx dx dy − y −2 = x −2 dx dy x −2 y2 = = − provided x ≠ 0. dx − y − 2 x2
This is a linear equation in
17.
d d x 2 + y 2 ) = ( ye x ) ( dx dx d 2 d 2 d d x + y = y ex +ex y dx dx dx dx dy dy 2x + 2 y = ye x + e x dx dx This is a linear equation in
dy dy . Solving for , we have dx dx
dy dy −ex = y e x − 2x dx dx = y e x − 2x ( 2 y − e x ) dy dx dy y e x − 2 x provided 2 y − e x ≠ 0. = dx 2y − e x
2y
19.
d ⎛ x y⎞ d x ⎜ + ⎟ = ( 6e ) dx ⎝ y x ⎠ dx d ⎤ ⎡ d d ⎤ ⎡ d ⎢ y dx x − x dx y ⎥ ⎢ x dx y − y dx x ⎥ x ⎢ ⎥+⎢ ⎥ = 6e 2 2 y x ⎢ ⎥ ⎢ ⎥ ⎣ ⎦ ⎣ ⎦ dy ⎤ ⎡ dy ⎡ ⎤ ⎢ y − x dx ⎥ ⎢ x dx − y ⎥ x ⎢ ⎥+⎢ ⎥ = 6e 2 2 ⎢ y ⎥ ⎢ x ⎥ ⎣ ⎦ ⎣ ⎦ dy dy x 2y − x3 + xy 2 − y3 dx dx = 6e x x2y 2 dy dy x2y − x3 + xy 2 − y 3 = 6 x 2 y 2e x dx dx
SECTION 4.7 259
dy dy . Solving for , we have dx dx dy − ( x 3 − xy 2 ) = 6 x 2 y 2e x − x 2 y + y 3 dx dy 6 x 2 y 2e x − x 2 y + y 3 = provided − x 3 + xy 2 ≠ 0. 3 2 dx − x + xy
This is a linear equation in
21.
d 2 d x = ( y 2 ln x ) dx dx d d 2 x = y 2 ln x + ln x y 2 dx dx 1 dy 2 x = y 2 ⋅ + ln x ⋅ 2 y x dx dy dy . Solving for , we have dx dx y2 dy 2x = + 2 y ln x x dx 2 y dy 2x − = 2 y ln x x dx 2 y 2x − 2 2 dy x = 2 x − y provided 2 xy ln x ≠ 0. = 2 xy ln x dx 2 y ln x
This is a linear equation in
23.
d d 2 (2x + 3y ) = ( x 2 + y 2 ) dx dx d dy 2 ( 2x + 3 y ) ( 2x + 3 y ) = 2x + 2 y dx dx dy ⎞ dy ⎛ 2 (2x + 3y ) ⎜ 2 + 3 ⎟ = 2x + 2 y dx ⎠ dx ⎝ dy dy ⎞ ⎛ 2 ( 2x + 3 y ) ⎜ 2 + 3 ⎟ = 2 x + 2 y dx dx ⎠ ⎝ dy dy dy 4x + 6x + 6 y + 9 y = x+ y dx dx dx This is a linear equation in 6x
dy dy . Solving for , we have dx dx
dy dy dy + 9y − y = x − 4x − 6 y dx dx dx dy (6x + 9 y − y ) = x − 4x − 6 y dx dy −3x − 6 y = provided 6 x + 8 y ≠ 0. dx 6 x + 8 y
260 SECTION 4.7 25.
2 d 2 d 3 x + y2 ) = ( x − y ) ( dx dx d 2 d 2 ( x2 + y 2 ) ( x2 + y 2 ) = 3 ( x − y ) ( x − y) dx dx dy ⎞ dy ⎞ 2⎛ ⎛ 2 ( x 2 + y 2 ) ⎜ 2 x + 2 y ⎟ = 3 ( x − y ) ⎜1 − ⎟ dx ⎠ ⎝ ⎝ dx ⎠ dy dy dy 4 x3 + 4 x 2 y + 4 xy 2 + 4 y 3 = 3 x 2 − 6 xy + 3 y 2 − ( 3 x 2 − 6 xy + 3 y 2 ) dx dx dx
dy dy . Solving for , we have dx dx dy dy dy 4 x2 y + 4 y3 + ( 3 x 2 − 6 xy + 3 y 2 ) = 3 x 2 − 6 xy + 3 y 2 − 4 x3 − 4 xy 2 dx dx dx dy ( 4 x 2 y + 4 y3 + 3x 2 − 6 xy + 3 y 2 ) dx = 3x 2 − 6 xy + 3 y 2 − 4 x3 − 4 xy 2 dy 3 x 2 − 6 xy + 3 y 2 − 4 x 3 − 4 xy 2 = dx 4 x 2 y + 4 y 3 + 3 x 2 − 6 xy + 3 y 2 provided 4 x 2 y + 4 y 3 + 3x 2 − 6 xy + 3 y 2 ≠ 0.
This is a linear equation in
27.
2 d 3 d x + y3 ) = ( x2 y 2 ) ( dx dx d d 2 d 2 ( x3 + y 3 ) ( x3 + y 3 ) = x 2 y + y 2 x2 dx dx dx dy ⎞ dy ⎛ 2 ( x3 + y 3 ) ⎜ 3x 2 + 3 y 2 ⎟ = x 2 ⋅ 2 y + y 2 ⋅ 2 x dx ⎠ dx ⎝ dy ⎞ dy ⎛ 2 ( x3 + y 3 ) ⎜ 3x 2 + 3 y 2 ⎟ = 2 x 2 y + 2 xy 2 dx ⎠ dx ⎝ dy dy dy 3x5 + 3x3 y 2 + 3x 2 y 3 + 3 y 5 = x 2 y + xy 2 dx dx dx
dy dy . Solving for , we have dx dx dy dy dy 3x3 y 2 + 3 y5 − x2 y = xy 2 − 3x5 − 3x 2 y 3 dx dx dx = xy 2 − 3x5 − 3 x 2 y 3 ( 3x3 y 2 + 3 y5 − x 2 y ) dy dx dy xy 2 − 3 x 5 − 3 x 2 y 3 = provided 3x3 y 2 + 3 y 5 − x 2 y ≠ 0. 3 2 5 2 dx 3 x y + 3 y − x y
This is a linear equation in
29.
d d 2 2 y = e x +y dx dx 2 2 d dy = e x +y x2 + y 2 ) ( dx dx
SECTION 4.7 261 2 2 ⎛ dy dy ⎞ = e x + y ⎜ 2x + 2 y ⎟ dx dx ⎠ ⎝ 2 2 2 2 dy dy = 2 xe x + y + 2 ye x + y dx dx
This is a linear equation in
dy dy . Solving for , we have dx dx
2 2 dy 2 2 dy − 2 ye x + y = 2 xe x + y dx dx 2 2 dy 2 2 1 − 2 ye x + y = 2 xe x + y dx
(
)
2
2
dy 2 xe x + y = 2 2 dx 1 − 2 ye x + y 31.
provided 1 − 2 ye x
2
+ y2
≠ 0.
x The first derivative is y′ = − , provided y ≠ 0 (from Problem 1). y The second derivative is d ⎛ dy ⎞ d ⎜ 2x + 2 y ⎟ = 0 dx ⎝ dx ⎠ dx d dy 2 x + 2 ( yy′ ) = 0 dx dx d ⎤ ⎡ d 2 + 2 ⎢ y y′ + y′ y ⎥ = 0 dx ⎦ ⎣ dx 2 + 2 [ yy′′ + y′y′] = 0 1 + yy′′ + ( y′ ) = 0 2
2
⎛ x⎞ x2 1 − − − +1 2 ⎜ ⎟ − ( y′ ) − 1 y⎠ x2 + y2 y2 ⎝ ′′ y = = =− =− y y y y3
provided y ≠ 0. 33.
The first derivative is d d xy + yx 2 ) = 2 ( dx dx d ⎤ ⎡ d 2 ⎡ d ⎤ 2 d ⎢⎣ x dx y + y dx x ⎥⎦ + ⎢⎣ y dx x + x dx y ⎥⎦ = 0 [ xy′ + y ] + ⎡⎣ y ⋅ 2 x + x 2 y′⎤⎦ = 0 xy′ + y + 2 xy + x 2 y′ = 0
( x + x ) y′ + (1 + 2 x ) y = 0 ( x + x ) y′ = − (1 + 2 x ) y 2
2
262 SECTION 4.7
y′ = −
(1 + 2 x ) y
provided x ≠ 0, x ≠ – 1.
x2 + x
Using the fifth line from above, we find the second derivative is d ⎡ d x + x 2 ) y′ + (1 + 2 x ) y ⎤⎦ = 0 ( ⎣ dx dx d d d ⎡ ⎡ ⎤ 2 d 2 ⎤ ⎢⎣( x + x ) dx y′ + y′ dx ( x + x ) ⎥⎦ + ⎢⎣(1 + 2 x ) dx y + y dx (1 + 2 x )⎥⎦ = 0 ⎡( x + x 2 ) y′′ + y′ (1 + 2 x ) ⎤ + ⎡⎣(1 + 2 x ) y′ + y ⋅ 2 ⎤⎦ = 0 ⎣ ⎦ 2 ( x + x ) y′′ + y′ (1 + 2 x ) + (1 + 2 x ) y′ + 2 y = 0
( x + x ) y′′ + 2 (1 + 2 x ) y′ + 2 y = 0 ( x + x ) y′′ = − 2 (1 + 2 x ) y′ − 2 y 2
2
y′′ = −
y′′ = − y′′ = − y′′ = −
x2 + x ⎛ y + 2 xy ⎞ 2 (1 + 2 x ) ⎜ 2 ⎟ + 2y x +x ⎠ ⎝ y′′ = − x2 + x 2 (1 + 2 x )( y + 2 xy ) + 2 y ( x 2 + x )
(x
The slope of the tangent line is
Solving for
(x
2
+ x)
+ x)
2
2 y + 10 xy + 10 x 2 y
(x
2
+ x)
dy , which is dx d d x2 + y2) = 5 ( dx dx dy 2x + 2 y =0 dx
dy , we have dx dy = − 2x dx dy −2 x x = =− provided y ≠ 0. dx 2 y y
2y
2
2
2 y + 8 xy + 8 x 2 y + 2 x 2 y + 2 xy
provided x ≠ 0, x ≠ – 1. 35.
2 (1 + 2 x ) y′ + 2 y
2
SECTION 4.7 263
1 The slope of the tangent line at the point (1, 2) is mtan = − . The equation of the tangent 2 line is y − y1 = m ( x − x1 ) 1 ( x − 1) 2 1 1 y−2= − x+ 2 2 1 5 y= − x+ 2 2 y−2= −
37. (Note: The problem as printed in the first printing of the text is incorrect. That equation dy has no tangent line at (0, 0).) The slope of the tangent line is , which is dx d xy d e = x dx dx d e xy ( xy ) = 1 dx d ⎞ ⎛ d e xy ⎜ x y + y ⎟ = 1 dx ⎠ ⎝ dx ⎛ dy ⎞ e xy ⎜ x + y ⎟ = 1 ⎝ dx ⎠ dy e xy x + ye xy = 1 dx dy 1 − ye xy = dx xe xy
The slope of the tangent line at the point (1, 0) is mtan = the tangent line is
1− 0 ⋅ e 0 = 1 . The equation of 1e 0
y − y1 = m ( x − x1 ) y − 0 = 1( x − 1)
y=x–1 39. The tangent line is horizontal when the slope is zero, that is when
dy = 0. dx
d d x2 + y2) = 4 ( dx dx dy 2x + 2 y =0 dx dy −2 x x = =− provided y ≠ 0. dx 2 y y
264 SECTION 4.7
dy = 0 when x = 0. dx When x = 0, y 2 = 4, or y = ± 2. So there are horizontal tangent lines at the points (0, 2) and (0, – 2). 41. The tangent line is horizontal when the slope is zero, that is when
dy = 0. dx
d d y 2 + 4 x 2 ) = 16 ( dx dx dy 2 y + 8x = 0 dx dy − 8 x 4x = =− provided y ≠ 0. dx 2 y y dy = 0 when x = 0. dx When x = 0, y 2 = 16, or y = ± 4. So there are horizontal tangent lines at the points (0, 4) and (0, – 4). 43. (a) The slope of the tangent line is
dy , which is dx
d d x + xy + 2 y 2 ) = 6 ( dx dx d d d d x + x y + y x + ( 2 y2 ) = 0 dx dx dx dx dy dy 1+ x + y + 4y =0 dx dx dy ( x + 4y) = − y −1 dx dy − y − 1 y +1 provided x + 4 y ≠ 0. = =− dx x + 4 y x + 4y
(b) At the point (2, 1) the slope of the tangent line is mtan = − an equation of the tangent line is
1+1 2 1 = − = − , and 2 + 4 ⋅1 6 3
y − y1 = m ( x − x1 )
1 ( x − 2) 3 1 5 y=− x+ 3 3
y −1 = −
(c) To find the coordinates of the points (x, y) at which the slope of the tangent line equals the slope of the tangent line at (2, 1), we need to solve the system of equations
SECTION 4.7 265
x + xy + 2 y 2 = 6 (1) y +1 1 − =− (2) x + 4y 3 Beginning with equation (2), we get 3y + 3 = x + 4y or 3 = x + y or x = 3 – y. Substituting x = 3 – y into equation (1), we get (3 − y ) + (3 − y ) y + 2 y2 = 6
3 − y + 3y − y2 + 2 y2 = 6 3 + 2 y + y2 = 6 y2 + 2 y − 3 = 0 ( y − 1)( y + 3) = 0 y =1 y = −3 Back substituting y = 1 into equation (2) we get x = 2. Back substituting y = – 3 into equation (2) we get x = 6. So the coordinates of the point which has the same slope as the tangent line at (2, 1) is (6, – 3). 45.
d ⎛ a ⎞ d C ⎜P+ 2 ⎟ = dP ⎝ V ⎠ dP V − b d d d −1 P+ aV − 2 = C (V − b ) dP dP dP dV −2 d 1 − 2aV − 3 = − C (V − b ) (V − b ) dP dP 2a dV C dV =− 1− 3 2 V dP (V − b ) dP ⎛ 2a ⎞ dV C ⎟ 1= ⎜ 3 − 2 ⎜V ⎟ dP − V b ( ) ⎝ ⎠ ⎛ 2a (V − b ) 2 − CV 3 ⎞ dV ⎜ ⎟ =1 2 3 ⎜ ⎟ dP V V − b ( ) ⎝ ⎠
V 3 (V − b ) dV = dP 2a (V − b ) 2 − CV 3 2
47.
d N (t ) d ⎛ 3t ⎞ e = ⎜ 430,163t + 2 ⎟ dt dt ⎝ t +2⎠
(a)
( d N t d e ( ) N ( t ) = 430,163t + dt dt e
N (t )
t 2 + 2)
d d ( 3t ) − 3t ( t 2 + 2 ) dt dt 2 2 (t + 2)
3 ( t + 2 ) − 6t dN = 430,163 + 2 dt (t 2 + 2) 2
2
266 SECTION 4.8 2 2 2 dN 430,163 ( t + 2 ) + 3 ( t + 2 ) − 6t = 2 dt (t 2 + 2) 2
e
N (t )
2 2 2 430,163 ( t 2 + 2 ) + 3 ( t 2 + 2 ) − 6t 2 dN 430,163 ( t + 2 ) + 3 ( t + 2 ) − 6t = = 2 N t 2 3t ⎞ 2 dt ⎛ e ( ) (t 2 + 2) ⎜ 430,163t + 2 ⎟ (t + 2) t +2⎠ ⎝ 2
2
430,163 ( t 2 + 2 ) + 3 ( t 2 + 2 ) − 6t 2 2
=
430,163 ( 2 2 + 2 ) + 3 ( 2 2 + 2 ) − 6 ⋅ 2 2
430,163t ( t 2 + 2 ) + 3t ( t 2 + 2 ) 2
2
(b) N (2) =
( 430,163 ⋅ 2 ) ( 2
2
+ 2) + (3 ⋅ 2) ( 2 + 2) 2
2
430,163 ( 4 2 + 2 ) + 3 ( 4 2 + 2 ) − 6 ⋅ 4 2
= 0.500
2
N (4) =
4.8
( 430,163 ⋅ 4 ) ( 4 2 + 2 )
2
+ (3 ⋅ 4) ( 4 2 + 2)
= 0.250
The Derivative of x p/q
1.
f′ (x) =
d 4 / 3 4 ( 4 / 3) − 1 4 1/ 3 x = x = x dx 3 3
3.
f′ (x) =
d 2 / 3 2 ( 2 / 3) − 1 2 − 1/ 3 2 = x = 1/ 3 x = x dx 3 3 3x
5.
f′ (x) =
d 1 d − 1/ 2 1 − 1/ 2 − 1 1 1 = = − x ( ) = − x −3 / 2 = − 3 / 2 x 1/ 2 2 2 2x dx x dx
7.
f′ (x) =
d 3 3 3/ 2 3 / 2 −1 d 1/ 2 1/ 2 ( 2 x + 3 ) = ( 2 x + 3 )( ) ( 2 x + 3) = ( 2 x + 3) ⋅ 2 = 3 ( 2 x + 3) dx dx 2 2
9.
f′ (x) =
3/ 2 1/ 2 1/ 2 (3 / 2) −1 d d 3 3 x 2 + 4) = ( x 2 + 4) x 2 + 4 ) = ( x 2 + 4 ) ⋅ 2 x = 3x ( x 2 + 4 ) ( ( dx dx 2 2
11. We first change each radical to its fractional exponent equivalent. 1/ 2 f ( x ) = ( 2 x + 3)
f′ (x) =
d 1 1 1 1/ 2 1/ 2 −1 d − 1/ 2 ( 2 x + 3) = ( 2 x + 3) ( ) ( 2 x + 3) = ( 2 x + 3) ⋅ 2 = 1/ 2 dx dx 2 2 ( 2 x + 3) =
1 2x + 3
266 SECTION 4.8 2 2 2 dN 430,163 ( t + 2 ) + 3 ( t + 2 ) − 6t = 2 dt (t 2 + 2) 2
e
N (t )
2 2 2 430,163 ( t 2 + 2 ) + 3 ( t 2 + 2 ) − 6t 2 dN 430,163 ( t + 2 ) + 3 ( t + 2 ) − 6t = = 2 N t 2 3t ⎞ 2 dt ⎛ e ( ) (t 2 + 2) ⎜ 430,163t + 2 ⎟ (t + 2) t +2⎠ ⎝ 2
2
430,163 ( t 2 + 2 ) + 3 ( t 2 + 2 ) − 6t 2 2
=
430,163 ( 2 2 + 2 ) + 3 ( 2 2 + 2 ) − 6 ⋅ 2 2
430,163t ( t 2 + 2 ) + 3t ( t 2 + 2 ) 2
2
(b) N (2) =
( 430,163 ⋅ 2 ) ( 2
2
+ 2) + (3 ⋅ 2) ( 2 + 2) 2
2
430,163 ( 4 2 + 2 ) + 3 ( 4 2 + 2 ) − 6 ⋅ 4 2
= 0.500
2
N (4) =
4.8
( 430,163 ⋅ 4 ) ( 4 2 + 2 )
2
+ (3 ⋅ 4) ( 4 2 + 2)
= 0.250
The Derivative of x p/q
1.
f′ (x) =
d 4 / 3 4 ( 4 / 3) − 1 4 1/ 3 x = x = x dx 3 3
3.
f′ (x) =
d 2 / 3 2 ( 2 / 3) − 1 2 − 1/ 3 2 = x = 1/ 3 x = x dx 3 3 3x
5.
f′ (x) =
d 1 d − 1/ 2 1 − 1/ 2 − 1 1 1 = = − x ( ) = − x −3 / 2 = − 3 / 2 x 1/ 2 2 2 2x dx x dx
7.
f′ (x) =
d 3 3 3/ 2 3 / 2 −1 d 1/ 2 1/ 2 ( 2 x + 3 ) = ( 2 x + 3 )( ) ( 2 x + 3) = ( 2 x + 3) ⋅ 2 = 3 ( 2 x + 3) dx dx 2 2
9.
f′ (x) =
3/ 2 1/ 2 1/ 2 (3 / 2) −1 d d 3 3 x 2 + 4) = ( x 2 + 4) x 2 + 4 ) = ( x 2 + 4 ) ⋅ 2 x = 3x ( x 2 + 4 ) ( ( dx dx 2 2
11. We first change each radical to its fractional exponent equivalent. 1/ 2 f ( x ) = ( 2 x + 3)
f′ (x) =
d 1 1 1 1/ 2 1/ 2 −1 d − 1/ 2 ( 2 x + 3) = ( 2 x + 3) ( ) ( 2 x + 3) = ( 2 x + 3) ⋅ 2 = 1/ 2 dx dx 2 2 ( 2 x + 3) =
1 2x + 3
SECTION 4.8 267 13. We first change each radical to its fractional exponent equivalent. f ( x ) = ( 9 x 2 + 1)
f′ (x) =
15.
f′ (x) =
1/ 2
1/ 2 − 1/ 2 (1/ 2 ) − 1 d d 1 1 9 x 2 + 1) = ( 9 x 2 + 1) 9 x 2 + 1) = ( 9 x 2 + 1) ⋅ 18 x ( ( 2 2 dx dx 9x 9x = = 1/ 2 2 9x 2 + 1 ( 9 x + 1)
d d d 3 x 5 / 3 − 6 x 1/ 3 ) = 3 x 5 / 3 − 6 x 1/ 3 ( dx dx dx 5 5 / 3 −1 1 1/ 3 − 1 =3⋅ x( ) −6⋅ x( ) 3 3 = 5x 2 / 3 − 2x − 2 / 3 = 5x 2 / 3 −
2 x 2/3
17. To find the derivative of function f we can either use the formula for the derivative of a product or multiply the factors and find the derivative of the sum. We chose to multiply first. f ( x ) = x 1/ 3 ( x 2 − 4 ) = x 7 / 3 − 4 x 1/ 3
f′ (x) =
d d 7/3 d 7 7 / 3 −1 1 1/ 3 − 1 x 7 / 3 − 4 x 1/ 3 ) = x − 4 x 1/ 3 = x ( ) − 4 ⋅ x ( ) ( dx dx dx 3 3 7 4/3 4 −2/3 7 4/3 4 = x − x = x − 3 3 3 3x 2 / 3
19. We first change each radical to its fractional exponent equivalent. x x f ( x) = = 1/ 2 2 2 x − 4 ( x − 4)
f′ (x) =
d x dx ( x 2 − 4 ) 1/ 2
1/ 2 d d x − x ( x 2 − 4) Use the formula for the dx dx = 2 derivative of a quotient. ⎡ x 2 − 4 ) 1/ 2 ⎤ ⎢⎣( ⎥⎦ 1/ 2 (1/ 2 ) − 1 d x 2 − 4) ( x 2 − 4 ) − x ⋅ 12 ( x 2 − 4 ) ( dx = 2 − x 4 ( )
=
(x
2
− 4)
1/ 2
(x
2
− 4)
1/ 2
(x =
2
− 4)
1/ 2
−1/ 2 1 2 x − 4) ⋅ 2x ( 2 ( x 2 − 4)
−x⋅
− x 2 ( x 2 − 4)
(x
( x − 4) − x = ( x − 4) 2
2
3/ 2
2
−1/ 2
− 4)
2
=−
4
( x 2 − 4)
3/ 2
Multiply by
(x (x
2
2
) − 4) −4
1/ 2
1/ 2
; simplify.
268 SECTION 4.8 21. We first change each radical to its fractional exponent equivalent. f ( x) = e x = (e x )
1/ 2
f′ (x) =
d x 1/ 2 1 x (1/ 2) − 1 d x 1 x − 1/ 2 x 1 x 1/ 2 ex e ) = (e ) e = (e ) e = (e ) = ( 2 2 2 2 dx dx
23. We first change each radical to its fractional exponent equivalent. 1/ 2 f ( x ) = ln x = ( ln x )
f′ (x) =
d 1 1 1 1 1/ 2 1/ 2 − 1 d − 1/ 2 1 ln x = ( ln x ) ⋅ = = ( ln x ) = ( ln x ) ( ) 1/ 2 2 2 dx dx x 2 x ( ln x ) 2 x ln x
25. We first change each radical to its fractional exponent equivalent. 1/ 3 3 f ( x) = e x = e x 1/ 3
3
3
1/ 3 d 1/ 3 1/ 3 d x 1/ 3 1 1/ 3 − 1 1 ex e x e x f′ (x) = =ex e x 1/ 3 = e x ⋅ x ( ) = x − 2 / 3 e x = 2 / 3 = 2 / 3 = 3 dx dx 3 3 3x 3x 3 x2
27. We first change each radical to its fractional exponent equivalent. 1/ 3 f ( x ) = 3 ln x = ( ln x )
f′ (x) =
d 1 1 1 1 1/ 3 1/ 3 − 1 d −2/3 1 ⋅ = = ln x = ( ln x ) ( ln x ) = ( ln x ) ( ) 2/3 2 dx dx x 3 x ( ln x ) 3 3 3x 3 ( ln x )
29. We first change each radical to its fractional exponent equivalent. f ( x ) = x e x = x 1/ 2 e x
f′ (x) =
d 1/ 2 x d d 1/ 2 x e ) = x 1/ 2 e x + e x x ( dx dx dx 1 1/ 2 − 1 = x 1/ 2 e x + e x ⋅ x ( ) 2 1 = x 1/ 2 e x + x − 1/ 2 e x 2 ex = x 1/ 2 e x + 2 x 1/ 2 =
Use the formula for the derivative of a product.
x 2 x e x + e x e ( 2 x + 1) = 2 x 1/ 2 2 x
31. We first change each radical to its fractional exponent equivalent. f ( x ) = e 2 x x 2 + 1 = e 2 x ( x 2 + 1)
(
)
1/ 2
1/ 2 1/ 2 1/ 2 d d 2x 2 d e ( x + 1) = e 2 x ( x 2 + 1) + ( x 2 + 1) e 2x dx dx dx 1/ 2 (1/ 2 ) − 1 d 1 d = e 2 x ( x 2 + 1) x 2 + 1) + ( x 2 + 1) e 2 x ( 2 x ) ( dx dx 2
f′ (x) =
Use the formula for the derivative of a quotient.
Use the Chain Rule.
SECTION 4.8 269 − 1/ 2 1/ 2 1 2 x + 1) ⋅ 2 x + ( x 2 + 1) e 2 x ⋅ 2 ( 2
= e 2x
= xe 2 x ( x 2 + 1) =
33.
(x
(
xe 2 x 2
+ 1)
d x+ dx 1 1 + 2 x 2 y 1 2 y
1/ 2
)
− 1/ 2
+ 2e 2 x ( x 2 + 1)
+ 2e 2 x ( x 2 + 1)
y =
1/ 2
=
Differentiate.
1/ 2
Simplify.
xe 2 x + 2e 2 x ( x 2 + 1)
(x
2
+ 1)
1/ 2
=
e 2 x ( 2x 2 + x + 2)
d 4 dx
dy =0 dx dy 1 =− dx 2 x
2 y y dy =− =− dx x 2 x 35.
x2 + y2 = ( x 2 + y 2 )
1/ 2
1/ 2 d d x2 + y2) = x ( dx dx (1/ 2 ) − 1 d 1 2 x + y2) x2 + y 2) =1 ( ( 2 dx − 1/ 2 1 2 dy ⎞ ⎛ x + y2) ( ⎜ 2x + 2 y ⎟ = 1 dx ⎠ 2 ⎝ 1/ 2 dy 2x + 2y = 2 (x2 + y 2) dx 1/ 2 dy =( x 2 + y 2 ) − x y dx 2 2 dy ( x + y ) = dx y
37.
d 1/ 3 d x + y 1/ 3 ) = 1 ( dx dx 1 ( 1/ 3) − 1 1 (1/ 3) − 1 x + y =0 3 3 dy =0 x− 2 / 3 + y − 2 / 3 dx dy = − x− 2 / 3 y −2/3 dx dy = − x− 2 / 3 y dx
2/3
=−
y 2/3 x 2/3
1/ 2
−x
=
x2 + y2 − x y
x 2 +1
270 SECTION 4.8
(
39.
d e dx e e
x
⋅
x
x
d x +e dx
1 2 x
d y =0 dx 1 dy ⋅ =0 2 y dx
y
+e
e
) = dxd 4
y
+e
x
x
y
+
e e
Use the Chain Rule. Differentiate.
y
dy =0 y dx
Simplify.
y
dy e x =− y dx x y ye dy e x =− ⋅ y =− dx x e xe
x y
=−
ye
x−
y
x
41. (a) The domain of f is {x | x ≥ 0} or on the interval [0, ∞).. 1 (b) f ′ ( x ) = 2 x (c) The domain of f′ (x) is {x | x > 0} or on the interval (0, ∞).
(d) f′ (x) is never equal to 0. (e) x = 0 is in the domain of f, but not in the domain of f′ (x). (f) f ′′ ( x ) =
d ⎛ 1 ⎞ 1 d − 1/ 2 1 ⎛ 1 ⎞ ( − 1/ 2) − 1 1 1 x = ⋅ ⎜− ⎟x = − x −3/ 2 = − ⎜ ⎟= dx ⎝ 2 x ⎠ 2 dx 2 ⎝ 2⎠ 4 4x 3 / 2
(g) The domain of f′ ′(x) is {x | x > 0} or on the interval (0, ∞). 43. (a) The domain of f is all real numbers or the interval (−∞, ∞).
(b) f ′ ( x ) =
d 2 / 3 2 ( 2 / 3) − 1 2 − 1/ 3 2 2 x = x = x = 1/ 3 = 3 dx 3 3 3x 3 x
(c) The domain of f′ (x) is all real numbers except x = 0, that is the set {x | x ≠ 0}. (d) f′ (x) is never equal to 0. (e) x = 0 is in the domain of f, but not in the domain of f′ (x). (f) f ′′ ( x ) =
d ⎛ 2 −1/ 3 ⎞ 2 ⎛ 1 ⎞ ( − 1/ 3) − 1 2 2 = − x −4/3 = − ⎜ x ⎟ = ⋅ ⎜− ⎟x dx ⎝ 3 9 9x 4 / 3 ⎠ 3 ⎝ 3⎠
(g) The domain of f′ ′(x) is all real numbers except x = 0, that is the set {x | x ≠ 0}.
SECTION 4.8 271 45. (a) The domain of f is all real numbers or the interval (−∞, ∞).
(b) f ′ ( x ) =
d 2 2 / 3 −1 1 1/ 3 − 1 x 2 / 3 + 2 x 1/ 3 ) = x ( ) + 2 ⋅ x ( ) ( dx 3 3 2 − 1/ 3 2 − 2 / 3 2 2 = x + x = + 1/ 3 3 3 3x 3x 2 / 3
(c) The domain of f′ (x) is all real numbers except x = 0, that is the set {x | x ≠ 0}. (d) f′ (x) is never equal to zero. (e) x = 0 is in the domain of f, but not in the domain of f′ (x). d ⎛ 2 − 1/ 3 2 − 2 / 3 ⎞ 2 ⎛ 1 ⎞ ( − 1/ 3) − 1 2 ⎛ 2 ⎞ ( − 2 / 3) − 1 + x + ⋅ ⎜− ⎟x ⎜ x ⎟ = ⋅ ⎜− ⎟x 3 3 ⎝ 3⎠ dx ⎝ 3 ⎠ 3 ⎝ 3⎠ 2 4 2 4 = − x −4/3 − x −5/3 = − 4/3 − 5/3 9 9 9x 9x
(f) f ′′ ( x ) =
(g) The domain of f′ ′(x) is all real numbers except x = 0, that is the set {x | x ≠ 0}. 47. (a) The domain of f is all real numbers or the interval (−∞, ∞).
(b) f ′ ( x ) =
2/3 − 1/ 3 ( 2 / 3) − 1 d d 2 2 x 2 − 1) = ( x 2 − 1) x 2 − 1) = ( x 2 − 1) ⋅ 2x ( ( dx dx 3 3 4x = 1/ 3 3 ( x 2 − 1)
(c) The domain of f′ (x) is all real numbers except x = 1 and x = – 1, that is the set {x | x ≠ 1 and x ≠ – 1}. (d) f′ (x) = 0 when x = 0. (e) x = 1 and x = – 1are in the domain of f, but not in the domain of f′ (x).
(f) f ′′ ( x ) =
d 4x 2 dx 3 ( x − 1)1/ 3
1/ 3 d 1/ 3 ⎤ d ⎡ 2 x − 1) x − x ( x 2 − 1) ⎥ ( ⎢ 4 dx dx = ⎢ ⎥ 2/3 2 3⎢ ( x − 1) ⎥ ⎣ ⎦ 1/ 3 1/ 3 − 1 ( ) d 2 1 ⎡ 2 ⎤ x − 1) − x ⋅ ( x 2 − 1) x − 1) ⎥ ( 4 ⎢( dx 3 = ⎢ ⎥ 2/3 2 3⎢ ( x − 1) ⎥ ⎣ ⎦
272 SECTION 4.8 1/ 3 −2/3 x 2 ⎡ 2 ⎤ − − − ⋅ 2x ⎥ x 1 x 1 ( ) ( ) 4⎢ 3 = ⎢ ⎥ 2/3 2 3⎢ ( x − 1) ⎥ ⎣ ⎦ 2 1/ 3 ⎡ 2 ⎤ 2x ⎢ ( x − 1) − ⎥ 2 / 3 3 ( x 2 − 1) 4⎢ ⎥ = ⎢ 2 / 3 ⎥ 3 ( x2 − 1) ⎢ ⎥ ⎢ ⎥ ⎣ ⎦ 2 2 ⎡ ⎤ 4 ( x 2 − 3) 4 ⎢ 3 ( x − 1) − 2 x ⎥ = = 3 ⎢ 3 ( x 2 − 1) 4 / 3 ⎥ 9 ( x 2 − 1) 4 / 3 ⎣ ⎦
(g) The domain of f′ ′(x) is all real numbers except x = 1 and x = – 1, that is the set {x | x ≠ 1 and x ≠ – 1}. 49. (a) Since 1 − x 2 ≥ 0 , 1 – x 2 ≥ 0 or x 2 ≤ 1. Solving for x, we get – 1 ≤ x ≤ 1, so the domain of f is the interval [– 1, 1].
(b) f ′ ( x ) =
)
(
1/ 2 1/ 2 1/ 2 d d d ⎡ d x 1− x 2 = x (1 − x 2 ) ⎤ = x (1 − x 2 ) + (1 − x 2 ) x ⎢ ⎥ ⎦ dx dx ⎣ dx dx 1/ 2 (1/ 2 ) −1 d 1 = x ⋅ (1 − x 2 ) 1 − x 2 ) + (1 − x 2 ) ⋅ 1 ( 2 dx − 1/ 2 1/ 2 x = (1 − x 2 ) ( −2 x ) + (1 − x 2 ) 2 − x 2 + (1 − x 2 ) 1 − 2x 2 −x 2 2 1/ 2 1 = + − x = = ( ) 1/ 2 1/ 2 1/ 2 (1 − x 2 ) (1 − x 2 ) (1 − x 2 )
(c) Since (1 − x 2 )
1/ 2
> 0, 1 – x 2 > 0 or x 2 < 1. Solving for x, we get – 1 < x < 1, so the
domain of f′ (x) is all real numbers or the interval (– 1, 1). (d) f′ (x) = 0 when 1 − 2x 2 = 0, or when x 2 =
1 1 or x = , or x = – 2 2
1 . 2
(e) The points x = – 1 and x = 1 are in the domain of f, but are not part of the domain of f′ .
(f) f ′′ ( x ) =
d 1 − 2x = dx (1 − x 2 ) 1/ 2 2
(1 − x )
2 1/ 2
1/ 2 d d 1 − 2 x 2 ) − (1 − 2 x 2 ) (1 − x 2 ) ( dx dx
((1 − x )
2 1/ 2
)
2
SECTION 4.8 273 ( ) d (1 − x ) ( −4 x ) − (1 − 2 x ) 12 (1 − x ) (1 − x ) dx = (1 − x ) 1 −4 x (1 − x ) − (1 − 2 x ) (1 − x ) ( −2 x ) 2 = (1 − x ) −4 x (1 − x ) + ( x − 2 x )(1 − x ) = (1 − x ) −4 x (1 − x ) + ( x − 2 x ) = (1 − x ) 2 1/ 2
2
2
1/ 2 − 1
2
2
2 1/ 2
2
2
− 1/ 2
2
2 1/ 2
3
2
− 1/ 2
2
2
3
3/ 2
2
=
(g) Since (1 − x 2 )
3/ 2
−4 x + 4 x 3 + x − 2 x 3
(1 − x ) 2
3/ 2
=
2 x 3 − 3x
(1 − x ) 2
3/ 2
> 0, 1 – x 2 > 0 or x 2 < 1. Solving for x, we get – 1 < x < 1, so the
domain of f′′ (x) is all real numbers or the interval (– 1, 1). 51.
(a) N ′ ( t ) =
(
)
d ⎛ 10,000 −1/ 2 ⎞ d − + 11,000 ⎟ = − 10,000 (1 + 0.1t ) + 11,000 ⎜ dx ⎝ 1 + 0.1t ⎠ dx d d −1/ 2 = − 10,000 (1 + 0.1t ) + 11,000 dx dx ( −1/ 2 ) − 1 d ⎛ 1⎞ = − 10,000 ⋅ ⎜ − ⎟ (1 + 0.1t ) (1 + 0.1t ) dx ⎝ 2⎠ 500 −3 / 2 = 5,000 (1 + 0.1t ) ( 0.1) = 3/ 2 (1 + 0.1t )
(b) In 10 years, t = 10. N ′ (10 ) =
500
(1 + 0.1(10 ) )
3/ 2
= 176.777 = 177 students
53. Since z is a constant, we write z = K. Then the production function becomes K = x 0.5y 0.4 dy We will find using implicit differentiation. dx d d K = ( x 0.5 y 0.4 ) dx dx d 0.4 d 0 = x 0.5 y + y 0.4 x 0.5 dx dx dy 0 = x 0.5 ⋅ 0.4 y 0.4 − 1 + y 0.4 ⋅ 0.5 x 0.5 − 1 dx
274 CHAPTER 4 REVIEW
dy + 0.5 y 0.4 x − 0.5 dx dy 0.5 y 0.4 + 0.5 dx x
0 = 0.4 x 0.5 y − 0.6 0.4 x 0.5 0= y 0.6
0.4 x 0.5 dy 0.5 y 0.4 = − y 0.6 dx x 0.5
dy 0.5 y 0.4 y 0.6 0.5 y 0.4 +.06 10 5y =− ⋅ = − ⋅ =− 0.5 0.5 0.5 + 0.5 dx x 0.4 x 0.4 x 10 4x 55. (a) The instantaneous rate of pollution is the derivative of the function A.
1/ 4 3 2 d 2 d 1 (1/ 4) − 1 3 ( t + 3) 1/ 4 1/ 4 3 3 A′ ( t ) = ( t 1/ 4 + 3) = 3 ( t 1/ 4 + 3) t + 3 = t + ⋅ t = ( ) ( ) 4 4t 3 / 4 dt dt
2
(b) After 16 years the rate of A′(16) = 57.
3 (16 1/ 4 + 3) 4 ⋅ 16 3 / 4
2
=
3 ⋅ 52 = 2.344 units per year. 32
(a) Velocity is the derivative of the distance function s. d 3 / 2 3 ( 3 / 2) −1 3t 1/ 2 = feet per second. v = s′ ( t ) = t = t dt 2 2 3 ⋅ 11/ 2 3 = = 1.5 feet per second. After 1 second the child has a velocity of v(1) = 2 2 (b) If the slide is 8 feet long, it takes t 3/ 2 = 8
(t )
3/ 2 2/3
= (8)
2/3
= 4 seconds to get down the slide and strike the ground.
The velocity the child when striking the ground is 3 ⋅ 4 1/ 2 v ( 4) = = 3 feet per second. 2
Chapter 4 Review TRUE-FALSE ITEMS 1. True
3. True
5. False
7. True
FILL-IN-THE-BLANKS 1. tangent
3. Power Rule; Chain Rule
5. zero
274 CHAPTER 4 REVIEW
dy + 0.5 y 0.4 x − 0.5 dx dy 0.5 y 0.4 + 0.5 dx x
0 = 0.4 x 0.5 y − 0.6 0.4 x 0.5 0= y 0.6
0.4 x 0.5 dy 0.5 y 0.4 = − y 0.6 dx x 0.5
dy 0.5 y 0.4 y 0.6 0.5 y 0.4 +.06 10 5y =− ⋅ = − ⋅ =− 0.5 0.5 0.5 + 0.5 dx x 0.4 x 0.4 x 10 4x 55. (a) The instantaneous rate of pollution is the derivative of the function A.
1/ 4 3 2 d 2 d 1 (1/ 4) − 1 3 ( t + 3) 1/ 4 1/ 4 3 3 A′ ( t ) = ( t 1/ 4 + 3) = 3 ( t 1/ 4 + 3) t + 3 = t + ⋅ t = ( ) ( ) 4 4t 3 / 4 dt dt
2
(b) After 16 years the rate of A′(16) = 57.
3 (16 1/ 4 + 3) 4 ⋅ 16 3 / 4
2
=
3 ⋅ 52 = 2.344 units per year. 32
(a) Velocity is the derivative of the distance function s. d 3 / 2 3 ( 3 / 2) −1 3t 1/ 2 = feet per second. v = s′ ( t ) = t = t dt 2 2 3 ⋅ 11/ 2 3 = = 1.5 feet per second. After 1 second the child has a velocity of v(1) = 2 2 (b) If the slide is 8 feet long, it takes t 3/ 2 = 8
(t )
3/ 2 2/3
= (8)
2/3
= 4 seconds to get down the slide and strike the ground.
The velocity the child when striking the ground is 3 ⋅ 4 1/ 2 v ( 4) = = 3 feet per second. 2
Chapter 4 Review TRUE-FALSE ITEMS 1. True
3. True
5. False
7. True
FILL-IN-THE-BLANKS 1. tangent
3. Power Rule; Chain Rule
5. zero
CHAPTER 4 REVIEW 275 REVIEW EXERCISES 1. f ′(x) = 2; f ′(2) = 2
3. f ′(x) = 2x; f ′(2) = 2 · 2 = 4
5. f ′(x) = 2x – 2; f ′(1) = 2 · 1 – 2 = 0 7.
9.
11.
f ′(x) = e 3 x
d ( 3x ) = 3e 3 x ; f ′(0) = 3e 0 = 3 dx
⎡ 4 ( x + h ) + 3⎤⎦ − [ 4 x + 3] 4 x + 4h + 3 − 4 x − 3 4h = lim 4 = 4 f ′ ( x ) = lim ⎣ = lim = lim h→0 h → 0 h → 0 h→ 0 h h h ⎡ 2 ( x + h ) 2 + 1⎤ − ⎡ 2 x 2 + 1⎤ 2 x 2 + 4 xh + 2h 2 − 2 x 2 − 1 ⎣ ⎦ ⎣ ⎦ ′ f ( x ) = lim = lim h →0 h→0 h h h ( 4 x + 2h ) = lim = lim ( 4 x + 2h ) = lim 4 x + lim 2h = 4 x + 0 = 4 x h →0 h→0 h→0 h→0 h
13. f ′(x) = 5x 4
15.
17. f ′(x) = 2 · 2x – 3 = 4x – 3
19.
f ′(x) =
1 d 4 1 x = ⋅ 4x 3 = x 3 4 dx 4
f ′(x) = 7
d ( x 2 − 4 ) = 7 ⋅ 2 x = 14 x dx
d d ⎡ ⎤ 21. f ′(x) = 5 ⎢( x 2 − 3 x ) ( x − 6 ) + ( x − 6 ) ( x 2 − 3 x ) ⎥ = 5 ⎡⎣ x 2 − 3 x + ( x − 6 )( 2 x − 3) ⎤⎦ dx dx ⎣ ⎦ 2 2 2 = 5 ( x − 3 x + 2 x − 15 x + 18 ) = 5 ( 3x − 18 x + 18 ) = 15 ( x 2 − 6 x + 6 ) d ⎡ d 12 x ( 8 x 3 + 2 x 2 − 5 x + 2 ) ⎤⎦ = 12 ( 8 x 4 + 2 x 3 − 5 x 2 + 2 x ) ⎣ dx dx 3 2 = 12 ( 32 x + 6 x − 10 x + 2 ) = 24 (16 x 3 + 3 x 2 − 5 x + 1)
23. f ′(x) =
d 2x + 2 ( 25. f ′(x) = = dx 5 x − 3 =
5 x − 3)
d d ( 2 x + 2 ) − ( 2 x + 2 ) ( 5 x − 3) dx dx 2 ( 5 x − 3)
( 5 x − 3) ⋅ 2 − ( 2 x + 2 ) ⋅ 5 = 10 x − 6 − 10 x − 10 = −16 2 2 2 ( 5 x − 3) ( 5 x − 3) ( 5 x − 3)
– 13 = – 24 x – 13 = − 27. f ′(x) = 2 · (– 12) x
29. f ′(x) =
24 x 13
d ⎛ 3 4 ⎞ d 3 8 −1 −2 −2 −3 ⎜ 2 + + 2 ⎟ = ( 2 + 3x + 4 x ) = − 3x − 8 x = − 2 − 3 dx ⎝ x x ⎠ dx x x
276 CHAPTER 4 REVIEW
31. f ′(x) = d ⎛ 3 x − 2 ⎞ = ⎜ ⎟ dx ⎝ x + 5 ⎠
d dx
d dx
( x + 5) ( 3x − 2 ) − ( 3x − 2 ) ( x + 5 ) Derivative of a quotient.
( x + 5) ( x + 5)( 3) − ( 3x − 2 )(1) = 3x + 15 − 3x + 2 = 17 = 2 2 2 ( x + 5) ( x + 5) ( x + 5)
33. f ′(x) =
2
5 4 d d 3x 2 − 2 x ) = 5 ( 3x 2 − 2 x ) 3x 2 − 2 x ) ( ( dx dx
= 5 ( 3x 2 − 2 x )
4
( 6 x − 2 ) = 10 ( 3x 2 − 2 x ) ( 3x − 1) 4
d dx
⎡7 x ( x 2 + 2 x + 1) 2 ⎤ ⎢⎣ ⎥⎦ 2 2 d d = 7 x ( x 2 + 2 x + 1) + ( x 2 + 2 x + 1) (7x) dx dx 2 d = 7 x ⋅ 2 ( x 2 + 2 x + 1) x 2 + 2 x + 1) + ( x 2 + 2 x + 1) ⋅ 7 ( dx
35. f ′(x) =
Use the Power Rule.
= 14 x ( x 2 + 2 x + 1) ( 2 x + 2 ) + 7 ( x 2 + 2 x + 1)
2
= 7 ( x 2 + 2 x + 1) ⎡⎣ 2 x ( 2 x + 2 ) + ( x 2 + 2 x + 1) ⎤⎦
The derivative of a product. Use the Power Rule. Differentiate. Factor.
= 7 ( x 2 + 2 x + 1) ⎡⎣ 4 x 2 + 4 x + x 2 + 2 x + 1⎤⎦
Simplify in the brackets.
= 7 ( x 2 + 2 x + 1) ⎡⎣ 5 x 2 + 6 x + 1⎤⎦
Simplify in the brackets.
= 7 ( x + 1) ⎡⎣( 5 x + 1)( x + 1) ⎤⎦ = 7 ( x + 1) ( 5 x + 1) 2
3
Factor.
2
37. f ′(x) = d ⎛⎜ x + 1 ⎞⎟ = 2 ⎛⎜ x + 1 ⎞⎟ d ⎛⎜ x + 1 ⎞⎟ dx ⎝ 3 x + 2 ⎠ ⎝ 3 x + 2 ⎠ dx ⎝ 3x + 2 ⎠ d d ⎡ ⎤ 3 x + 2 ) ( x + 1) − ( x + 1) ( 3 x + 2 ) ⎥ ( ⎢ ⎛ x +1 ⎞ dx dx = 2⎜ ⎥ ⎟⎢ 2 ⎝ 3x + 2 ⎠ ⎢ ( 3x + 2 ) ⎥ ⎣ ⎦ ⎛ x + 1 ⎞ ⎡ ( 3 x + 2 ) ⋅ 1 − ( x + 1) ⋅ 3 ⎤ = 2⎜ ⎥ ⎟⎢ 2 ⎝ 3 x + 2 ⎠ ⎢⎣ ( 3x + 2 ) ⎥⎦ 2 ( x + 1)( 3x + 2 − 3x − 3) = 3 ( 3x + 2 )
=−
2 ( x + 1)
( 3x + 2 )
3
Use the Power Rule.
Derivative of a quotient.
Differentiate.
Simplify.
Chapter 4 – The Derivative of a Function Section 4.1 The Definition of a Derivative In Problems 45–54, find the derivative of each function at the given number using a graphing utility. 45.
f ( x ) = 3 x3 − 6 x 2 + 2 at −2 .
We can find the derivative of a function at a point using the nDeriv( function on the graphing calculator. The format of the command is nDeriv(function, variable, value)
The nDeriv( function is found under the menu.
}}}
Select the nDeriv( function. Enter the expression for f ( x ) , the name of the variable, and the limit point, separated by commas. Find the value of the derivative. . Í„›Â¹¸„¡ÃÁ¢ „¢ÌÁ¤Í
79
Thus, rounding the result obtained on the calculator, we have f ' ( −2 ) ≈ 60 .
47.
− x3 + 1 f ( x) = 2 at 8 x + 5x + 7
Use nDeriv( to find the value of the derivative at the given point.
Thus, f ' ( 8 ) ≈ −0.859 .
49.
f ( x ) = xe x at 0
Use nDeriv( to find the value of the derivative at the given point.
Thus, rounding the result obtained on the calculator, we have f ' ( 0 ) ≈ 1 .
51.
f ( x ) = x 2 e x at 1
Use nDeriv( to find the value of the derivative at the given point.
80
Thus, f ' (1) ≈ 8.155 . 53.
f ( x ) = xe − x at 1
Use nDeriv( to find the value of the derivative at the given point.
Recall that the output 1.22625E-7 is the number 1.22625 ×10−7 . Thus, rounding the result obtained on the calculator, we have f ' (1) ≈ 0 . ________________________________________________________________________
Section 4.5 The Derivatives of the Exponential and Logarithmic Functions; the Chain Rule 73.
Maximizing Profit At the Super Bowl, the demand for game-day t-shirts is given by x p = 50 − 4 ln + 1 100 where p is the price of the shirt in dollars and x is the number of shirts demanded. (k)
Use the TABLE feature of a graphing utility to find the quantity x that maximizes profit.
81
Thus, f ' (1) ≈ 8.155 . 53.
f ( x ) = xe − x at 1
Use nDeriv( to find the value of the derivative at the given point.
Recall that the output 1.22625E-7 is the number 1.22625 ×10−7 . Thus, rounding the result obtained on the calculator, we have f ' (1) ≈ 0 . ________________________________________________________________________
Section 4.5 The Derivatives of the Exponential and Logarithmic Functions; the Chain Rule 73.
Maximizing Profit At the Super Bowl, the demand for game-day t-shirts is given by x p = 50 − 4 ln + 1 100 where p is the price of the shirt in dollars and x is the number of shirts demanded. (k)
Use the TABLE feature of a graphing utility to find the quantity x that maximizes profit.
81
x The profit function is given by P ( x ) = 46 x − 4 x ln + 1 . Enter the formula for the 100 profit function into Y1 in the function editor. Experiment with a viewing rectangle until your graph shows the maximum.
Go to TABLE and enter values for x to find the maximum profit.
While there appears to be no change in the y-values when x is between 3,300,000 and 4,000,000, the table window is rounding all values to three significant digits. Move the cursor to the second column to investigate the y-values.
82
Thus, the value of x that maximizes the profit is between 3,500,00 and 3,700,000. Trying values for x in this interval we obtain the following table.
Thus, the value of x that maximizes the profit is between 3,600,00 and 3,700,000. Trying values for x in this interval we obtain the following table.
Thus, the value of x that maximizes the profit is between 3,620,00 and 3,640,000. Trying values for x in this interval we obtain the following table.
83
Thus, the value of x that maximizes the profit is between 3,631,00 and 3,633,000. We can continue to narrow down the x-interval that contains the maximum, but this is a rather tedious process. A quicker way uses the maximum function found in the menu.
84
Sell 3,631,550 t-shirts in order to maximize the profit. (l)
What price should be charged for a t-shirt to maximize the profit?
Substitute the value 3,631,550 in for x in the demand equation.
Rounding to the nearest cent, the price per t-shirt should be $8.00.
________________________________________________________________________
85
Summary The command introduced in this chapter was: nDeriv(
86
Chapter 5 Applications: Graphing Functions; Optimization 5.1 Horizontal and Vertical Tangent Lines; Continuity and Differentiability 1.
f ′ ( x ) = 2x − 4
Horizontal tangent lines occur where f ′(x) = 0. 2x – 4 = 0 2x = 4 x=2 Evaluate the function f at x = 2: f (2) = (2) 2 – 4(2) = – 4 The tangent line to the graph of f is horizontal at the point (2, f (2)) = (2, – 4). There is no vertical tangent line since f′ (x) is never unbounded. 3. f ′(x) = – 2x + 8 Horizontal tangent lines occur where f ′(x) = 0. – 2x + 8 = 0 – 2x = – 8 x=4 Evaluate the function f at x = 4: f (4) = – (4) 2 + 8(4) = 16 The tangent line to the graph of f is horizontal at the point (4, f (4)) = (4, 16). There is no vertical tangent line since f′ (x) is never unbounded. 5. f ′(x) = – 4x + 8 Horizontal tangent lines occur where f ′(x) = 0. – 4x + 8 = 0 – 4x = – 8 x=2 Evaluate the function f at x = 2: f (2) = – 2(2) 2 + 8(2) + 1 = – 8 + 16 + 1 = 9 The tangent line to the graph of f is horizontal at the point (2, f (2)) = (2, 9). There is no vertical tangent line since f′ (x) is never unbounded. 7.
2 2 ⋅ 3x − 1 / 3 = 1 / 3 3 x f ′(x) is never zero, so there is no horizontal tangent line.
f ′(x) =
286 SECTION 5.1
f ′(x) is unbounded at x = 0. We evaluate f at 0: f (0) = 3 · 0 2/3 + 1 = 1. The tangent line to the graph of f is vertical at the point (0, 1). 9. f ′(x) = – 3x 2 + 3 Horizontal tangent lines occur where f ′(x) = 0. – 3x 2 + 3 = 0 x2 – 1 = 0 (x – 1)(x + 1) = 0 x – 1 = 0 or x + 1 = 0 x = 1 or x=–1
Evaluate f at 1 and – 1. f (1) = − 1 3 + 3 (1) + 1 = 3
f ( − 1) = − ( − 1) + 3 ( − 1) + 1 = − 1 3
There are two horizontal lines tangent to the graph of f , one at (1, 3) and the other at (– 1, – 1). There is no vertical tangent line since f′ (x) is never unbounded. 11.
3 3 ⋅ 4x −1 / 4 = 1 / 4 4 x f ′(x) ≠ 0 so there is no horizontal tangent line. 3/ 4 f ′(x) is unbounded at x = 0. We evaluate f (0) = 4 ( 0 ) − 2 = − 2. f ′ ( x) =
The tangent line to the graph of f at (0, – 2) is vertical. 13.
f ′ ( x ) = 5 x 4 − 40 x 3
Horizontal tangents occur where f ′(x) = 0. 5 x 4 − 40 x 3 = 0 5x 3 ( x − 8 ) = 0
5x 3 = 0 x=0
or or
x–8=0 x=8
Evaluate f at 0 and 8. 5 4 f ( 0 ) = ( 0 ) − 10 ( 0 ) = 0
f ( 8 ) = ( 8 ) − 10 ( 8 ) = − 8192 5
4
We conclude that the graph of f has 2 horizontal tangent lines, one at (0, 0), the other at (8, – 8192). There is no vertical tangent line. 15.
f ′ ( x ) = 15 x 4 + 60 x 2
Horizontal tangent lines occur where f ′(x) = 0. 15 x 4 + 60 x 2 = 0
SECTION 5.1 287
15 x 2 ( x 2 + 4 ) = 0
15x 2 = 0 or x=0
x2 + 4 = 0
Evaluate f at 0: f ( 0 ) = 3 ( 0 ) + 20 ( 0 ) − 1 = − 1 5
3
The graph of f has a horizontal tangent line at (0, – 1); f ′ is never unbounded, so f has no vertical tangent. 17.
2 −1/ 3 1 2 2 + ⋅ 2 x − 2 / 3 = 1/ 3 + 2 / 3 x 3 3 3x 3x 1/ 3 1/ 3 2 x 2 2x 2 2 x 1/ 3 + 2 = 1/ 3 ⋅ 1/ 3 + 2 / 3 = 2 / 3 + 2 / 3 = 3x x 3x 3x 3x 3x 2 / 3
f ′( x ) =
Horizontal tangent lines occur where f′ (x) = 0. 2 x 1/ 3 + 2 =0 3x 2 / 3 2 x 1/ 3 + 2 = 0 x 1/ 3 = – 1 x=–1 Vertical tangent lines occur where f′ (x) is unbounded. f′ (x) is unbounded at 0. Evaluate f at 0 and – 1: 1/ 3 f ( 0) = 0 2 / 3 + 2 ( 0) = 0
f ( −1) = ( −1)
2/3
+ 2 ( −1)
1/ 3
= −1
We conclude that the graph of f has a horizontal tangent line at (– 1, – 1) and a vertical tangent line at (0, 0). 19.
f ( x ) = x 2 / 3 ( x − 10 ) = x 5 / 3 − 10 x 2 / 3
5 2 5 x 2/3 f ′ ( x ) = x 2 / 3 − ⋅ 10 x −1/ 3 = 3 3 3 5 x 2/3 = 3
20 3x 1/ 3 x 1/ 3 20 5x 20 5 x − 20 ⋅ 1/ 3 − 1/ 3 = 1/ 3 − 1/ 3 = 3x 3x 3x 3x 1/ 3 x −
Horizontal tangent lines occur where f′ (x) = 0. 5 x − 20 =0 3x 1/ 3 5x – 20 = 0
288 SECTION 5.1
5x = 20 x=4 Vertical tangent lines occur where f′ (x) is unbounded. f′ (x) is unbounded at 0. Evaluate f at 0 and 4: f ( 0 ) = 0 2 / 3 ( 0 − 10 ) = 0 f ( 4 ) = 4 2 / 3 ( 4 − 10 ) = − 6 ⋅ 4 2 / 3 = − 6 ⋅ ( 8 ⋅ 2 )
1/ 3
= − 6 ⋅ 2 ⋅ 2 1/ 3 = − 12 ⋅ 2 1/ 3 = − 12 3 2
We conclude that the graph of f has a horizontal tangent line at (4, −12 3 2 ) and a vertical tangent line at (0, 0). 21.
f ( x ) = x 2 / 3 ( x 2 − 16 ) = x 8 / 3 − 16 x 2 / 3 8 2 8 x 5/3 32 − 1/ 3 f ′ ( x ) = x 5 / 3 − ⋅ 16 x −1/ 3 = 3 3 3 3x =
8 x 5 / 3 x 1/ 3 32 8x2 32 8 x 2 − 32 ⋅ 1/ 3 − 1/ 3 = 1/ 3 − 1/ 3 = 3 3x 3x 3x 3 x 1/ 3 x
Horizontal tangent lines occur where f′ (x) = 0. 8 x 2 − 32 =0 3x 1/ 3 8x 2 – 32 = 0 8(x 2 – 4) = 0 (x – 2)(x + 2) = 0 x – 2 = 0 or x + 2 = 0 x = 2 or x=–2 Vertical tangent lines occur where f′ (x) is unbounded. f′ (x) is unbounded at 0. Evaluate f at 2, – 2, and 0: f ( 2 ) = 2 2 / 3 ( 2 2 − 16 ) = 2 2 / 3 ( −12 ) = −12 ⋅ 2 2 / 3 = −12 3 4 f ( − 2) = ( − 2)
2/3
(( − 2)
2
f ( 0 ) = 0 2 / 3 ( 0 2 − 16 ) = 0
)
− 16 = 2 2 / 3 ( −12 ) = −12 ⋅ 2 2 / 3 = −12 3 4
(
)
We conclude that the graph of f has horizontal tangent lines at 2, − 12 3 4 and
( − 2, − 12 4 ) and a vertical tangent line at (0, 0). 3
23.
f ( x) =
x 2/3 x−2
x≠2
SECTION 5.1 289
2 −1/ 3 x ⋅ ( x − 2) − x 2 / 3 ⋅ 1 ⎡ 2 ( x − 2) ⎤ 1 f ′( x) = 3 = − x 2/3 ⎥ 2 2 ⎢ 1/ 3 ( x − 2) ( x − 2 ) ⎣ 3x ⎦ =
= =
1
( x − 2)
2
1
( x − 2)
2
1
( x − 2)
2
1/ 3 ⎡ 2 ( x − 2) ⎤ 2 / 3 3x x − ⋅ ⎢ 1/ 3 1/ 3 ⎥ 3x ⎦ ⎣ 3x
⎡ 2 ( x − 2) 3x ⎤ − 1/ 3 ⎥ ⎢ 1/ 3 3x ⎦ ⎣ 3x −x−4 ⎡ 2 x − 4 − 3x ⎤ 2 ⎢⎣ 3 x 1/ 3 ⎥⎦ = 1/ 3 3x ( x − 2 )
Horizontal tangent lines occur where f′ (x) = 0. −x−4 =0 2 1/ 3 3x ( x − 2 ) –x–4=0 x=–4
Vertical tangent lines occur where f′ (x) is unbounded. f′ (x) is unbounded if x = 0 or x = 2, but x = 2 is not part of the domain of f, so we disregard it. Evaluate f at – 4 and 0: 3 ( − 4 ) 2 / 3 16 1/ 3 2 ⋅ 2 1/ 3 2 1/ 3 2 f ( − 4) = = =− =− =− 6 3 3 ( − 4) − 2 − 6 f ( 0) =
0 2/3 =0 0−2
3 ⎛ 2⎞ We conclude that the graph of f has a horizontal tangent line at ⎜ − 4, − ⎟ and 3 ⎝ ⎠ a vertical tangent line at (0, 0).
25.
x 1/ 3 x≠1 x −1 1 −2 / 3 x ( x − 1) − x 1/ 3 ⋅ 1 1 = f ′( x ) = 3 2 2 ( x − 1) ( x − 1)
f ( x) =
= =
1
( x − 1)
2
1
( x − 1)
2
⎡ x −1 1/ 3 ⎤ ⎢⎣ 3x 2 / 3 − x ⎥⎦ 2/3 ⎡ x −1 ⎤ 1/ 3 3 x − ⋅ x ⎢ 3x 2 / 3 2/3 ⎥ 3x ⎦ ⎣
3x ⎤ ⎡ x −1 ⎢⎣ 3 x 2 / 3 − 3 x 2 / 3 ⎥⎦
290 SECTION 5.1
=
1
( x − 1)
2
− 2x −1 ⎡ x − 1 − 3x ⎤ 2 ⎢⎣ 3 x 2 / 3 ⎥⎦ = 2 / 3 3 x ( x − 1)
Horizontal tangent lines occur where f′ (x) = 0. −2 x − 1 =0 2 2/3 3x ( x − 1) – 2x – 1 = 0 1 x=– 2 Vertical tangent lines occur where f′ (x) is unbounded. f′ (x) is unbounded if x = 0 or x = 1, but x = 1 is not part of the domain of f, so we disregard it. 1 Evaluate f at – and 0: 2 1/ 3 1/ 3 ⎛ 1⎞ ⎛ 1⎞ 1/ 3 ⎜− ⎟ ⎜− ⎟ 1 2 2 2/3 3 4 2⎠ ⎛ 1⎞ ⎝ 2⎠ ⎛ 1⎞ ⎛ 2⎞ ⎝ f ⎜− ⎟ = = = ⎜ − ⎟ ⋅ ⎜ − ⎟ = 1/ 3 ⋅ = = 3 2 3 2 3 3 3 ⎝ 2 ⎠ − 1 −1 ⎝ ⎠ ⎝ ⎠ − 2 2 1/ 3 0 f ( 0) = =0 0 −1 ⎛ 1 34⎞ We conclude that the graph of f has a horizontal tangent line at ⎜⎜ − , ⎟⎟ and ⎝ 2 3 ⎠ a vertical tangent line at (0, 0). 27. (a)
f ( 0 ) = 0 2 / 3 = 0 . The one-sided limits are lim f ( x ) = lim− x 2 / 3 = 0
x → 0−
x→0
lim f ( x ) = lim+ x 2 / 3 = 0
x → 0+
x→0
Since lim f ( x ) = f ( 0 ) the function is continuous at 0. x→0
(b) The derivative of f at 0 is f ′ ( 0 ) = lim
x→0
f ( x ) − f (0) x−0
x 2/3 − 0 = lim . x→0 x−0
We look at the one-sided limits: x 2/3 − 0 x 2/3 1 lim− = lim− = lim− 1/ 3 = − ∞ x→0 x→0 x→0 x x−0 x 2/3 2/3 x −0 x 1 lim+ = lim+ = lim+ 1/ 3 = ∞ x→0 x→0 x→0 x x−0 x Since the one-sided limits are not equal, f′ (0) does not exist.
SECTION 5.1 291
(c) The derivative is unbounded at x = 0, so there is a vertical tangent line at 0. 29. (a) f (1) is not defined, so f is not continuous at x = 1.
(b) Since f is not continuous at x = 1, f′ (1) does not exist. 31. (a) f (0) = 0 2 = 0. The one sided limits are lim− f ( x ) = lim− 3 x = 0 lim+ f ( x ) = lim+ x 2 = 0 x→ 0
x→ 0
x→ 0
x→ 0
Since lim f ( x ) = f ( 0 ) , the function f is continuous at 0. x→ 0
(b) The derivative of f at 0 is f′ (0) = lim
f ( x ) − f (0) x−0
x→ 0
= lim
f ( x) − 0
x→ 0
x−0
.
We look at the one-sided limits: f ( x) − 0 3x − 0 3x lim− = lim− = lim− = lim 3 = 3 x→ 0 x→ 0 x−0 x − 0 x→ 0 x x→ 0 − f ( x) − 0 x2 −0 x2 lim+ = lim+ = lim+ = lim x = 0 x→ 0 x→ 0 x−0 x − 0 x→ 0 x x→ 0 + Since the one-sided limits are not equal, f′ (0) does not exist. (c) There is no tangent line at x = 0. (d)
33.
(a) f ( 2 ) = 4 ⋅ 2 = 8 . The one-sided limits are lim− f ( x ) = lim− 4 x = 4 ⋅ 2 = 8
x →2
x →2
lim+ f ( x ) = lim+ x 2 = 2 2 = 4
x→2
x →2
Since the one-sided limits are unequal, the lim f ( x ) doesn’t exist and the function is not x→2
continuous at x = 2.
292 SECTION 5.2
(b) The function is not continuous at x = 2, so f′ (2) does not exist.
(d)
(c) Does not apply to this problem.
35. (a) f (0) = 0 3 = 0. The one-sided limits are: lim− f ( x ) = lim− x 2 = 0 2 = 0 x→0
x→0
lim f ( x ) = lim+ x 3 = 0 3 = 0
x →0 +
x →0
The one-sided limits are equal so lim f ( x ) exists, and since lim f ( x ) = f (0), the function x →0
x →0
is continuous at x = 0. (b) The derivative of f at 0 is f ′ ( 0 ) = lim
x→0
f ( x ) − f (0) x−0
= lim
x→0
f ( x) − 0 x−0
= lim
x→0
f ( x) x
.
We look at one-sided limits. f ( x) f ( x) x3 x2 lim− = lim− = lim− x = 0 lim+ = lim+ = lim+ x 2 = 0 0 0 x →0 x → x → x → x → 0 0 x x x x x →0 We conclude that f′ (0) = 0, and that there is a horizontal tangent line at 0. (c) Does not apply to this problem.
(d)
5.2 Increasing and Decreasing Functions; the First Derivative Test 1. The domain of f is [x1, x9].
3. The graph of f is increasing on (x1, x4), (x5, x7), and (x8, x9).
5. f′ (x) = 0 for x4, x7, and x8.
7. f has a local maximum at (x4, y4) and at (x7, y7).
292 SECTION 5.2
(b) The function is not continuous at x = 2, so f′ (2) does not exist.
(d)
(c) Does not apply to this problem.
35. (a) f (0) = 0 3 = 0. The one-sided limits are: lim− f ( x ) = lim− x 2 = 0 2 = 0 x→0
x→0
lim f ( x ) = lim+ x 3 = 0 3 = 0
x →0 +
x →0
The one-sided limits are equal so lim f ( x ) exists, and since lim f ( x ) = f (0), the function x →0
x →0
is continuous at x = 0. (b) The derivative of f at 0 is f ′ ( 0 ) = lim
x→0
f ( x ) − f (0) x−0
= lim
x→0
f ( x) − 0 x−0
= lim
x→0
f ( x) x
.
We look at one-sided limits. f ( x) f ( x) x3 x2 lim− = lim− = lim− x = 0 lim+ = lim+ = lim+ x 2 = 0 0 0 x →0 x → x → x → x → 0 0 x x x x x →0 We conclude that f′ (0) = 0, and that there is a horizontal tangent line at 0. (c) Does not apply to this problem.
(d)
5.2 Increasing and Decreasing Functions; the First Derivative Test 1. The domain of f is [x1, x9].
3. The graph of f is increasing on (x1, x4), (x5, x7), and (x8, x9).
5. f′ (x) = 0 for x4, x7, and x8.
7. f has a local maximum at (x4, y4) and at (x7, y7).
SECTION 5.2 293 9.
f (x) = – 2x 2 + 4x – 2 STEP 1 The domain of f is all real numbers. STEP 2 Let x = 0. Then y = f (0) = – 2. The y-intercept is (0, – 2). Now let y = 0. Then – 2x 2 + 4x – 2 = 0 x 2 – 2x + 1 = 0 (x – 1) 2 = 0 x–1=0 x=1 The x-intercept is (1, 0). STEP 3 To find where the graph is increasing or decreasing, we find f′ (x): f′ (x) = – 4x + 4
The solution to f′ (x) = 0 is – 4x + 4 = 0 x=1 Use 1 to separate the number line into 2 parts, and use x = 0 and x = 2 as test numbers. – ∞ < x < 1 and 1 < x < ∞
f′ (0) = 4
f′ (2) = – 4(2) + 4 = – 4
We conclude that the graph of f is increasing on the interval (– ∞, 1); f is decreasing on the interval (1, ∞). STEP 4 Since f is increasing to the left of 1 and decreasing to the right of 1, we conclude that there is a local maximum at (1, f (1)) = (1, 0). STEP 5 We found f′ (1) = 0, indicating that there is a horizontal tangent at (1, 0). The first derivative is never unbounded, so there is no vertical tangent line. STEP 6 Since f is a polynomial function, its end behavior is that of y = – 2x 2. Polynomial functions have no asymptotes.
294 SECTION 5.2 11.
f ( x ) = x 3 − 9 x 2 + 27 x − 27
STEP 1 The domain of f is all real numbers. STEP 2 Let x = 0. Then y = f (0) = – 27. The y-intercept is (0, – 27). Now let y = 0. Then x 3 − 9 x 2 + 27 x − 27 = 0
( x − 3)
3
=0
x–3=0 x=3 The x-intercept is (3, 0). STEP 3 To find where the graph is increasing or decreasing, we find f′ (x): f ′ ( x ) = 3 x 2 − 18 x + 27
The solution to f′ (x) = 0 is 3x 2 − 18 x + 27 = 0 3( x 2 − 6 x + 9) = 0 3 ( x − 3) = 0 2
x–3=0 x=3 Use 3 to separate the number line into 2 parts, and use 0 and 4 as test numbers. –∞<x 3}
49. [2, ∞) = { x| x ≥ 2}
51. (−7, ∞) = { x| x > −7}
53.
55. (−∞, −20) = { x| < x−20}
57.
59. [3,5] = { x|3 ≤ x ≤ 5}
61.
63.
65. (−6, 0) = { x|−6 < x < 0}
67.
69.
71. (−1, 3) = { x|−1 < x < 3}
73. (−3, 3) = { x|−3 < x < 3}
75. (−∞, −4) (3, ∞) = { x| x < −4 or x > 3}
77. (−∞, −3) (4, ∞) = { x| x < −3 or x > 4}
79. {}
81. (1, ∞) = { x| x > 1}
83. (−∞, 1) (2, 3) = { x| x < 1 or 2 < x < 3}
85. (−2, 0) (4, ∞) = { x|−2 < x < 0 or x > 4}
87. (−1, 0) (1, ∞) = { x|−1 < x < 0 or x > 1}
89. (1, ∞) = { x| x > 1}
91. (−∞ −1) (1, ∞) = { x| x < −1 or x > 1}
93. (−∞, −1) (4, 1) = { x| x < −1 or 0 < x < 1}
95. (−1, 1) [2, ∞) = { x|−1 < x < 1 or x ≥ 2}
97. (−∞, 2) = { x| x < 2}
99. (−∞, −3) (−1, 1) (2, ∞) = { x| x < −3 or −1 < x < 1 or x > 2}
101. The solution is 74 ≤ x < 124, but assuming that the highest possible test score is 100, the range of possible exam scores that will enable you to earn a B is from 74 to 100.
103. The range of possible commissions is $45, 00 to $95, 00. The commission varies from 5% of the selling price to 8.6% of the selling price.
105. The amount withheld varies from $81.35 to $131.35.
107. Usage ranged from 657.41 kilowatt-hours to 2500.91 kilowatt-hours.
109. The dealer’s cost range from $7457.63 to $7857.14.
111. Answers will vary.
113. Answers will vary.
Exercise 0.6 1. 3
3. −2
5.
7.
9. x3 y2
11. x2 y
13.
15.
17.
19.
21.
23.
25.
27.
29.
31.
33.
35.
37.
39.
41.
43. {3}
45. 4
47. −3
49. 64
51.
53.
55.
57. x7/12
59. xy2
61. x4/3 y5/3
63.
65.
67.
69.
71.
73.
75.
77.
79.
81. 2x1/2(x + 1)(3x − 4)
83. (x2 + 4)1/3(11x2 + 12)
85. (2x + 3)1/2(3x + 5)1/3(17x + 27)
87.
89.
Exercise 0.7
1. c = 13
3. c = 26
5. c = 25
7. This is a right triangle. The hypotenuse is the side of length 5.
9. This is not a right triangle.
11. This is a right triangle. The hypotenuse is the side of length 25.
13. This is not a right triangle.
15. A = 8 inches2
17. A = 4 inches2
19. A = 25πm2, C = 10 m
21. V = 224 ft3, S = 232 ft2
23.
25. V = 648 π inches3, S = 306 π inches2
27. The area is π units2.
29. The area is 2 π units2.
31. The wheel travels 64 π inches after four revolutions.
33. The area is 64 ft2.
35. The area of the window is 24 + 2 π ft2 ≈ 30.28 ft2. 16 + 2π ≈ 22.28 ft of wood frame are needed to enclose the window.
37. You can see a distance of 28,920 ft, which is about 5.478 miles.
39. A person can see 64,667 ft or 12.248 miles from the deck. A person can see 79,200 ft or 15.0 miles from the bridge.
41. The areas of the rectangular pools vary from 0 ft2 to 62,500 ft2. The shape of the rectangle of largest area is a square with side length 250 ft. The area of the circular pool is ft2 ≈ 79, 577 ft2. The best choice for a pool of largest area would be the circular pool.
Exercise 0.8 1. (a) Quadrant II
(b) Positive x-axis
(c) Quadrant III
(d) Quadrant I
(e) Negative y-axis
(f)
Quadrant IV
3. The points will be on a vertical line that is 2 units to the right of the y-axis.
5.
7.
9.
11.
13.
15.
17.
19.
21.
23.
25. (2, 2); (2, −4)
27. (0, 0); (8, 0)
29.
31.
33.
35. (a) First base (90, 0); second base (90, 90); third base (0, 90)
(b)
(c)
37. d = 50 t
Exercise 0.9 1. x 0 −2 2 −2
y 4
0 8
4 −4
0 12 −4
3. x
0 3
2
−2 4
−4
y −6 0 −2 −10 2 −14
5. (a) Vertical line: x = 2
(b) Horizontal line: y = −3
7. (a) Vertical line: x = −4
(b) Horizontal line: y = 1
9. A slope of
means that for every 2 unit change in x, y changes 1 unit.
11. m = −1; A slope of −1 means that for every 1 unit change in x, y changes by (−1) units.
13. m = 3; A slope of 3 means that for every 1 unit change in x, y will change 3 units.
15. A slope of
means that for every 2 unit change in x, y will change (−1) unit.
17. m = 0; A slope of 0 means that regardless how x changes, y remains constant.
19. The slope is not defined.
21.
23.
25.
27.
29. x − 2y = 0
31. x+y=2
33. 2x − y = −9
35. 2x + 3y = −1
37. x − 2y = −5
39. 2x + y = 3
41. 3x − y = −12
43. 4x − 5y = 0
45. x − 2y = 2
47. x=1
49. y=4
51. slope: m = 2; y-intercept: (0, 3)
53. slope: m = 2; y-intercept: (0, −2)
55. slope:
y-intercept: (0, −2)
57. slope: m = 1; y-intercept: (0, 1)
59. Slope is not defined; there is no y-intercept.
61. slope: m = 0; y-intercept: (0, 5)
63. slope: m = 1; y-intercept: (0, 0)
65. slope:
; y-intercept: (0, 0)
67. y = −3
69. C = 0.122x
71. (a) C = 0.08275x + 7.58, 0 ≤ x ≤ 400
(b)
(c) The monthly charge for using 100 KWH is $15.86.
(d) The monthly charge for using 300 KWH is $32.41.
(e) The slope indicates that for every extra KWH used (up to 400 KWH), the electric bill increases by 8.275 cents.
73. w = 4h − 129
75. C = 0.53x + 1,070,000
77. (a)
(b) C = 20°
79. (a)
(b) A = 52.74 billion gallons
(c) The slope tells us that the reservoir loses 1 billion gallons of water every 75 days.
(d) A = 52.194 billion gallons
(e) In 10.89 years the reservoir will be empty.
(f)
Answers will vary.
81. Window: X min = −10; X max = 10 Y min = −10; Y max = 10 x-intercept: (1.67, 0); y-intercept: (0, 2.50)
83. Window: X min = −10; X max = 10 Y min = −10; Y max = 10 x-intercept: (2.52, 0); y-intercept: (0, −3.53)
85. Window: X min = −10; X max = 10 Y min = −10; Y max = 10 x-intercept: (2.83, 0); y-intercept: (0, 2.56)
87. Window: X min = −10; X max = 10; Y min = −10; Y max = 10 x-intercept: (0.78, 0); y-intercept: (0, −1.41)
89. (b)
91. (d)
93. y = x + 2 or x − y = −2
95.
97. (b), (c), (e), (g)
99. y=0
101. Answers will vary.
103. No; answers will vary; No; answers will vary.
105. The lines are identical.
107. Two lines can have the same y-intercept and the same x-intercept but different slopes only if their y-intercept is the point (0, 0).
CHAPTER 1 Functions and Their Graphs Exercise 1.1 1. (a) (3, −4)
(b) (−3, 4)
(c) (−3, −4)
3. (a) (−2, −1)
(b) (2, 1)
(c) (2, −1)
5. (a) (1, −1)
(b) (−1, 1)
(c) (−1, −1)
7. (a) (−3, 4)
(b) (3, −4)
(c) (3, 4)
9. (a) (0, 3)
(b) (0, −3)
(c) (0, 3)
11. (a) The x-intercepts are (−1, 0) and (1, 0). There is no y-intercept.
(b) The graph is symmetric with respect to the x-axis, y-axis, and the origin.
13. (a) The x-intercepts are
and
. The y-intercept is (0, 1).
(b) The graph is symmetric with respect to the y-axis.
15. (a) The x-intercept and the y-intercept are both (0, 0).
(b) The graph is symmetric with respect to the x-axis.
17. (a) The x-intercept is (1, 0). There is no y-intercept.
(b) The graph is not symmetric with respect to either axis or to the origin.
19. (a) The x-intercepts are (−1, 0) and (1, 0). The y-intercept is (0, −1).
(b) The graph is symmetric with respect to the y-axis.
21. (a) There are no intercepts.
(b) The graph is symmetric with respect to the origin.
23. The point (0, 0) is on the graph. The points (1, 1) and (−1, 0) are not on the graph.
25. The point (0, 3) is on the graph. The points (3, 0) and (−3, 0) are not on the graph.
27. The points (0, 2) and
are on the graph. The point (−2, 2) is not on the graph.
29. The x-intercept and the y-intercept are (0, 0). The graph is symmetric with respect to the y-axis.
31. The x-intercept and the y-intercept are (0, 0). The graph is symmetric with respect to the origin.
33. The x-intercepts are (−3, 0) and (3, 0). The y-intercept is (0, 9). The graph is symmetric with respect to the y-axis.
35. The x-intercepts are (−2, 0) and (2, 0). The y-intercepts are (0, −3) and (0, 3). The graph is symmetric with respect to the x-axis, the y-axis, and the origin.
37. The x-intercept is (3, 0). The y-intercept is (0, −27). The graph is not symmetric with respect to either axis or to the origin.
39. The x-intercepts are (−1, 0) and (4, 0). The y-intercept is (0, −4). The graph is not symmetric with respect to either axis or to the origin.
41. The x-intercept and the y-intercept are both (0, 0). The graph is symmetric with respect to the origin.
43. The x-intercept and the y-intercept are both (0, 0). The graph is symmetric with respect to the y-axis.
45.
47.
49. a = −1
51.
53. (a)
The graphs of
and y = | x| are the same.
55. Let (x, y) be a point on the graph of the equation. Assume that the graph is symmetric with respect to both axes. Because of symmetry with respect to the y-axis, the point (− x, y) is on the graph. Similarly, because of symmetry with respect to the x-axis, the point (− x, − y) is also on the graph. Thus, the graph is symmetric with respect to the origin. Assume that the graph is symmetric with respect to the x-axis and to the origin. Because of symmetry with respect to the x-axis, (x, − y) is on the graph. Because of symmetry with respect to the origin, (− x, y) is also on the graph. Thus, the graph is symmetric with respect to the y-axis. Assume that the graph is symmetric with respect to the y-axis and to the origin. Because of
symmetry with respect to the y-axis, (− x, y) is on the graph. Because of symmetry with respect to the origin, (x, − y) is also on the graph. Thus, the graph is symmetric with respect to the x-axis.
Exercise 1.2 1. (a) f(0) = −4
(b) f(1) = 1
(c) f(−1) = −3
(d) f(− x) = 3x2 − 2x − 4
(e) − f(x) = −3x2 − 2x + 4
(f)
f(x + 1) = 3x2 + 8x + 1
(g) f(2x) = 12x2 + 4x − 4
(h) f(x + h) = 3x2 + 6xh + 2x + 3h2 + 2h − 4
3. (a) f(0) = 0
(b)
(c)
(d)
(e)
(f)
(g)
(h)
5. (a) f(0) = 4
(b) f(1) = 5
(c) f(−1) = 5
(d) f(− x) = | x| + 4
(e) − f(x) = −| x| − 4
(f)
f(x + 1) = | x + 1| + 4
(g) f(2x) = 2| x| + 4
(h) f(x + h) = | x + h| + 4
7. (a)
(b)
9. 4
11. 2x + h − 1
13. 3x2 + 3xh + h2
15. 4x3 + 6hx2 + 4h2 x + h3
17. function
19. function
21. not a function
23. not a function
25. function
27. not a function
29. all real numbers
31. all real numbers
33. { x| x ≠ −4, x ≠ 4}
35. { x| x ≠ 0
37. { x| x ≥ 4} or the interval [4, ∞)
39. { x| x > 9} or the interval (9, ∞)
41. { x| x > 1} or the interval (1, ∞)
43.
45. A = −4, f is undefined at x = −2
47. A = 8, f is undefined at x = 3
49. G(x) = 10x
51.
53.
55. 28,027 thousand acres of wheat will be planted in 2010.
57. The expected Mathematics SAT score would be 456 in 2010.
59. (a) The cost per passenger is $222.
(b) The cost per passenger is $225.
(c) The cost per passenger is $220.
(d) The cost per passenger is $230.
61. (a) Yes
(b) No
(c) No
(d) No
63. Answers will vary.
Exercise 1.3 1. This is not the graph of a function.
3. This is the graph of a function.
(a) The domain is { x|−p ≤ x ≤ p} or the interval [−π, π]. The range is { y|−1 ≤ y ≤ 1} or the interval [−1, 1].
(b) The x-intercepts are
and
. The y-intercept is (0, 1).
(c) The graph is symmetric with respect to the y-axis.
5. This is not the graph of a function.
7. This is the graph of a function.
(a) The domain is { x| x > 0} or the interval (0, ∞). The range is all real numbers or the interval (−∞, ∞).
(b) The x-intercept is (1, 0). There is no y-intercept.
(c) This graph does not have symmetry with respect to the x-axis, y-axis, or the origin.
9. This is the graph of a function.
(a) The domain is all real numbers or the interval (−∞, ∞). The range is { y| y ≤ 2} or the interval (−∞, 2].
(b) The x-intercepts are (−3, 0) and (3, 0). The y-intercept is (0, 2).
(c) The graph is symmetric with respect to the y-axis.
11. This is the graph of a function.
(a) The domain is all real numbers or the interval (−∞, ∞). The range is { y| y ≥ −3} or the interval [−3, ∞).
(b) The x-intercepts are (1, 0) and (3, 0). The y-intercept is (0, 9).
(c) This graph does not have symmetry with respect to the x-axis, y-axis, or the origin.
13. (a)
f(0) = 3, f(−6) = −3
(b)
f(6) = 0, f(11) = −1
(c)
f(3) is positive.
(d)
f(−4) is negative.
(e)
x = −3, x = 6, and x = 10
(f)
f(x) > 0 on the intervals (−3, 6) and (10, 11].
(g)
The domain of f is { x|−6 ≤ x ≤ 11} or the interval [−6, 11].
(h)
The range of f is { y|−3 ≤ y ≤ 4} or the interval [−3, 4].
(i)
The x-intercepts are (−3, 0), (6, 0), and (10, 0).
(j)
The y-intercept is (0, 3).
(k)
The line
intersects the graph 3 times.
(l)
The line x = 5 intersects the graph once.
(m) f(x) = 3 when x = 0 and x = 4.
(n)
f(x) = −2 when x = −5 and x = 8.
15. (a) Yes
(b) f(−2) = 9; the point (−2, 9) is on the graph of f.
(c) x = 0 or
; the points (0, −1) and
are on the graph of f.
(d) The domain of f is all real numbers or the interval (−∞, ∞).
(e) The x-intercepts are
(f)
and (1, 0).
The y-intercept is (0, −1)
17. (a) No
(b) f(4) = −3; the point (4, −3) is on the graph of f.
(c) x = 14; the point (14, 2) is on the graph of f.
(d) The domain is the set { x| x ≠ 6}.
(e) The x-intercept is (−2, 0).
(f)
The y-intercept is
.
19. (a) Yes
(b)
; the point
is on the graph of f.
(c) x = −1 or x = 1; the points (−1, 1)and (1, 1) are on the graph of f.
(d) The domain is the set of all real numbers or the interval (−∞ ∞).
(e) The x-intercept is (0, 0).
(f)
The y-intercept is (0, 0).
21. Yes
23. No
25. f is increasing on the intervals (−8, −2), (0, 2), and (5, ∞) or for −8 < x < 2, 0 < x < 2 and x > 5.
27. There is a local maximum at x = 2. The local maximum is f(2) = 10.
29. f has local maxima at x = −2 and x = 2. The local maxima are f(−2) = 6, and f(2) = 10.
31. (a) The x-intercepts are (−2, 0) and (2, 0). The y-intercept is (0, 3).
(b) The domain is { x|−4 ≤ x ≤ 4} or the interval [−4, 4]. The range is { y|0 ≤ y ≤ 3} or [0, 3].
(c) The function is increasing on (−2, 0) and (2, 4) or for −2 < x < 0 and 2 < x < 4. The function is decreasing on (−4, −2) and (0, 2) or for −4 < x < −2 and 0 < x < 2.
(d) The function is even.
33. (a) The y-intercept is (0, 1). There is no x-intercept.
(b) The domain is the set of all real numbers. The range is set of positive numbers or { y| y > 0} or the interval (0, ∞).
(c) The function is increasing on (−∞, ∞) or for all x-values.
(d) The function is neither even nor odd.
35. (a) The x-intercepts are (−π, 0), (0, 0) and (π, 0). The y-intercept is (0, 0).
(b) The domain is [−π, π]. The range is [−1, 1].
(c) The function is increasing on function is decreasing on and .
(d) The function is odd.
or for and
. The or for
37. (a) The x-intercepts are
and
. The y-intercept is
.
(b) The domain is { x|−3 ≤ x ≤ 3} or the interval [−3, 3]. The range is { y|−1 ≤ y ≤ 2} or the inerval [−1, 2].
(c) The function is increasing on (2, 3) or for 2 < x < 3. The function is decreasing on (−1, 1) or for − 1 < x < 1. The function is constant on (−3, −1) and (1, 2) or for −3 < x < 1 and 1 < x < 2.
(d) The function is neither even nor odd.
39. (a) The function has a local maximum of 3 at x = 0.
(b) The function has local minima of 0 at x = −2 and x = 2.
41. (a) The function has a local maximum of 1 at
(b) The function has a local minimum of −1 at
43. (a)
(b)
(c)
.
.
45. (a)
(b)
(c) y = 5x
47. (a)
(b)
(c) y = −3x + 1
49. (a)
(b)
(c) y = x − 2
51. (a)
(b)
(c) y = 6x − 6
53. (a)
(b)
(c)
55. (a)
(b)
(c)
57. odd
59. even
61. odd
63. neither even nor odd
65. even
67. odd
69.
The function has a local minimum of 0 at x = 1 and a local maximum of 4 at x = −1. The function is increasing on (−2, −1) and (1, 2) and is decreasing on (−1, 1).
71.
The function has a local minimum of −0.19 at x = 0.77 and a local maximum of 0.19 at x = −0.77. The function is increasing on (−2, −0.77) and (0.77, 2) and is decreasing on (−0.77, 0.77).
73.
The function has a local minimum of −18.89 at x = −3.77 and a local maximum of −1.91 at x = 1.77. The function is increasing on (−3.77, 1.77) and it is decreasing on (−6, −3.77) and (1.77, 4).
75.
The function has a local minimum of 0.95 at x = −1.87, a local maximum of 3 at x = 0, and a local minimum of 2.65 at x = 0.97. The function is increasing on (−1.87,0) and (0.97, 2) and is decreasing on (−3, −1.87) and (0, 0.97).
77. (a)
(b)
(c)
(d)
(e)
(f)
(g) The secant lines are approaching the tangent line to the graph of f at x = 0.
(h) The slopes of the secant lines are approach the slope of the tangent line to the graph of f at x = 0, which is 0.
79. (a) 81.07 ft.
(b) 129.59 ft.
(c) 26.63 ft.
(d) The golf ball was hit 528.13 feet.
(e)
(f)
(g)
The ball is at a height of 90 feet when it has traveled 115.07 feet and 413.05 feet.
(h) The ball travels about 275 feet before reaching its maximum height. The maximum height of the ball is 132 feet.
(i)
The ball travels 264 feet before reaching its maximum height.
81. (a) V(x) = x(24 − 2x)2 = 4x3 − 96x2 + 576x
(b) The volume is 972 in3.
(c) The volume is 160 in3.
(d)
The volume V is largest when x = 4 in.
83. (a)
(b) Producing 10 riding lawn mowers minimizes average cost.
(c) The minimum average cost is $239.
85. Reasons will vary. (a) II (b) V (c) IV (d) III (e) I
87. Answers will vary.
89. Answers will vary.
91. Answers will vary.
93. Answers will vary.
Exercise 1.4 1. C.
3. E.
5. B.
7. F.
9.
11.
13.
15.
17. (a) f(−2) = 4
(b) f(0) = 2
(c) f(2) = 5
19. (a) f(1.2) = 2
(b) f(1.6) = 3
(c) f(−1.8) = −4.
21. (a) The domain is all real numbers, or the interval (−∞, ∞).
(b) There is no x-intercept. The y-intercept is (0, 1).
(c)
(d) The range is { y| y ≠ 0} or the intervals (−∞, 0) and (0, ∞).
23. (a) The domain is all real numbers.
(b) There is no x-intercept. The y-intercept is (0, 3).
(c)
(d) The range is { y| y ≥ 1} or the interval [1, ∞).
25. (a) The domain is the set of real numbers greater than or equal to −2, { x| x ≥ −2} or [−2, ∞).
(b) The x-intercept is (2, 0). The y-intercept is (0, 3).
(c)
(d)
27. (a) The domain is set of all real numbers or the interval (−∞, ∞).
(b) The x-intercepts are (−1, 0) and (0, 0). The y-intercept is (0, 0).
(c)
(d) The range is all real numbers or the interval (−∞, ∞).
29. (a) The domain is { x| x ≥ −2} or the interval [−2, ∞).
(b) There is no x-intercept. The y-intercept is (0, 1).
(c)
(d) The range is { y| y > 0} or the interval (0, ∞).
31. (a) The domain is all real numbers.
(b) The x-intercepts lie in the interval [0, 1). The y-intercept is (0, 0).
(c)
(d) The range is the set of even integers.
33.
35.
37. (a) $39.99
(b) $43.74
(c) $40.24
39. (a) $59.33
(b) $396.04
(c)
(d)
41. (a) 10 °C.
(b) 3.98 °C.
(c) −2.67 °C.
(d) −3.70 °C.
(e) For wind speeds under 1.79 m/sec, the wind chill factor is simply the air temperature.
(f)
43.
45.
For wind speeds above 20 m/sec, the wind chill factor is a function of the air temperature.
If a>0, the graph of y = x2 + a is a vertical shift of the graph of y = x2 upward by a units. If a −2} or the interval (−2,∞).
(b) The x-intercept and y-intercept are both (0, 0).
(c)
(d) The range is { y| y > −6} or the interval (−6,∞).
55. (a) The domain is { x| x − 4} or the interval [−4,∞).
(b) The y-intercept is (0, 1). There is no x-intercept.
(c)
(d) The range is the intervals [−4, 0) and (0,∞).
57.
The x-intercepts are (−4, 0) and (4, 0). The y-intercept is (0,−4). The domain is all real numbers or the interval (−∞,∞), and the range is { y| y ≥ −4} or the interval [−4,∞).
59.
The x-intercept is (1, 0). There is no y-intercept. The domain is { x| x ≥ 1} or [1, ∞), and the range is { y| y ≥ 0} or the interval [0, ∞).
61.
The x-intercept is (1, 0). The y-intercept is (0, 1). The domain is { x| x ≤ 1} or the interval (−∞, 1], and the range is { y| y ≥ 0} or the interval [0, ∞).
63.
There is no x-intercept. The y-intercept is (0, 3). The domain is (−∞,∞), and the range is { y| y ≥ 2} or [2, ∞).
65. (a)
(b)
(c)
(d)
67. f(x) = −2x + 3
69. A = 11
71. V(r) = 2π r3
73. (a)
(b) The revenue is $13,333.33.
75. (a)
(b) The revenue is $255.
77. (a) The total cost C in dollars is given by
(b) The cost is $16.03.
(c) The cost is $29.13.
(d)
The cost is least when the radius is 3.758 cm.
Mathematical Questions from Professional Exams 1. (d) 0 ≤ y < 4
2. (d) 87
3. (c) [−1, 0] [1,∞)
.
CHAPTER 2 Classes of Functions Exercise 2.1 1. C.
3. F.
5. G.
7. H
9.
The graph opens upward. The vertex is located at the point (−1,−1). The axis of symmetry is the line x = −1. The y-intercept is (0,0). The x-intercepts are (−2, 0) and (0, 0). The domain is the interval (−∞,∞), and the range is the interval [−1,∞). The function is increasing on the interval (−1,∞), and is decreasing on the interval (−∞,1).
11.
The graph opens downward. The vertex is located at the point (−3,9). The axis of symmetry is the line x = −3. The y-intercept is (0,0). The x-intercepts are (−6,0) and (0,0). The domain
is the interval (−∞,∞), and the range is the interval (−∞,9]. The function is increasing on the interval (−∞,−3), and is decreasing on the interval (−3,∞).
13.
The graph opens upward. The vertex is located at the point (2, −8). The axis of symmetry is the line x = 2. The y-intercept is (0,0). The x-intercepts are and (0,0) and (4,0). The domain is the interval (−∞,∞), and the range is the interval [−8,∞]. The function is increasing on the interval (2,∞), and is decreasing on the interval (−∞,2).
15.
The graph opens upward. The vertex is located at the point (−1,−9). The axis of symmetry is the line x = −1. The y-intercept is (0,−8). The x-intercepts are (−4,0) and (2,0). The domain is the interval (−∞,∞), and the range is the interval [−9,∞]. The function is increasing on the interval (−1,∞), and is decreasing on the interval (−∞,−1).
17.
The graph opens upward. The vertex is located at the point (−1,0). The axis of symmetry is the line x = −1. The y-intercept is (0,1). The x-intercept is (−1,0). The domain is the interval (−∞,∞), and the range is the interval [0,∞). The function is increasing on the interval (−1,∞), and is decreasing on the interval (−∞,−1).
19.
The graph opens upward. The vertex is located at the point the line
. The axis of symmetry is
. The y-intercept is (0,2). There is no x-intercept is (−1,0). The domain is the
interval (−∞,∞), and the range is the interval [ interval (
,∞). The function is increasing on the
), and is decreasing on the interval (−∞, )
21.
The graph opens downward. The vertex is located at the point symmetry is the line
. The y-intercept is (0,−3). There is no x-intercept. The domain is
the interval (−∞,∞), and the range is the interval on the interval
. The axis of
, and is decreasing on the interval
. The function is increasing .
23.
The graph opens upward. The vertex is located at the point (−1,−1). The axis of symmetry is the line x = −1. The y-intercept is (0,2). The x-intercepts are
and
. The domain is the interval (−∞,∞), and the range is the interval [−1,∞). The function is increasing on the interval (−1,∞), and is decreasing on the interval (−∞,−1).
25.
The graphs opens downward. The vertex is located at the point symmetry is the line
. The axis of
. The y-intercept is (0,2). The x-intercepts are and
(−∞,∞), and the range is the interval
. The domain is the interval . The function is increasing on the interval
, and is decreasing on the interval
.
27. The function has a minimum value of f(−3) = −18.
29. The function has a minimum value of f(−3) = −21.
31. The function has a maximum value of f(5) = 21.
33. The function has a maximum value of f(2) = 13.
35. (a) f(x) = (x + 3)(x − 1), f(x) = 2(x + 3)(x − 1), f(x) = −2(x + 3)(x − 1), f(x) = 5(x + 3)(x − 1)
(b) The x-intercepts are unaffected. The y-coordinate of the y-intercept changes from that of y = (x + 3)(x − 1) by a factor of a.
(c) The value of a has no effect on the axis of symmetry.
(d) The x-coordinate of the vertex is the same for each function. The y-coordinate of the vertex changes from that of y = (x + 3)(x − 1) by a factor of a.
(e) The x-coordinate of the vertex is equal to the x-coordinate of the midpoint of the x-intercepts.
37. A unit price of $500 should be established to maximize revenue. The maximum revenue is $1,000,000.
39. (a)
(b) The revenue is $13, 333.33.
(c) A quantity of 300 units maximizes revenue. The maximum revenue is $15, 000.
(d) The company should charge $50.
41. (a)
(b) The revenue is $255.
(c) A quantity of 50 units maximizes revenue. The maximum revenue is $500.
(d) The company should charge $10.
43. (a)
(b) x = 100
(c) The maximum area is 10, 000 square yards.
45. The largest area is 2,000,000 m2.
47. (a) The projectile is 39.0625 feet horizontally from the base of the cliff when it achieves is maximum height.
(b) The maximum height is 219.53125 feet above the water.
(c) The projectile strikes the water 170.024 feet from the base of the cliff.
(d)
(e) The projectile is 135.698 feet from the cliff.
49. A depth of 3 inches will provide the maximum cross-sectional area.
51. 100 meters by
by 31.8 meters
53. (a) There are the most hunters at the income level of $56,584. At this income level, there are about 3685 hunters.
(b)
The number of hunters is increasing between the $20,000 and $40,000 income levels.
55. (a) The number of 23-year old male murder victims is 1795.
(b) The number of male murder victims at age 28 years is about 1456.
(c)
(d) The number of male murder victims decreases with age until age 70 and then begins to increase.
57.
59.
61.
63. A = 25 units2
65. Answers will vary.
67.
Each parabola opens upward, has its vertex at
, and has the line
as its
axis of symmetry.
69. Write the formula for a quadratic function as is zero if minimum at
and is positive otherwise. Hence the function has a , and the graph opens upward. If a < 0, then
and is negative otherwise. Hence the function has a maximum at the graph opens downward.
Exercise 2.2 1. Answers will vary. Possible answers include (−1,−1), (0, 0) and (1, 1).
3. origin
5.
. If a > 0, then is zero if , and
7.
9.
11. This is a polynomial function of degree 3.
13. This is a polynomial function of degree 2.
15. This is not a polynomial function because the term
has the monomial x in the denominator.
17. This is not a polynomial function because the exponent of the term x3/2 is not a nonnegative integer.
19. This is a polynomial function of degree 4.
21. This is a polynomial function of degree 4.
23. y = 3x4
25. y = −2x5
27. y = 5x3
29.
31.
33.
35.
37. (−∞,∞)
39.
41. (a) The percentage of union membership in the labor force in 2000 was 13.2%.
(b) u(75) = 18.02. The percentage of union membership in the labor force in 2005 will be 18.02%.
Exercise 2.3 1. (a) 11.2116
(b) 11.5873
(c) 11.6639
(d) 11.6648
3. (a) 8.8152
(b) 8.8214
(c) 8.8244
(d) 8.8250
5. (a) 21.2166
(b) 22.2167
(c) 22.4404
(d) 22.4592
7. 3.3201
9. 0.4274
11. This is not an exponential function.
13. This is an exponential function with base a = 4.
15. This is an exponential function with base a = 2.
17. This is not an exponential function.
19. B.
21. D.
23. A.
25. E.
27.
Domain = (−∞,∞), Range = (1,∞); the horizontal asymptote is y = 1.
29.
Domain = (−∞,∞), Range = (−2,∞); the horizontal asymptote is y = −2.
31.
Domain = (−∞,∞), Range = (0,∞); the horizontal asymptote is y = 0.
33.
Domain = (−∞,∞), Range = (−1,∞); the horizontal asymptote is y = −1.
35.
37.
39.
41. {0}
43. {4}
45.
47. {1, 2}
49.
51.
53. y = 3x
55. y = −6x
57. (a) 74.1% of light will pass through 10 panes.
(b) 47.2% of light will pass through 25 panes.
59. (a) There will be 44.35 watts after 30 days.
(b) There will be 11.61 watts after one year.
61. There will be 3.35 mg of the drug in the bloodstream after 1 hour. There will be 0.45 mg of the drug in the bloodstream after 6 hours.
63. (a) The probability that a car will arrive within 10 minutes of 12:00 PM is 0.632.
(b) The probability that a car will arrive within 40 minutes of 12:00 PM is 0.982.
(c) F(t) approaches 1 as t becomes unbounded in the positive direction.
(d)
(e) About 6.931 minutes are needed for the probability to reach 50%.
65. (a) The probability that 15 cars will arrive between 5:00 PM and 6:00 PM is 5.2%.
(b) The probability that 20 cars will arrive between 5:00 PM and 6:00 PM is 8.9%.
67. (a) A 3-year old Civic DX Sedan costs $12,123.27.
(b) A 9-year old Civic DX Sedan costs $6442.80.
69. (a) 5.4 amperes, 7.6 amperes, 10.4 amperes
(b)
(c) The maximum current is 12 amperes.
(d) 3.3 amperes, 5.3 amperes, 9.4 amperes
(e)
(f)
The maximum current is 24 amperes.
71. n
Difference
4
2.7083333333
0.0099484951
6
2.7180555556
0.0002262729
8
2.7182787698
0.0000030586
10
2.7182818011
0.0000000273
73.
75.
77. (a) The relative humidity is 71%.
(b) The relative humidity is 73%.
(c) The relative humidity is 100%.
79. (a)
(b)
81. It took 59 minutes to fill half of the container.
83. There is no power function that increases more rapidly than an exponential function whose base is greater than 1. Explanations will vary.
85. The graphs are identical because y = a− x and
Exercise 2.4 1. log3 9 = 2
3. loga 1.6 = 2
5. log1.1 M = 2
7. log2 7.2 = x
9.
11. ln 8 = x
13. 23 = 8
15. a6 = 3
17. 3x =2
represent the same function.
19. 21.3 = M
21.
23.
25. 0
27. 2
29. −4
31.
33. 4
35.
37.
39.
41.
43.
45. 0.511
47. 30.099
49.
51.
53.
55. B.
57. D.
59. A.
61. E.
63.
Domain (−4, ∞), Range = (−∞,∞); the vertical asymptote is x = −4.
65.
Domain = (0,∞), Range = (−∞,∞); the vertical asymptote is x = 0.
67.
Domain = (4,∞), Range = (−∞,∞); the vertical asymptote is x = 4.
69.
Domain = (0,∞), Range = (−∞,∞); the vertical asymptote is x = 0.
71. {9}
73.
75. {2}
77. {5}
79. {3}
81. {2}
83.
85.
87.
89. {−1}
91. (a) 1
(b) 2
(c) 3
(d) The pH increases as the hydrogen ion concentration decreases.
(e) [H+ ] = 0.000316
(f)
[H+ ] = 3.981 × 10−8
93. (a) The aircraft is 5.965 km above sea level.
(b) The height of the mountain is 0.900 km above sea level.
95. (a) It will take 6.931 minutes for the probability to reach 50%.
(b) It will take 16.094 minutes for the probability to reach 80%.
(c) No, the probability cannot equal 100% because the exponential term e−0.1t is never equal to zero.
97. The time between injections is about 2 hours, 17 minutes and 27 seconds.
99. It takes 0.269 seconds to achieve a current of 0.5 ampere and 0.896 seconds to achieve a current of 1.0 ampere.
101. The population will be 309,123,927 people.
103. 50 decibels
105. 110 decibels
107. The magnitude of the earthquake was 8.1 on the Richter scale.
109. (a)
(b) The risk is 90.9%.
(c) A blood alcohol concentration of 0.175 corresponds to a risk of 100%.
(d) A driver should be arrested with a blood alcohol concentration of 0.08 or greater.
(e) Answers will vary.
111. Explanations will vary.
Exercise 2.5 1. 71
3. −4
5. 7
7. 1
9. 1
11. 2
13.
15. 4
17. a+b
19. b−a
21. 3a
23.
25. 2 log55 − log5 x
27. 3 log2 z
29. ln e + ln x
31. ln x + x ln e
33. 2 log u + 3 log a v
35.
37. 3 log2 x − log2(x − 3)
39. logx + log(x + 2) − 3 log(x + 3)
41.
43.
45. log5 u3 v4
47. log3 x7/2
49.
51.
53. log2 x(3x − 2)4
55.
57.
59. 2.771
61. −3.880
63. 5.615
65. 0.874
67.
69.
71.
73. y = Cx
75. y = C(x2 + x)
77. y = Ce3x
79. y = Ce−4x + 3
81.
83. 3
85. 1
87.
89.
91.
93.
95.
97. The functions are not equivalent. Explanations will vary.
Exercise 2.6 1. $1127.50
3. $580.92
5. $98.02
7. $466.20
9. The amount is $1020.20, and the interest is $20.20.
11. (a) $4434.60
(b) $3933.14
13. A 23.1% interest rate is required.
15. It will take approximately 11 years for the investment to triple.
17. $913.93 is needed to get $1000 in 1 year. $835.27 is needed to get $1000 in 2 years.
19. They should invest $35,476.82.
21. A 22.0% interest rate is required.
23. (a) The Rule of 70 approximation is 70 years, which is greater than the actual solution of 69.3147 years by about 0.685 year.
(b) The Rule of 70 approximation is 14 years, which is greater than the actual solution of 13.8629 years by about 0.137 year.
(c) The Rule of 70 approximation is 7 years, which is greater than the actual solution of 6.9315 years by about 0.069 year.
CHAPTER 2 Review True-False Items 1. True
2. False
3. True
4. True
5. False
6. False
7. False
8. True
9. False
Fill In The Blanks 1. parabola
2. axis of symmetry
3.
4.
5. 1
6. 4
7. (0, ∞)
8.
9. 1
10. r loga M
Review Exercises 1.
The graph opens upward. The vertex is (2,2). The axis of symmetry is the linex = 2. The y-intercept is (0,6). There is no x-intercept.
3.
The graph opens up. The vertex is (0,−16). The axis of symmetry is the linex = 0. The y-intercept is (0,−16). The x-intercepts are (−8,0) and (8,0).
5.
The graph opens downward. The vertex is
. The axis of symmetry is the
. The
y-intercept is (0,0). The x-intercepts are (0,0) and (1,0).
7.
The graph opens upward. The vertex is
. The axis of symmetry is the
.
The y-intercept is (0,1). There is no x-intercept.
9.
The graph opens upward. The vertex is The y-intercept is (0,−1). The x-intercepts are (−1.55, 0) and (0.22, 0)
11. Minimum value = 1
. The axis of symmetry is the and
.
or approximately
13. Maximum value = 12
15. Maximum value = 16
17. Answers will vary. Possibilities include (−1, 1), (0, 0), and (1, 1).
19.
21.
23. This is a polynomial function of degree 5.
25. This is not a polynomial function because the exponent in the term 5x1/2 is not a nonnegative integer.
27. y = −2x4
29. The domain is
.
31. The domain is
33. (a) 81
(b) 2
(c)
(d) −3
35. log5 z = 2
37. 513 = u
39.
41.
43. −3
45. 4
47. 2
49.
.
51. 0.4
53. log3 u + 2log3 v − log3 w
55.
57.
59. log4 x25/4
61.
63.
65. 2.124
67.
Domain = (−∞,∞), Range = (0,∞); the x-axis is an asymptote.
69.
Domain = (−∞,∞) , Range = (−∞,1) ; the line y = 1 is an asymptote.
71.
Domain = (0,∞), Range = (−∞,∞); the y-axis is an asymptote.
73.
75.
77.
79. {4}
81. {11}
83. $125.23
85. $923.12
87. It will take almost 11.6 years to double.
89. The Piper Cub is 3229.5 meters above sea level.
91. 50 feet by 50 feet
93. 25 feet by
by 15.92 feet
95. (a) The limiting magnitude is 11.77.
(b) A diameter of 9.56 inches is required.
97. (a) The annual interest rate was 10.436%.
(b) The actual value will be $32,249.24.
99. (a) 63 clubs should be manufactured.
(b) The marginal cost is $151.90.
Mathematical Questions from Professional Exams 1. (e) 2b + log627
2. (e) I, II, and III
3. (c)
4. (e) 125
5. (b) 1
6. (a) 2
7. (a) 3
CHAPTER 3 Review True-False Items 1. True
2. False
3. True
4. True
5. True
6. True
7. True
Fill In The Blanks 1.
2. equals
3. not exist
4. continuous
5. ≠
6. equals
7. y=2, horizontal
Review Exercises 1. 12
3.
5. 9
7. 25
9. 4
11. 0
13. 64
15. −16
17.
19.
21. 0
23.
25.
27.
29. −∞
31. −∞
33. ∞
35. Continuous
37. Discontinuous
39. Discontinuous
41. Continuous
43. The line y = 0 is a horizontal asymptote. The lines x = − 1 and x = 1 are vertical asymptotes.
45. The y = 5 is a horizontal asymptote. The line x = −2 is a vertical asymptote.
47. (a)
(−∞,2) (2,5) (5,∞)
(b)
(−∞,∞)
(c)
(−2,0),(0,0),(1,0)(6,0)
(d)
(0,0)
(e)
f(−6) = 2, f(−4) = 1
(f)
f(−2) = 0, f(6) = 0
(g)
(h)
(i)
(j)
does not exist.
(k)
does not exist.
(l)
No
(m) No
(n)
No
(o)
No
(p)
Yes
(q)
No
(r)
(−6,−4) (−2, 0) (6,∞)
(s)
(−∞,−6) (0, 2) (2, 5) (5, 6)
(t)
(u)
There is no local maximum. There is a local minimum of 2 at x = −6, a local minimum of 0 at x = 0, and a local minimum of 0 at x = 6.
(v)
The line y = 2 is a horizontal asymptote. The line x = 2 is a vertical asymptote.
49. −11
51.
53. The graph has a hole at x = −4 because graph has a vertical asymptote at x = 4 because
, but R is not defined at x = −4. The and
55. R(x) is undefined at and x = 2 and x = 9. A hole appears at x = 2, and a vertical asymptote appears at x = 9.
57. Answers will vary. One possibility is graphed below.
.
59. (a)
(b) For larger and larger advertising expenditures, the sales level will eventually level off at 571 units.
Mathematical Questions from Professional Exams 1. (b)
2. (c) III
3. (d)
4. (e) x ≠ −1
CHAPTER 3 The Limit of a Function Exercise 3.1 1. 0.9 0.99 0.999
x
f(x) = 2x 1.8 1.98 1.998
1.1 1.01 1.001
x
f(x) = 2x 2.2 2.02 2.002
3. x
−0.1 −0.01
−0.001
f(x) = x2 + x 2.01 2.0001 2.000001
x
0.1
0.01
0.001
f(x)= x2+ x 2.01 2.0001 2.000001
5. x
−2.1 −2.01 −2.001
−4.1 −4.01 −4.001
x
−1.9 −1.99 −1.999
−3.9 −3.99 −3.999
7. x
−1.1 −1.01
−1.001
3.31 3.0301 3.003001
x
−0.9 −0.99
−0.999
2.71 2.9701 2.997001
9. 32
11. 1
13. 4
15. 2
17. 3
19. 4
21. The limit does not exit.
23.
25.
27.
29.
31.
33.
35.
The limit does not exist.
37.
39.
41. 0.67
43. 1.6
45. 0
Exercise 3.2
1. 5
3. 4
5. 8
7. 8
9. −1
11. 8
13. 3
15. −1
17. 32
19. 2
21.
23. 3
25. 0
27.
29.
31. 0
33. 5
35. 6
37. −1
39. 0
41. −1
43.
45. 10
47. 8
49.
51. 10
Exercise 3.3 1.
3. (−8,0),(−5,0)
5. f(−8) = 0, f(−4) = 2
7. 3
9. 2
11. 1
13. Yes,
15. No.
17. Yes
19. No.
21. 5
23. 7
25. 1
27. 4
29.
31.
33. Continuous
35. Continuous
37. Discontinuous
39. Discontinuous
41. Discontinuous
43. Continuous
45. Discontinuous
47. Continuous
49. f is continuous on the interval (−∞, ∞).
51. f is continuous on the interval (−∞, ∞).
53. f is continuous on the interval (0, ∞). f is not discontinuous for any numbers in its domain.
55. f is continuous on the interval (−∞, ∞).
57. f is continuous for all numbers in the set { x| x≠−2, x≠2}. f is discontinuous at x=−2 and x=2.
59. f is continuous on the intervals (0,1) and (1,∞). f is discontinuous at x=1.
61. f is continuous for all number in the set { x| x≠0}. f is discontinuous at x=0.
63. (a)
(b)
(c) The function C(x) is continuous at x=350.
(d) Answers will vary.
65. (a)
(b)
(c)
(d)
(e)
(f)
W is not continuous at v=1.79 since 10≠10.00095.
(g)
, W(1.79)=10.00; W is continuous at v=1.79.
(h)
Answers will vary.
(i)
(j)
(k)
(l)
W is not continuous at v=20 since −3.70332≠−3.7034.
(m)
, W(20)=−3.70; W is continuous at v=20.
(n)
Exercise 3.4 1. 1
3. 2
5. 3
7. 0
9. ∞
Answers will vary.
11. −∞
13. ∞
15. ∞
17. ∞
19. ∞
21. The horizontal asymptote is y=3. The vertical asymptote is x=0.
23. The horizontal asymptote is y=2. The vertical asymptote is x=1.
25. The horizontal asymptote is y=1. The vertical asymptotes are x=−2 and x=2.
27. (a)
{ x| x≠6}
(b)
[0,∞)
(c)
The x-intercepts are (−4, 0), and (0, 0). The y-intercept is (0, 0).
(d)
f(−2)=2
(e)
x=4 or x=8
(f)
f is discontinuous at x=6.
(g)
x=6
(h)
y=4
(i)
There is a local maximum of 2 at x=−2.
(j)
There are local minima of 0 at x=−4 and x=0 and a local minimum of 4 at x=8.
(k)
f is increasing on (−4, −2) (0, 6) (4,∞)
(l)
f is decreasing on (−∞, −4) (−2, 0) (6, 8)
(m) 4
(n)
∞
(o)
∞
(p)
∞
29. R(x)=−∞, and x=−1.
, R(x)=∞, so the graph of R will have a vertical asymptote at
, but R is not defined at x=1, so the graph of R will have a hole at
.
31. , but R is not defined at x=1, so the graph of R will have a hole at and
.
, so the graph of R will have a vertical asymptote at
x=1.
33. R(x) is undefined at x=1, where a hole appears, and at x=−2, where a vertical asymptote occurs.
35. R(x) is undefined at x=2, where a hole appears, and at x=−3, where a vertical asymptote occurs.
37. R(x) is undefined at x=−1, where a hole appears.
39. (a) C(x)=79,000+10x
(b) { x| x≥0}
(c)
(d) { x| x>0}.
(e)
; the average cost of producing a few calculators is very high due to the fixed costs. Therefore, the average cost of producing no calculators is unbounded.
(f)
; the more calculators that are produced, the closer the average cost gets to $10 per calculator.
41. (a)
(b) No, it is not possible to remove 100% of the pollutant. Explanations will vary.
43. Answers will vary. One possibility is graphed below.
CHAPTER 4 Review True-False Items 1. True
2. False
3. True
4. True
5. False
6. False
7. True
Fill In The Blanks 1. tangent
2. marginal cost
3. chain rule
4. velocity
5. 0
6. implicit
Review Exercises 1. f ′(2) = 2
3. f ′(2) = 4
5. f ′(1) = 0
7. f ′(0) = 3
9. f ′(x) = 4
11. f ′(x) = 4x
13. f ′(x) = 5x4
15. f ′(x) = x3
17. f ′(x) = 4x − 3
19. f ′(x) = 14x
21. f ′(x) = 15(x2 − 6x + 6)
23. f ′(x) = 24(16x3 + 3x2 − 5x + 1)
25.
27.
29.
31.
33. f ′(x) = 10x4(3x − 1)(3x − 2)4
35. f ′(x) = 7(x + 1)3(5x + 1)
37.
39.
41.
43. f ′(x) = 3e x + 2x
45. f ′(x) = 3e3x + 1
47. f ′(x) = e x (2x2 + 11x + 7)
49.
51.
53.
55. f ′(x) = x + 2x ln x
57.
59. f ′(x) = (ln 2)2 x + 2x
61.
63.
65. f ′(x) = 5x2/3
67.
69.
71.
73.
75. f ′(x) = 3x2, f″(x) = 6x
77. f ′(x) = −3e−3x , f″(x) = 9e−3x
79.
81.
83.
85. mtan = −1, y = − x − 9
87. mtan = 1, y = x + 1
89. (a) The average rate of change is 7.
(b) The instantaneous rate of change at 2 is f ′(2) = 15.
91. (a) 2.5 seconds elapse before the stone hits the water.
(b) The average velocity is −40 feet per second.
(c) The instantaneous velocity is −80 feet per second.
93. (a) The ball reaches its maximum height 4 seconds after it is thrown.
(b) The maximum height that the ball reaches is 262 feet.
(c) The total distance the ball travels is 518 feet.
(d) The velocity of the ball at any time t is v(t) = 128 − 32t feet per second.
(e) The velocity of the ball is zero at t = 4 seconds, which is the time at which the ball’s velocity changes from upward motion to downward motion.
(f)
The ball is in the air for 8.0 seconds.
(g) The velocity of ball is −129.5 feet per second when it hits the ground.
(h) The acceleration at any time t is −32 feet per second per second.
(i)
The velocity of ball is 64 feet per second when it has been in the air for 2 seconds and is −64 feet per second when it has been in the air for 6 seconds.
(j)
Answers will vary.
95. (a) R(x) = −0.50x2 + 75x
(b) R′(x) = −1.00x + 75
(c) C″(x) = 15
(d) The break even points are x = 10 and x = 110.
(e) The marginal revenue and the marginal cost are equal when x = 60 units are produced.
Mathematical Questions from Professional Exams 1. (e)
2. (d) x≠−3
3. (d)
4. (a)
5. (e) 15
6. (c) t = 2
CHAPTER 4 The Derivative of a Function Exercise 4.1 1. mtan = 3, y = 3x + 5
3. mtan = −2, y = −2x + 1
5. mtan = 12, y = 12x − 12
7. mtan = 5, y = 5x − 2
9. mtan = −4, y = −4x + 2
11. mtan = 1, y = x + 1
13. f ′(3) = −4
15. f ′(0) = 0
17. f ′(1) = 7
19. f ′(0) = 4
21. f ′(1) = 3
23. f ′(1) = −1
25. f ′(x) = 2
27. f ′(x) = −2
29. f ′(x) = 2x
31. f ′(x) = 6x − 2
33. f ′(x) = 3x2
35. f ′(x) = m
37. (a) The average rate of change is 3.
(b) The instantaneous rate of change at x = 1 is 3.
39. (a) The average rate of change is 12.
(b) The instantaneous rate of change at x = 1 is 6.
41. (a) The average rate of change is 6.
(b) The instantaneous rate of change at x = 1 is 4.
43. (a) The average rate of change is 7.
(b) The instantaneous rate of change at x = 1 is 3.
45. f ′(−2) = 60
47.
49. f ′(0) = 1
51. f ′(1) = 3e≈8.155
53. f ′(1) = 0
55. No
57. The pilot should release the bomb at the point (2, 4).
59. (a) The average rate of change in sales is 74 tickets per day.
(b) The average rate of change in sales is 94 tickets per day.
(c) The average rate of change in sales is 110 tickets per day.
(d) The instantaneous rate of change of sales on day 5 is 90 tickets per day.
(e) The instantaneous rate of change of sales on day 10 is 130 tickets per day.
61. (a) The farmer is willing to supply 4500 crates for $10 per crate.
(b) The farmer is willing to supply 7800 crates for $13 per crate.
(c) The average rate of change in supply is 1100 crates per dollar.
(d) The instantaneous rate of change is 950 crates per dollar.
(e) Answers will vary.
63. (a) R′(x) = 8 − 2x
(b) C′(x) = 2
(c) The break-even points are x = 1 and x = 5.
(d) x = 3
(e)
65. (a) R(x) = − 10x2 + 2000x
(b) R′(x) = −20x + 2000
(c) R′(100) = $0
(d) The average rate of change in revenue is −10 dollars per ton.
67. (a) R(x) = 90x − 0.02x2
(b) R′(x) = 90x − 0.04x
(c) C′(x) = 10
(d) The break-even points are x = 0 and x = 4000.
(e) The marginal revenue equals marginal cost at a production level of 2000 units.
69. The instantaneous rate of change of the volume with respect to the radius is 18π ≈ 56.55 cubic feet per foot.
Exercise 4.2 1. f ′(x) = 0
3. f ′(x) = 3x2
5. f ′(x) = 12x
7. f ′(t) = t3
9. f ′(t) = 2x + 1
11. f ′(x) = 3x2 − 2x
13. f ′(t) = 4t − 1
15. f ′(x) = 4x7 + 3
17.
19. f ′(x) = 2ax + b
21.
23.
25.
27.
29. f ′(−3) = −24
31. f ′(4) = 15
33. f ′(3) = −4
35. f ′(1) = 1
37.
39.
41.
43.
45.
47.
49.
51. x=2
53. x = −1, x = 1
55. There are no such values of x.
57.
59. y = − 4x + 1, y = 4x − 7
61. (a) The average cost is $45.00.
(b) C′(x) = 0.4x + 3
(c) The marginal cost of production level of 100 pairs of eyeglasses is $43.00
(d) We can interpret C′(100) to be the cost of producing the 101st pair of eyeglasses.
63. (a) V′(R) = 4kR3
(b) V′(0.3) = 0.108k cm3/cm
(c) V′(0.4) = 0.256k cm3/cm
(d) The amount of blood flowing through the artery increases by about 0.0175k cm3.
65. (a) The daily cost of producing 40 microwave ovens is $3920.
(b) The marginal cost function is C′(x) = 50 − 0.1x.
(c) C′(40) = 46. The marginal cost of producing 40 microwave ovens may be interpreted as the cost to produce the 41st microwave oven.
(d) The cost of producing 41 microwave ovens is approximately $3966.
(e) The actual cost of producing 41 microwave ovens is $3965.95. The actual cost is $0.05 less than the estimated cost.
(f)
The actual cost of producing the 41st microwave oven is $45.95.
(g) The average cost function is
.
(h) The average cost of producing 41 microwave ovens is $96.73.
(i)
The average cost of producing 41 microwave ovens is $50.78 greater than the actual cost of producing the 41st microwave oven, and $50.73 greater than the estimated cost of producing the 41st microwave oven. Explanations will vary.
67. (a) The marginal price of beans in 1995 was −$2.431.
(b) The marginal price of beans in 2002 was −$9.634.
(c) Answers will vary.
69. V′(2) = 16π≈50.27 cubic feet per foot
71. A′(t) = 3a3 t2 + 2a2 t + a1
73. Let f(x) = x n .
Exercise 4.3 1. f ′(x) = 16x − 2
3. f ′(t) = 4t3 − 6t
5. f ′(x) = 18x2 − 20x + 3
7. f ′(x) = 24x7 + 40x4 + 9x2
9.
11.
13.
15.
17.
19.
21. mtan = 3, y = 3x − 1
23.
25.
27. x = −2, x = 0
29. y′ = 9x2 − 4x
31.
33.
35.
37.
39.
41.
(a) The average rate of change is −$1000 per year.
(b)
dollars per year.
(c) The instantaneous rate of change after two years is −$2500 per year.
(d) The instantaneous rate of change after five years is −$400 per year.
(e) Answers will vary.
43. (a) R(x) = 10x + 40
(b) R′(x) = 10
(c) R′(4) = 10
(d) R′(6) = 10
45. (a)
47. (a) The population is growing at a rate of 38.820 bacteria per hour.
(b) The population is growing at a rate of 35.503 bacteria per hour.
(c) The population is growing at a rate of 30.637 bacteria per hour.
(d) The population is growing at a rate of 24.970 bacteria per hour.
49. The rate of change of the intensity is −2 units per meter.
51. (a)
(b) The marginal cost is −$0.044 dollars per mph.
(c) The marginal cost is −$0.019 dollars per mph.
(d) The marginal cost is −$0.078 dollars per mph.
53. (a)
(b) Answers will vary.
Exercise 4.4 1. f ′(x) = 8(2x − 3)3
3. f ′(x) = 6x(x2 + 4)2
5. f ′(x) = 12x(3x2 + 4)
7. f ′(x) = (4x + 1)(x + 1)2
9. f ′(x) = 8x(6x + 1)(2x + 1)3
11. f ′(x) = 3x2(2x − 1)(x − 1)2
13.
15.
17.
19.
21.
23.
25.
27.
29. (a) The car is depreciating at a rate of $7733.33 per year.
(b) The car is depreciating at a rate of $4793.39 per year.
(c) The car is depreciating at a rate of $3017.69 per year.
(d) The car is depreciating at a rate of $1972.79 per year.
31. (a)
(b)
(c)
(d) R′(10) = 39.44 dollars per pound, R′(40) = 6.11 dollars per pound
(e) Answers will vary.
33. (a) The average rate of change in mass is −3.5 grams per hour.
(b) M′(0) = −7 grams per hour
(c) Answers will vary.
Exercise 4.5 1. f ′(x) = 3x2 − e x
3. f ′(x) = e x (x2 + 2x)
5.
7.
9.
11.
13.
15.
17.
19.
21.
23.
25.
27.
29.
31.
33.
35.
37.
39.
41.
43.
45.
47.
49.
51.
53.
55.
57.
59.
61.
63.
65. (a) The reaction rate is 1.1 per unit.
(b) The reaction rate is 0.55 per unit.
(c) Answers will vary.
67. The rate of change of the pressure with respect to the height is −1.130 kilograms per square meter per meter at a height of 500 meters and is −1.103 kilograms per square meter per meter at a height of 700 meters.
69. (a) A′(t) = 18.9e−0.21t percent of the market per year
(b) A′(5) = 6.614 percent of the market per year. In the sixth year, DVD players will penetrate approximately an additional 6.614 percent of the market.
(c) A′(10) = 2.314 percent of the market per year. In the eleventh year, DVD players will penetrate approximately an additional 2.314 percent of the market
(d) A′(30) = 0.035 percent of the market per year. In the thirty-first year, DVD players will penetrate approximatenly an additional 0.035 percent of the market.
71. (a)
dollars in sales per thousands of dollars of advertising cost
(b) S′(10) = 40,000 dollars in sales per thousands of dollars of advertising cost
(c) S′(20) = 20,000 dollars in sales per thousands of dollars of advertising cost
(d) Answers will vary.
73. (a) 1000 t-shirts can be sold at $40.41.
(b) 5000 t-shirts can be sold at $34.27.
(c) p′(1000) = − $0.0036. This means that another t-shirt will be demanded if the price were reduced by $0.0036.
(d) p′(5000) = −$0.00078. This means that another t-shirt will be demanded if the price were reduced by $0.00078.
(e)
(f)
R′(1000) = $36.77. The revenue received for selling the 1001st t-shirt is $36.77.
(g) R′(5000) = $30.35. The revenue received for selling the 5001st t-shirt is $30.35.
(h)
(i)
P(1000) = $36,408.42
(j)
P(5000) = $151,363.49
(k) For x = 3,631,550, the profit is the greatest.
(l)
A price of $8.00 should be charged to maximize profit.
75. (a)
dollars per year
(b) p′(5) = 0.0052 dollars per year
(c) p′(10) = 0.0026 dollars per year
(d) Answers will vary.
77. Let y = ln u and u = g(x). Thus y(u(x)) = ln(g(x)). Thus .
Exercise 4.6 1. f ′(x) = 2, f″(x) = 0
3. f ′(x) = 6x + 1, f″(x) = 6
5. f ′(x) = −12x3 + 4x, f″(x) = −36x2 + 4
7.
9.
11.
13.
15.
17.
19.
21.
23.
25. (a) The domain is the interval (−∞,∞).
(b) f ′(x) = 2x
(c) The domain is the interval (−∞,∞).
(d) x = 0
(e) The derivative f ′(x) exists for all values of x.
(f)
f″(x) = 2
(g) The domain is the interval (−∞,∞).
27. (a) The domain is the interval (−∞,∞).
(b) f ′(x) = 3x2 − 18x + 27
(c) The domain is the interval (−∞,∞).
(d) x = 3
(e) The derivative f ′(x) exists for all values of x.
(f)
f″(x) = 6x − 18
(g) The domain is the interval (−∞,∞).
29. (a) The domain is the interval (−∞,∞).
(b) f ′(x) = 12x3 − 36x2
(c) The domain is the interval (−∞,∞).
(d) x = 0, x = 3
(e) The derivative f ′(x) exists for all values of x.
(f)
f″(x) = 36x2 − 72x
(g) The domain is the interval (−∞,∞).
31. (a) The domain is the set { x| x≠−2, x≠2}.
(b)
(c) The domain is the set { x| x≠−2, x≠2}.
(d) The derivative f ′(x) is never zero.
(e) The derivative f ′(x) does not exist for x = − 2 and x = 2.
(f)
(g) The domain is the set { x| x≠−2, x≠2}.
33. f (4)(x) = 0
35. f (20)(x) = 0
37. f (8)(x) = 5040
39. v = 32t + 20, a = 32
41. v = 9.8t + 4, a = 9.8
43.
45.
47.
49.
51.
53.
55. y″ − 4y = 0
57. f″(x) = x2 g″(x) + 4xg′(x) + 2g(x)
59. (a) The velocity is 16 feet per second.
(b) The ball will reach its maximum height 2.5 seconds after it is thrown.
(c) The maximum height of the ball is 106 feet.
(d) The acceleration is −32 feet per second per second.
(e) The ball is in the for air 5.074 seconds.
(f)
The velocity of the ball is −82.365 feet per second upon impact.
(g) The total distance traveled by the ball is 206 feet.
61. v(1) = 3 meters per second, a(t) = 6t − 12 meters per second.
63. (a) It takes 4.24 seconds for the rock to hit the ground.
(b) The average velocity is −20.8 meters per second.
(c) The average velocity is −14.7 meters per second.
(d) The velocity is −41.6 meters per second when the rock hits the ground.
Exercise 4.7 1.
3.
5.
7.
9.
11.
13.
15.
17.
19.
21.
23.
25.
27.
29.
31.
33.
35.
37. There is no tangent line at (0, 0).
39. (0, −2) and (0, 2)
41. (0, −4) and (0, 4)
43. (a)
(b)
(c) (2,1)
45.
47. (a)
(b)
(c) Answers will vary.
Exercise 4.8
1.
3.
5.
7. f ′(x) = 3(2x + 3)1/2
9. f ′(x) = 3x(x2 + 4)1/2
11.
13.
15.
17.
19.
21.
23.
25.
27.
29.
31.
33.
35.
37.
39.
41. (a) The domain is the interval [0,∞).
(b)
(c) The domain is the interval [0,∞).
(d) The derivative f ′(x) is nonzero on its domain.
(e) x = 0
(f)
(g) The domain is the interval (0,∞).
43. (a) The domain is the interval (−∞,∞).
(b)
(c) The domain is the set { x| x≠0}.
(d) The derivative f ′(x) is nonzero on its domain.
(e) x = 0
(f)
(g) The domain is the set { x| x≠0}.
45. (a) The domain is the interval (−∞,∞).
(b)
(c) The domain is the set { x| x≠0}.
(d) x = −1
(e) x = 0
(f)
(g) The domain is the set { x| x≠0}.
47. (a) The domain is the interval (−∞,∞).
(b)
(c) The domaon is the set { x| x≠−1, x≠1}.
(d) x = 0
(e) x = −1, x = 1
(f)
(g) The domain is the set { x| x≠−1, x≠1}.
49. (a) The domain is the interval [−1, 1].
(b)
(c) The domain is the interval (−1, 1).
(d)
(e) x = − 1, x = 1
(f)
(g) The domain is the interval (−1, 1).
51. (a)
(b) Student enrollment will be increasing at the rate of 176.78 students per year.
53.
55. (a)
units per year
(b) The instantaneous rate of change of pollution is 2.34 units per year.
57. After 1 second, the child’s velocity is 1.5 feet per second. The child strikes the ground with a velocity of 3 feet per second.
CHAPTER 5 Applications: Graphing Functions; Optimization Exercise 5.1 1. There is a horizontal tangent at (2, 0).
3. There is a horizontal tangent at (4, 16).
5. There is a horizontal tangent at (2, 9).
7. There is a vertical tangent at (0, 1).
9. There are horizontal tangents at (−1,−1) and (1, 3).
11. There is a vertical tangent at (0,−2).
13. There are horizontal tangents at (0, 0) and (8, −8192).
15. There is a horizontal tangent at (0,−1).
17. There is a horizontal tangent at (−1, −1), and there is a vertical tangent at (0, 0).
19. There is a horizontal tangent at (0, 0).
, and (
), there is a vertical tangent at
21. There are horizontal tangents at tangent at (0, 0).
and
, and there is a vertical
23. There is a horizontal tangent at
, and there is a vertical tangent at (0, 0).
25. There is a horizontal tangent at
27. (a) Yes
(b) No
(c) vertical tangent at (0, 0)
29. (a) No
(b) No
(c) f is not continuous at x = 1.
31. (a) Yes
(b) No
(c) no tangent at (0, 0)
(d)
, and a vertical tangent at (0, 0).
33. (a) No
(b) No
(c) f is not continuous at x = 2.
(d)
35. (a) Yes
(b) Yes, f (0) =
(c) f (0) does exist.
(d)
Exercise 5.2 1. The domain is the interval [x1,x9]
3. The graph is increasing on the intervals (x1,x4), (x5,x7), and (x8,x9).
5. x = x4, x = x6, x = x7, and x = x8
7. f has a local maximum at (x4,y4) and at (x7,y7).
9. Step 1: The domain is the interval (−∞,∞). Step 2: The x-intercept is (1,0), and the y-intercept is (0,−2). Step 3: The graph is increasing on the interval (−∞,1) and is decreasing on the interval (1,∞). Step 4: There is a local maximum at (1,0).
Step 5: The tangent line is horizontal at (1,0). Step 6: The end behavior is y = −2x2.
11. Step 1: The domain is the interval (−∞,∞). Step 2: The x-intercept is (3,0), and the y-intercept is (0,−27). Step 3: The graph is increasing on the interval (−∞,∞). Step 4: There are no local extrema. Step 5: The tangent line is horizontal at (3,0). Step 6: The end behavior is y = x3.
13. Step 1: The domain is the interval (−∞,∞). Step 2: The x-intercept and the y-intercept are both (0,0). Step 3: The graph is increasing on the intervals (−∞,2) and (3,∞) and is decreasing on the interval (2,3). Step 4: There is a local maximum at (2,28), and there is a local minimum at (3,27). Step 5: The tangent line is horizontal at the points (2,28) and (3,27). Step 6: The end behavior is y = 2x3.
15. Step 1: The domain is the interval (−∞,∞). Step 2: The y-intercept is (0,−1). Step 3: The graph is increasing on the interval (−1,1) and is decreasing on the intervals (−∞,−1) and (1,∞). Step 4: There is a local minimum at (−1,−3), and there is a local maximum at (1,1). Step 5: The tangent line is horizontal at the points (−1,−3) and (1, 1).
Step 6: The end behavior is y = − x3.
17. Step 1: The domain is the interval (−∞,∞). Step 2: The y-intercept is (0,2). Step 3: The graph is increasing on the interval (3,∞)and is decreasing on the interval (−∞,3). Step 4: There is a local minimum at (3,−79). Step 5: The tangent line is horizontal at the points (0,2) and (3,−79). Step 6: The end behavior is y = 3x4.
19. Step 1: The domain is the interval (−∞,∞).
21. Step 1: The domain is the interval (−∞,∞). Step 2: The y-intercept is (0,1). Step 3: The graph is increasing on the intervals (−∞,−2) and (2,∞) and is decreasing on (−2,2). Step 4: There is a local maximum at (−2,65), and there is a local minimum at (2,−63). Step 5: The tangent line is horizontal at (−2,65),(0,1), and (2,−63). Step 6: The end behavior is y = 3x5.
23. Step 1: The domain is the interval (−∞,∞). Step 2: The x-intercepts are (0,0) and (0,−8), and the y-intercept is (0,0). Step 3: The graph is increasing on the interval (−1,∞), and is decreasing on the interval (−∞,−1). Step 4: There is a local minimum at (−1,−1). Step 5: The tangent line is vertical at (0,0). Step 6: The end behavior is y = x2/3.
25. Step 1: The domain is the interval (−∞,∞).
Step 2: The x-intercepts are (−1,0) and (1,0), and the y-intercept is (0,1). Step 3: The graph is increasing on the intervals (−1,0) and (1,∞) and is decreasing on the intervals (−∞,−1) and (0,1). Step 4: There is a local maximum at (0,1), and there are local minima at (−1,0) and (1,0). Step 5: The tangent line is horizontal at (0,1) and is vertical at (−1,0) and (1, 0). Step 6: The end behavior is y = x4/3.
27. Step 1: The domain is the set { x| x ≠ −4, x ≠ 4}. Step 2: The y-intercept is (0,−½). Step 3: The graph is increasing on the intervals (−∞,−4) and (−4,0) and is decreasing on the intervals (0,4) and (4,∞). Step 4: There is a local maximum at (0,−½). Step 5: The tangent line is horizontal at (0,−½). Step 6: The end behavior is the horizontal asymptote y = 0. The lines x = −4 and x = 4 are vertical asymptotes.
29. Step 1: The domain is the set { x| x ≠ −3, x ≠ 3}. Step 2: The x-intercept and the y-intercept are both (0,0). Step 3: The graph is decreasing on the intervals (−∞,−3), (−3,3), and (3,∞). Step 4: There are no local extrema. Step 5: There are no horizontal tangents or vertical tangents. Step 6: The end behavior is the horizontal asymptote y = 0. The lines x = −3 and x = 3 are vertical asymptotes.
31. Step 1: The domain is the set { x| x ≠ −2, x ≠ 2}. Step 2: The x-intercept and y-intercept are both (0,0). Step 3: The graph is increasing on the intervals (−∞,−2) and (2,0), and is decreasing on the intervals (0,2) and (2,∞).
Step 4: There is a local maximum at (0,0). Step 5: There is a horizontal tangent at (0, 0). Step 6: The end behavior is the horizontal asymptote y = 1. The lines x = −2 and x = 2 are vertical asymptotes.
33. Step 1: The domain is the set { x| x > 0}. Step 2: The x-intercept is (1,0). Step 3: The graph is increasing on the interval (0.37,∞) and is decreasing on the interval (0.037). Step 4: There is a local minimum at (0.37,−0.37). Step 5: There is a horizontal tangent at (0.37,−0.37).
Step 6: The end behavior is y = x ln x.
and
35. S(x) = 8x + 50 > for 1 ≤ x ≤ 10. Therefore S is an increasing function.
37. (a) The graph of R is increasing on (0,2000) and is decreasing on (2000,∞).
(b) 2000 trucks need to be sold.
(c) The maximum revenue is $20,000.
(d)
39. (a) The function is increasing on the interval
.
(b) The acreage of wheat planted from 2004 to 2008 will be decreasing.
41. (a) The yield will be increasing for amounts of nitrogen between 0 units and units.
(b) The yield will be decreasing for amounts of nitrogen greater than units.
43. c=½
45. c=0
47.
49.
Exercise 5.3 1. The domain is
3. The graph of f is increasing on the intervals (x1,x3), (0,x4), and (x4,x6).
5. x = 0 and x = x6
7. (x3,y3),(x6,y6)
9. The graph of f is concave up on the intervals (x1,x3) and (x3,x4).
11. The line x = 4 is a vertical asymptote.
13. The graph is concave down on (−∞,2) and is concave up on (2,∞). The point (2,−15) is the only inflection point.
15. The graph is concave down on (0,1) and is concave up on (−∞,0) and (1,∞). The points (−0,1) and (1,4) are the inflection points.
17. The graph is concave down on (−∞,1) and is concave up on (1,∞). The point (1,1) is the only
inflection point.
19. The graph is concave down on (−∞,−1) and (0,1) and is concave up on (−1,0) and (1,∞). The points (−1,7),(0,10),and (1,13) are the inflection points.
21. The graph is concave down on (−∞, 1) and is concave up on (1, ∞). The point (1, −5) is the only inflection point.
23. The graph is concave down on (0, ∞) and is concave up on (−∞, 0). The point (0, 2) is the only inflection point.
25. The graph is concave down on (−∞, −2) and is concave up on (−2, 0) and (0, ∞). The point is the inflection point.
27. The graph is concave up on (−∞, 0) and (0, ∞). There are no inflection points.
29. Step 1: The domain is the interval (−∞, ∞). Step 2: The y-intercept is (0, 1). Step 3: The graph is increasing on the intervals (−∞, 0)and (4, ∞) and is decreasing on the interval (0, 4). Step 4: There is a local maximum at (0, 1), and there is a local minimum at (4, −31). Step 5: The tangent line is horizontal at (0, 1) and (4, −31). Step 6: The end behavior is y = x3. Step 7: The graph is concave up on the interval (2, ∞) and is concave down on the interval (−∞, 2). The point (2, −15) is the only inflection point.
31. Step 1: The domain is the interval (−∞, ∞). Step 2: The x-intercepts are (−1, 0) and (1, 0), and the y-intercept is (0,1). Step 3: The graph is increasing on the intervals (−1, 0) and (1, ∞) and is decreasing on the intervals (−∞, −1) and (0, 1). Step 4: There is a local maximum at (0, 1), and there are local minima at (−1, 0) and (1, 0). Step 5: The tangent line is horizontal at (−1, 0), (0, 1), and (1, 0). Step 6: The end behavior is y = x4.
Step 7: The graph is concave up on the intervals concave down on the interval
. The points
and and
inflection points.
33. Step 1: The domain is the interval (−∞, ∞). Step 2: The x-intercepts are (10, 0) and (0, 0), and the y-intercept is (0, 0).
and is are the
Step 3: The graph is increasing on the intervals (−∞, 0) and (8, ∞) and is decreasing on (0, 8). Step 4: There is a local maximum at (0, 0), and there is a local minimum at (8, −8192). Step 5: The tangent line is horizontal at (0, 0) and (8, −8192). Step 6: The end behavior is y = x5. Step 7: The graph is concave up on the interval (6, ∞) and is concave down on the interval (−∞, 6). The point (6, −5184) is the only inflection point.
35. Step 1: The domain is the interval (−∞, ∞). Step 2: The x-intercepts are (3, 0) and (0,0), and the y-intercept is (0, 0).
Step 3: The graph is increasing on the interval (−∞, 0) and
and is decreasing on the intervals
.
Step 4: There is a local minimum at
.
Step 5: The tangent line is horizontal at (0, 0) and
Step 6: The end behavior is y = x6.
.
Step 7: The graph is concave up on the intervals (−∞, 0), and (2, ∞) and is concave down on the interval (0, 2). The points (0, 0) and (2, −32) are the inflection points.
37. Step 1: The domain is the interval (−∞, ∞). Step 2: The x-intercepts are (4, 0) and (0, 0), and the y-intercept is (0, 0). Step 3: The graph is increasing on the interval (3, ∞) and is decreasing on the intervals (−∞, 0) and (0, 3). Step 4: There is a local minimum at (3, −81). Step 5: The tangent line is horizontal at (0, 0) and (3, −81). Step 6: The end behavior is y = 3x4. Step 7: The graph is concave up on the intervals (−∞, 0) and (2, ∞), and is concave down on the interval (0, 2). The points (0, 0) and (2, −48) are the inflection points.
39. Step 1: The domain is the interval (−∞, ∞). Step 2: The y-intercept is (0, 4).
Step 3: The graph is increasing on the intervals (−∞, 0) and the interval
and is decreasing on
.
Step 4: There is a local maximum at (0, 4), and there is a local minimum at .
Step 5: The tangent line is horizontal at (0, 4) and
.
Step 6: The end behavior is y = x5. Step 7: The graph is concave up on the interval (1,∞) and is concave down on the interval (−∞,1). The point (1,−5) is the only inflection point.
41. Step 1: The domain is the interval (−∞,∞). Step 2: The x-intercepts are (10,0), and (0,0) and the y-intercept is (0,0). Step 3: The graph is increasing on the intervals (−∞,0) and (4,∞) and is decreasing on the interval (0,4).
Step 4: There is a local maximum at (0,0), and there is a local minimum at
Step 5: The tangent line is horisontal at
.
and is vertical at (0,0).
Step 6: The end behavior is y = x5/3. Step 7: The graph is concave up on the intervals (−2,0), (0,∞) and is concave down on the interval (−∞,−2). The point is the only inflection point.
43. Step 1: The domain is the interval (−∞,∞). Step 2: The x-intercepts are (−4,0),(4,0), and (0,0), and the y-intercept is (0,0). Step 3: The graph is increasing on the intervals (−2,0) and (2,∞) and is decreasing on the intervals (−∞,−2) and (0,2).
Step 4: There is a local maximum at (0,0), and there are local minima at and
.
Step 5: The tangent line is horizontal at at (0,0). Step 6: The end behavior is y = x8/3.
and
and is vertical
Step 7: The graph is concave up on the intervals (−∞,0) and (0,∞). There is no inflection point.
45. Step 1: The domain is the interval (−∞,∞). Step 2: The x-intercept and the y-intercept are both (0,0). Step 3: The graph is increasing on the interval (−1,∞) and is decreasing on the interval (−∞,−1).
Step 4: There is a local minimum at
Step 5: The tangent line is horizontal at
Step 6: The end behavior is
.
.
and.
. The line y = 0 is a
horizontal asymptote. Step 7: The graph is concave up on the interval (−2,∞) and is concave down on the interval (−∞,−2). The point (2,−15) is the only inflection point.
47. There is a local maximum at (−1,4) and a local minimum at (1,0).
49. There is a local minimum at (−1,−4).
51. There is a local maximum at (0,2) and a local minimum at (4,−254).
53. There is a local maximum at (−1,−2) and a local minimum at (1,2).
55. Answers will vary.
57. Answers will vary.
59. a = −3, b = 9
61. (a)
(b) The minimum average cost per item is $20.
(c) C(x) = 4x
(d)
Point of intersection is (5, 20).
(e) The minimum average costs occurs at the production level for which the average cost equals the marginal cost.
63. (a)
(b) The minimum average cost per item is $12.
(c)
(d)
Point of intersection is (500, 12).
(e) The minimum average costs occurs at the production level for which the average cost equals the marginal cost.
65. (a) The domain of N is the interval [0,∞].
(b) The N-intercept is (0,1). There is no t-intercept.
(c) N is increasing on the interval (0,∞).
(d) N is concave up on the interval (0,ln(49,999)) and is concave down on the interval (ln(49,999),∞).
(e) The inflection point is (ln(49,999),25,000)).
(f)
(g) The rumor is spreading at its greatest rate at t = ln(49,999)≈10.82 units of time since the rumor began to spread.
67. (a)
(b) The maximum growth rate occurs at t = ln (9) ≈ 2.20 days.
(c) The equilibrium population is 8000.
(d)
69. (a) The sales rate is a maximum at x = ln (50) ≈ 3.91 months.
(b)
Exercise 5.4 1. The absolute maximum is 15, and the absolute minimum is −1.
3. The absolute maximum is 1, and the absolute minimum is −39.
5. The absolute maximum is 16, and the absolute minimum is −4.
7. The absolute maximum is 1, and the absolute minimum is 0.
9. The absolute maximum is 1, and the absolute minimum is 0.
11. The absolute maximum is 4, and the absolute minimum is 2.
13. The absolute maximum is , and the absolute minimum is
.
The absolute maximum is 0, and the absolute minimum is
.
15.
17. The absolute maximum is 98
, and the absolute minimum is 0.
19. The absolute maximum is
, and the absolute minimum is
.
21. The absolute maximum is 10e10, and the absolute minimum is
.
23. The absolute maximum is , and the absolute minimum is 0.
25. The dimensions are 2 cm × 8 cm × 8 cm.
27. The dimensions are
.
29. The radius is
, and the height is
31. The company should connect the telephone line 1.98 km from the box.
33. The most economical speed is 40 miles per hour.
35. The dimensions are 7 inches by 14 inches.
.
37.
Demand for the product decreases as the tax rate increases. The optimal tax rate is 12%, and the revenue generated by this tax rate is 16.97 monetary units.
39. Let r be the radius, h be the height, and S be the surface area of the cylinder of volume V. We know that V = π r 2 h, and hence . The formula for is . Now
. The only critical number of S is
. Furthermore, and S(r)>0 if r>rc . Therefore, by the First Derivative Test, the surface area is least when . The height of the cylinder when r = rc is
, as desired.
41. The concentration is greatest two hours after the injection.
Exercise 5.5 1. (a) x = 4000 − 100p
(b)
(c) E(5) = −0.143, The demand decreases by approximately 1.43%.
(d) E(15) = −0.6, The demand decreases by approximately 6%.
(e) E(20) = −1, The demand decreases by approximately 10%.
3. (a) x = 10,000 − 200p
(b)
(c) E(10) = −0.25, The demand decreases by approximately 1.25%.
(d) E(25) = −1, The demand decreases by approximately 5%.
(e) E(35) = −2.333, The demand decreases by approximately 11.665%.
5. , The demand is inelastic.
7. , The demand is inelastic.
9. , The demand is inelastic.
11. , The demand is inelastic.
13. , The demand is elastic.
15. , The demand is inelastic.
17. E(4) = −1.23
19. E(5) = −1.18
21. E(5) = −39
23. E(2) = −0.125
25. E(100) = −3
27. (a) The demand is elastic.
(b) The revenue will decrease.
29. (a) The demand is inelastic.
(b) The revenue will decrease.
31. (a)
(b) E(18) = − 0.667
(c) The demand will decrease by approximately 3.33%.
(d) The revenue will increase.
Exercise 5.6
1.
3.
5.
7.
9. The volume is increasing at a rate of 900 cubic centimeters per second.
11. The length of the leg of side length is decreasing at a rate of per minute.
13. The surface area is shrinking at a rate of −0.75 square meters per minute.
15. The water level is rising at a rate of
meter per minute.
17. The area of the spill is increasing at a rate of 316.67 square feet per minute.
19. (a) The rate of change in daily cost is $500 per day.
(b) The rate of change in daily revenues is $1480 per day.
(c) Revenue is increasing.
(d)
centimeters
(e) The rate of change in daily profit is $980 per day.
21. Revenues are increasing at a rate of $260,000 per year.
Exercise 5.7 1. dy = (3x2 − 2)dx
3.
5.
7.
9.
11.
13. d(x3 − x − 4) = (3x2 − 1)dx
15. y = 2x − 3
17.
19. y=x+1
21. (a) The change is approximately 0.006.
(b) The change is approximately 0.00125.
23. The approximate increase in surface area is
square centimeters.
25. The approximate increase in volume is
cubic meters.
27. The percentage error is 6%.
29. The approximate loss
is cubic centimeters.
31. The estimated height is 30 meters, and the percentage error of this estimate is 0.9%.
33. The clock will lose 70.28 minutes each day.
CHAPTER 5 Review True-False Items 1. False
2. True
3. False
4. False
5. True
6. False
Fill In The Blanks 1. decreasing
2. decreasing, increasing
3. concave up
4. horizontal
5. concavity
6. f ′(x)dx
7. linear approximation
Review Exercises 1. The graph has no horizontal or vertical tangent lines.
3. The graph has a horizontal tangent line at
and a vertical tangent line at (0, 0).
5. (a) Yes
(b) No
(c) vertical tangent line
7. (a) Yes
(b) No
(c) no tangent line
9. (a) The graph is increasing on the intervals (−∞,−2) and (2,∞) and is decreasing on the interval (−2,2).
(b) There is a local maximum at .
and a local minimum at
11. (a) The graph is increasing on the intervals and is decreasing on the intervals
and and
.
(b) The graph has a local maximum at (0,0).
13. (a) The graph is decreasing on (−∞,∞).
(b) The graph has no local maxima or minima.
15. (a) The domain is the interval (−∞,∞).
(b) The x-intercept is (1,0), and the y-intercept is (0,−1).
(c) The graph is increasing on the interval (−∞,∞).
(d) The graph has no local maxima or minima.
(e) The graph has a horizontal tangent at (1,0).
(f)
The end behavior is y = x3.
(g) The graph is concave down on the interval (−∞,1) and is concave up on the interval (1,∞). The point (1,0) is the only inflection point.
17. (a) The domain is the interval (−∞,∞).
(b) The x-intercepts are
,(0,0), and
, and the
y-intercept is (0,0).
(c) The graph is increasing on the intervals (−∞,−1) and (1,∞) and is decreasing on the interval (−1,1).
(d) The graph has a local maximum at (−1,4) and a local minimum at (1,−4).
(e) The tangent lines to the graph are horizontal at (−1,4) and (1,−4).
(f)
The end behavior is y = x5.
(g) The graph is concave down on the interval (−∞,0) and is concave up on the interval (0,∞). The point (0,0) is the only inflection point.
19. (a) The domain is the interval (−∞,∞).
(b) The x-intercepts are (−4,0) and (0,0). The y-intercept is (0,0).
(c) The graph is increasing on the interval (−1,∞) and is decreasing on the interval (−∞,−1).
(d) The graph has a local minimum at (−1,−3).
(e) The graph has a horizontal tangent at (−1,−3) and a vertical tangent at (0,0).
(f)
The end behavior is y = x4/3.
(g) The graph is concave up on the intervals (−∞,0) and (2,∞) and is concave down on the interval (0, 2).
The points (0,0) and
are the inflection points.
21. (a) The domain is the interval (−∞,∞).
(b) The x-intercept and the y-intercept are both (0,0).
(c) The graph is increasing on the interval (−1,1) and is decreasing on the intervals (−∞,−1) and (1,∞).
(d) The graph has a local maximum at (1,1) and a local minimum at (−1,−1).
(e) The graph has horizontal tangent lines at (−1,−1) and (1,1).
(f)
The end behavior is y = 0, which is a horizontal asymptote.
(g) The graph is concave down on the intervals and is concave up on the intervals The points
,
, (0,0) and
,
and and are the inflection
points.
23. There is a local maximum at
,1 ) and a local minimum at
, −1 ).
25. There is a local maximum at (0,0) and local minima at (−1,−1) and (1,−1).
27. There is a local minimum at ( − 1,
.
.
29. The absolute maximum is 8, and the absolute minimum is −1.
31. The absolute maximum is 9, and the absolute minimum is 0.
33. The absolute maximum is 8, and the absolute minimum is −3.
35. , The demand is elastic at p = 20.
37. , The demand is inelastic at p = 10.
39. (a) The demand function is elastic.
(b) Revenue will decrease if the price is raised. dy = (12x3 − 6x2 + 1)dx
41.
43.
45. y = 6x − 18
47.
49.
51. The surface area is increasing at a rate of
square meters per minute.
53. (a)
(b) The minimum average cost is $150.
(c) C(x) = 10x
(d)
Intersection point is (15,150).
(e) The minimum average cost occurs at the production level where the average cost equals the marginal cost.
55. 98 units need to be sold.
57. The radius is
centimeters, and the height is
centimeters.
59. The decrease in area is approximately
square millimeters.
61. (a) The demand decreases by approximately 1800 pounds.
(b) The demand decreases by approximately 9000 pounds.
63. (a) The concentration increases by approximately 0.0071 units.
(b) The concentration decreases by approximately 0.0208 units.
65. (b)
Mathematical Questions from Professional Exams 1. (b)6X2 + 8X + 3
2. (d)
3. (a) The average cost function multiplied by X.
4. (c) Substitute the solution(s) in the second derivative equation, and a positive solution indicates a minimum.
5. (a)
6. (b) Dec, 96
7. (d) 8,645 per year
8. (d) 1, −2
9. (c)(0,2)
10. (c)
11. (c)
12. (b)
13. (a)
14. (b) 0
15. (d) 54
16. (b)
CHAPTER 6 Review True-False Items 1.
2.
3.
3.
4.
5.
6.
7.
8.
9.
10.
True
False
False
False
False
False
True
True
True
False
True
Fill In The Blanks 1.
F(x) = f(x)
2.
3.
4.
5.
6.
∫ f(x)dx
integration by parts
lower, upper limits, integration
0
F(b) − F(a)
7.
Review Exercises 1.
F(x) = x6 + K
3.
5.
7.
9.
11.
13.
7x + K
15.
17.
19.
21.
23.
25.
27.
29.
31.
33. ; the cost is minimum at a production level of zero.
35. (a) R(x) = 500x − 0.005x2
(b) 50,000 televisions need to be sold.
(c) The maximum revenue is $12,500,000.
(d) The increase in revenue is $625,000.
37.
39.
41.
43.
45.
47.
49.
51.
53.
1
−10
15
55.
57.
59.
61.
63.
8
The monthly profit stays the same. The increase in profit is 0.
65. (a) 1 time unit
(b) 6 monetary units
67. (a)
(b) A ≈ 28
(c) A ≈ 20
(d) A ≈ 26
(e) A ≈ 22
(f)
(g) A = 24
69.
90.38
71.
73.
75.
y = e10x
77.
79.
There will be 23,689 bacteria.
The burial ground is 7403 years old.
81. (a)
(b) The economy will total $18.33 million.
83.
Margo should allow for 5432 labor-hours.
85. (a) The market price is $10, and the demand level is 100 units.
(b) The consumer’s surplus is $100, and the producer’s surplus is $250.
(c)
87. (a)
(b) A price of $389.71 should be charged.
(c) A price of $325.58 should be charged.
Mathematical Questions from Professional Exams 1.
(c) 1
2. (d)
3.
4.
5.
6.
(c) x2 e x + 2xe x
(e) None of the above
(c) 62,208
(c) (ln x)2 + C
CHAPTER 6 The Integral of a Function and Applications Exercise 6.1 1.
3. f(x)= x2 + 3x + K
5. F(x) = 4 ln | x| + K
7.
9. 3x + K
11.
13.
15.
17. 2x1/2 + K
19.
21.
23.
25. x − ln | x| + K
27. 2e x − 3 ln | x| + K
29.
31.
33.
35.
37. 4x + e x + K
39. R(x) = 600x
41. R(x) = 10x2 + 5x
43. C(x) = 7x2 − 2800x + 4300, The cost is a minimum when x = 200.
45. C(x) = 10x2 − 8000x + 500, The cost is a minimum when x = 400.
47. (a)
(b) R(x) = 3400x
(c)
(d) A sales volume of 60 units yields maximum profit.
(e) The profit is $99,000.
(f)
49. There will be 1,142,462 inmates in 2008.
51. The population will be 20,055 people in 10 months.
53. There will be 22,700 voting citizens in 3 years.
55. 2 − 2 ln 2 ≈ 0.614 milligrams were produced.
57. The reservoir will be empty after 500.28 days.
Exercise 6.2
1.
3.
5.
7.
9.
11.
13.
15.
17.
19.
21.
23.
25.
27.
29.
31.
33.
35. The value of the car after two years is $19,414.55. The value of the car after four years is $16,624.02.
37. (a) B(t) = 68.6e0.025t
(b) The budget will exceed $100 billion when t = 15.08, which is during January, 2016.
39. (a) N(t) = 2000 e0.01t − 1600
(b) The work force will reach 800 employees in 18.23 years.
41. Let u = ax + b. Then du = adx, so
du = dx. Substituting, we have
Exercise 6.3 1.
3.
5.
7.
9.
11.
13.
15.
17. The population is 5389 ants after four days and is 6012 ants after one week.
19. The car is worth $14,061.64 after 2 years. The car is worth $9,640.02 after 4 years.
Exercise 6.4 1.
3. e
5.
7.
9.
11.
13.
15.
17. 0
19.
21.
23.
25. ln 2
27.
29.
31.
33. 5 ln 5 − 4
35. 0
37. 0
39. 2
41. 64
43. 36
45. 12
47. The cost increase is 567,000.
49. (a) The total number of labor-hours needed is 15,921.
(b) The total number of labor-hours needed is 35,089.
(c) The quantity represents the difficulty of learning a new skill. The closer is to 0, the longer the time required to master the new skill.
51. The projected deficit is $197 billion.
53. The total sales during the first year were $13,450.01.
55. The total number of labor-hours needed is 3660.
57. (a)
(b)
Exercise 6.5 1. 56
3.
5.
7.
9. e−1
11.
13.
15.
17.
19.
21.
23.
25. 8
27.
29. (a) The integral represents area below the line y = 3x + 1, above the the x-axis, and between the vertical lines x = 0 and x = 4.
(b)
(c) 28
31. (a) The integral represents area below the curve y = x2 − 1, above the the x-axis, and between the vertical lines x = 2 and x = 5.
(b)
36
33. (a) The integral represents area below the curve y = e x , above the the x-axis, and between the vertical lines x = 0 and x = 2.
(b)
(c) e2 − 1
35. The consumer’s surplus is
, and the producer’s surplus is
.
37. The operation should continue for 16 years. The profit that can be generated during this period is $85.33 million.
39. (a)
(b) c = 2
41. (a) x2
(b)
(c)
Exercise 6.6 1. A≈3
3. A ≈ 56
5. (a)
(b) A ≈ 36
(c) A ≈ 72
(d) A ≈ 45
(e) A ≈ 63
(f)
A ≈ 54
7. (a)
(b) A ≈ 18
(c) A ≈ 9
(d)
(e)
(f)
9. (a)
(b) A ≈ 22
(c)
(d)
(e)
11. (a)
(b) A ≈ 36
(c) A ≈ 49
(d)
(e) A = 64
13. (a)
(b)
(c)
(d)
(e) A = ln 5
15. (a)
(b) A ≈ 11.475
(c) A ≈ 15.197
(d)
(e)
17. 1.46
19. 38.29
21. (a)
(b) A ≈ 1.42
(c) A ≈ 1.52
(d)
(e) A = 1.57
(f)
The area is .
Exercise 6.7 1.
3.
5.
7. y=ex +3
9.
11. There are 12,975 bacteria after one hour. There are 147,789 bacteria after 90 minutes. It will take 113.58 minutes to reach 1,000,000 bacteria.
13. There will be 7.68 grams of radium
15. The tree died 9727 years ago.
17. The population size is 6944 mosquitoes.
19. There are 55,418 bacteria.
21. (a) P(t) = 10,000 5 t/10 bacteria
(b) There were 250,000 bacteria.
(c) There were 20,000 bacteria after
23. The half-life is 4,620,981 years.
25. (a)
(b) A price of $106.07 should be charged.
(c) A price of $89.19 should be charged.
minutes.
CHAPTER 7 Other Applications and Extensions of the Integral Exercise 7.1 1. This integral is improper because the upper limit of integration is infinite.
3. This integral is improper because the integrand is discontinuous at the lower limit of integration.
5. This integral is improper because the integrand is discontinuous at the lower limit of integration.
7.
9. The improper integral has no value.
11.
13. The improper integral has no value.
15. Area = 2 square units
17. The capital value of the apartment is $102,480.
19. (a) The area is equal to the integral, reaction.
(b) This total reaction is
units.
which is the total
21. (a) The improper integral has no value.
(b)
Exercise 7.2 1.
3.
5. 9
7. −26
9.
11. The average population would be 8.22 · 109 people.
13. The average temperature is 37.5 degrees Celsius.
15. The average speed is 12 meters per second.
17. The average annual revenue is $207.32 billion.
19. The average rainfall is 0.188118 inches.
Exercise 7.3
1. on [0, 2].
.
3. on [0, 1].
.
5. on [0. 5]
.
7. on [1, e]
9.
11.
13.
15.
17. E(x) = 1
19.
21.
23.
25. The probability is .
.
27. The probability is
.
29. The probability is 0.865.
31. (a) The probability is 0.0737.
(b) The probability is 0.2865.
33. (a) The probability is 0.5276.
(b) The probability is 0.0490.
(c) The probability is 0.3867.
(d) Answers will vary.
35. The probability is 0.2865.
37. (a)
(b) The probability is 0.1954.
(c) The probability is 0.2231.
(d) The probability is 0.7769.
39. The expected waiting time is 9 minutes.
41. The contractor can be expected to be off by 2.29% on average.
43. (a) The probability is 0.242.
(b) The probability is 0.516.
45. (a) Let x be the cost of a new car, in thousands of dollars.
(b) The probability is
.
(c)
(d) The expected price is about $16,000.
(e) The expected price is $15,666.67.
(f)
Answers will vary.
47. (a) Answers will vary.
(b) Pr (0.6 ≤ X < 0.9) = 0.3.
49. Let
on [a, b]. Then
CHAPTER 7 Review True-False Items 1. False
2. True
3. False
Fill-In-The-Blanks 1. average value
2. random variable
3. probability density
4.
5.
Review Exercises 1. 1
3. 6
5. The improper integral has no value.
7. Area = 1 square unit.
9. 5
11.
13. 4
15. (a)
(b) E(x) = 1
17. (a)
.
(b)
19. The average yearly sales is 1255 units.
21. The average price is $14.47.
23. (a) The probability is .
(b) The probability is
.
(c) E(X) = 4
25. (a) .
(b) The probability is
.
(c) The probability is
.
(d) The man’s expected age of death is 78.52 years.
27. (a)
.
(b) The probability is
.
(c) The probability is
.
(d) The probability is
.
(e)
.
29. (a) The probability is
(b) The probability is
.
.
(c) The expected wait time is 7.5 minutes.
31. The probability is 0.0821.
33. (a) The probability is 0.1055.
(b) The probability is 0.3189.
35. The probability is 0.5134.
Mathematical Questions from Professional Exams 1.
2. (e) 0
3.
4.
5.
CHAPTER 8 Calculus of Functions of Two or More Variables Exercise 8.1 1.
3.
5.
7. (0, 0, 3), (0, 1, 0), (0, 1, 3), (2, 0, 0), (2, 0, 3), (2, 1, 0)
9. (1, 2, 5), (1, 4, 3), (1, 4, 5), (3, 2, 3), (3, 2, 5), (3, 4, 3)
11. (− 1, 0, 5), (− 1, 2, 2), (− 1, 2, 5), (4, 0, 2), (4, 0, 5), (4, 2, 2)
13. The plane through the point (0, 3, 0) that is parallel to the xz-plane
15. The yz-plain
17. The plain through the point (0, 0, 5) that is parallel to the xy-plane
19. 2
units
2
units
2
units
21.
23.
25.
27.
29. Center = (−1, 1, 0), radius = 2
31. Center = (−2, −2, −1), radius = 3
33. Center = (2, 0, −1), radius = 2
35. x2 + (y − 3)2 + (z − 6)2 = 17
Exercise 8.2 1. f(2, 1) = 5
3.
5.
7. f(2, 1) = 3
9. f(2, 1) = 0
11. (a) f(1, 0) = 3
(b) f(0, 1) = 2
(c) f(2, 1) = 10
(d)
(e)
13. (a) f(0, 0) = 0
(b) f(0, 1) = 0
(c) f(a2, t2) = at + a2
(d)
(e)
15. (a) f(1, 2, 3) = 14
(b) f(0, 1, 2) = 2
(c) f(− 1, −2, −3) = −14
17. The domain is the set {(x, y) | x ≥ 0 and y ≥ 0}. This set is the first quadrant together with its border.
19. The domain is the set {(x, y | x2 + y2 ≤ 9}. This set is the circle of radius 3 centered at the origin and its interior.
21. The domain is the set {(x, y)| x > 0 and y > 0 and y ≠ 1}. This set is the first quadrant.
23. The domain is the set {(x, y) | x2 + y2 ≠ 4}. This set is the union of the regions inside and outside of the circle of radius 2 centered at the origin.
25. The domain is the set {(x, y) | (x, y) ≠ (0, 0)}. This set is the entire xy-plane except for the origin.
27. The domain is the set {(x, y, z)| x2 + y2 + z2 ≥ 16}. This set is the region on and outside of the sphere of radius 4 centered at the origin.
29. The domain is the set {(x, y, z)| x2 + y2 + z2 ≠ 0}. This set is the set of all points in space except for the origin (0, 0, 0).
31. (a)
(b)
(c)
(d)
33. C(r, h) = 600π r2 + 1000π rh dollars
35. (a) 6.75
(b) 18
(c) 2
(d) 1.5
37. The total monthly bill is $159.99.
39. (a) The heat index is 105.2°F.
(b) The relative humidity is 43%.
(c) The relative humidity is 55%.
Exercise 8.3 1.
3.
5.
7.
9.
11.
13.
15.
17. f xy = 0 = f yx
19. f xy = 24x3 y + 14x = f yx
21.
23.
25.
27.
29.
31. The slope is 20.
33. The slope is
35. The slope is 1.
37. The slope is 1.
39. We have that
and
. Now
41. We have that
and
. Then
and
Now
. 43. (a)
(b) If the average price per pound of margarine remains fixed and the average price per pound of butter is increased by $1, is the change in demand for butter. If the average price per pound of butter remains fixed and the average price per pound of margarine is increased by $1 is the change in demand for margarine.
45. (a)
(b)
.
(c) If he pitched 217 innings and he gave up 79 earned runs, his earned run average would increase by 0.041. If he pitched 218 innings and gave up 78 earned runs, his earned run average would decrease by 0.015.
.
47. (a)
(b)
is the change in the heat index with respect to temperature, given a fixed humidity.
(c)
(d)
is the change in the heat index with respect to humidity, given a fixed temperature.
49. No. Explanations will vary.
Exercise 8.4 1. (−1, 0), (0, 0), (1, 0)
3. (−1, −1), (0, 0), (1, 1)
5. (0, 0)
7. The point (0, 0) is a local minimum.
9. The point is
b a local minimum.
11. The point (− 2, 4) is a saddle point.
13. The point (2, −1) is a local minimum.
15. The point (4, −2) is a local minimum.
17. The point (0, 0) is a saddle point.
19. The point (0, 0) is a saddle point, and the point (2, 2) is a local minimum.
21. The point (0, 0) is a critical point that is neither a saddle point nor a local extremum, and the point is a saddle point.
23. The funtion has no critical points.
25. The maximize profits, the quantities sold should be
units and
units. The corresponding prices are
and
. The maximum profit is
.
27. The manufacturer should produce 15,250 tons of grade A and 4100 tons of grade B to maximize profit.
29. For a fixed amount of the first drug, an amount of
of the second drug maximizes the
reaction. For a fixed amount of the second drug, an amount of the reaction. If the amounts of both drugs are variable, of the second drug maximize the reaction.
31. There are no such values of x and t that will maximize y.
of the first drug maximizes
units of the first drug and
units
33. (a) The dimensions are
inches by
inches by
inches.
(b) For the cylinder of maximum volume, the radius is inches and the height is
inches.
35. x = 50 tons, y = 1 ton per week
Exercise 8.5 1. The maximum value is 15.
3. The minimum value is .
5. The maximum value is 528.
7. The minimum value is 612.
9. The maximum value is 16,000.
11. The minimum value is
.
13. The two numbers are both 50.
15. The three numbers are all
.
17. To minimize cost, the factory should produce 18 units of type x and 36 units of type y.
19. The dimensions are
inches by
inches by
.
21. (a)
units of capital 750 and units of labor will maximize the total production.
(b) The maximum number of units of production is 529.14 units.
23. The dimensions are feet by feet by 5.593 feet by 5.593 feet.
feet by
feet, which is approximately 5.593
25. The dimensions are 2.29 feet by 3.43 feet.
Exercise 8.6 1.
3. 18x2 + 4x
5.
7. 8x − 22
9.
11. e y (e2 − 1)
feet by
feet by
feet, which is approximately 2.29 feet by
13.
15. (e4 − 1) e−4y
17. 8
19.
21.
23. 22
25. 12
27. 24
29. 21
31.
CHAPTER 8 Review True-False Items 1. True
2. False
3. False
4. False
Fill In The Blanks 1. surface
2.
3. x = x0
4. saddle point
Review Exercises 1. 3 units
3.
5. 5 units
7. The radius is 3 units.
9. (x + 6)2 + (y − 3)2 + (z − 1)2 = 4
11. The center is the point (1, −3, −8), and the radius is 5.
13. (a) (x − 1)2 + (y + 4)2 + (z − 3)2 = 36
(b) The center is the point (1, −4, 3), and the radius is 6.
15. (a) f(1, −3) = 11
(b) f(4, − 2) = −8
17. (a)
(b) f(4, −2) = 0
19. The domain is the entire xy-plane.
21. The domain is the set {(x, y)| y > x2 + 4}, which is the set of points above the parabola y = x2 + 4.
23. The domain is the set {(x, y)| (x + 2)2 + y2 ≥ 9}, which is the set of points on or outside of the circle of radius 3 centered at the point (−2, 0).
25.
27.
29.
31.
33.
35. The slope is 12.
37. The slope is 1.
39. (a) The only critical point (− 4, −2).
(b) The point (−4, −2) is a local maximum.
41. (a) The only critical point is (1, 2).
(b) The point (1, 2) is a local maximum.
43. (a) The only critical point is a
(b) The point a
45. The maximum value is
.
47. The minimum value is
49.
51. 24x2 + 8
.
.
is a local minimum.
53. 51
55. 448
57. 32
59.
61. The volume is 672 cubic units.
63. (a)
(b)
.
(c) The factory should increase the use of labor. Explanations will vary.
65. 67. Cx (x, y) = 40, Cy (x, y) = 45, If the number of deluxe vacuum cleaners produced remains fixed, increasing the production of standard vacuum cleaners by one will increase cost by $40. If the number of standard vacuum cleaners produced remains fixed, increasing the production of deluxe vacuum cleaners by one will increase cost by $45.
69. (a)
(b)
If the demand for deluxe vacuum cleaners produced remains fixed, an increase of one in the demand for standard vacuum cleaners will change revenue by Rx dollars. If the demand for standard vacuum cleaners produced remains fixed, an increase of one in the demand for deluxe vacuum cleaners will change revenue by Ry dollars.
71. (a)
(b)
, If the demand for deluxe vacuum cleaners produced remains at 30 vacuum cleaners, increasing the demand of standard vacuum cleaners from 50 to 51 will decrease profit by $160. If the demand for standard vacuum cleaners produced remains at 50 vacuum cleaners, increasing the demand of deluxe vacuum cleaners from 30 to 31 will increase profit by $70.
73. (a) 4000 units of brand x at a price of $4,000 and 5000 units of brand y at a price of $11,000 will maximize profit.
(b) The maximum profit is a loss of $159,000.
75. (a) $15,300 should be allocated to capital, and $35,700 should be allocated to labor.
(b) The maximum number of units is 3409 units.
APPENDIX A Graphing Utilities Exercise Appendix A.1 1. (−1, 2) quadrant II
3. (3, 1) quadrant I
5. X min = −6
X max = 6
X scl = 2
Y min = −4
Y max = 4
Y scl = 2
7. X min = −6
X max = 6
X scl = 2
Y min = −1
Y max = 3
Y scl = 1
9. X min = 3
X max = 9
X scl = 1
Y min = 2
Y max = 10
Y scl = 2
11. X min = −12
X max = 6
X scl = 1
Y min = −4
Y max = 8
Y scl = 1
13. X min = −30
X max = 50
X scl = 10
Y min = −100
Y max = 50
Y scl = 10
15. X min = −10
X max = 110
X scl = 10
Y min = −20
Y max = 180
Y scl = 20
Exercise Appendix A.2 1. (a)
(b)
(c)
(d)
3. (a)
(b)
(c)
(d)
5. (a)
(b)
(c)
(d)
7. (a)
(b)
(c)
(d)
9. (a)
(b)
(c)
(d)
11. (a)
(b)
(c)
(d)
13. (a)
(b)
(c)
(d)
15. (a)
(b)
(c)
(d)
17. (a)
19. (a)
21. (a)
23. (a)
25. (a)
27. (a)
29. (a)
31. (a)
Exercise Appendix A.3 1. Yes
3. Yes
5. No
7. Yes
9. Y min = 1
Y max = 9
Y scl = 1
Exercise Appendix A.4 1. The smaller of the two x-intercepts is −3.41.
3. The smaller of the two x-intercepts is −1.71.
5. The smaller of the two x-intercepts is −0.28.
7. The positive x-intercept is 3.
9. The positive x-intercept is 4.5.
11. The positive x-intercepts are 0.32 and 12.3.
13. The positive x-intercepts are 1 and 23.
15. (a) The x-intercepts are −1 and 1. The y-intercept is −1.
(b) The graph is symmetric with respects to the y-axis.
17. (a) The graph has no intercepts.
(b) The graph is symmetric with respects to the origin.