Student Solutions Manual to accompany Calculus for Business, Economics, and the Social and Life Sciences Tenth Edition, Brief Laurence D. Hoffman Smith Barney Gerald L. Bradley

Student Solutions Manual
to accompany


Calculus for Business,
Economics, and the
Social and Life Sciences
Tenth Edition, Brief

Laurence D. Hoffman
Smith Barney

Gerald L. Bradley
Claremon McKenna College




Prepared by
Devilyna Nichols
Purdue University

,CONTENTS




Chapter 1 Functions, Graphs, and Limits 1
1.1 Functions 1
1.2 The Graph of a Function 6
1.3 Linear Functions 14
1.4 Functional Models 19
1.5 Limits 26
1.6 One-Sided Limits and Continuity 30
Checkup for Chapter 1 33
Review Problems 36


Chapter 2 Differentiation: Basic Concepts 43
2.1 The Derivative 43
2.2 Techniques of Differentiation 52
2.3 Product and Quotient Rules; Higher-Order Derivatives 57
2.4 The Chain Rule 64
2.5 Marginal Analysis; Approximations Using Increments 72
2.6 Implicit Differentiation and Related Rates 75
Checkup for Chapter 2 82
Review Problems 84


Chapter 3 Additional Applications of the Derivative 93
3.1 Increasing and Decreasing Functions; Relative Extrema 93
3.2 Concavity and Points of Inflection 103
3.3 Curve Sketching 114
3.4 Optimization 124
3.5 Additional Applied Optimization 132
Checkup for Chapter 3 141
Review Problems 148


Chapter 4 Exponential and Logarithmic Functions 159
4.1 Exponential Functions 159
4.2 Logarithmic Functions 165
4.3 Differentiation of Logarithmic and Exponential Functions 173
4.4 Additional Exponential Models 182
Checkup for Chapter 4 199
Review Problems 205
iii

,iv Contents




Chapter 5 Integration 219
5.1 Antidifferentiation; the Indefinite Integral 219
5.2 Integration by Substitution 226
5.3 The Definite Integral and the Fundamental Theorem of Calculus 233
5.4 Applying Definite Integration: Area Between Curves and Average Value 238
5.5 Additional Applications to Business and Economics 245
5.6 Additional Applications to the Life and Social Sciences 252
Checkup for Chapter 5 259
Review Problems 262

Chapter 6 Additional Topics in Integration 273
6.1 Integration by Parts; Integral Tables 273
6.2 Introduction to Differential Equations 284
6.3 Improper Integrals; Continuous Probability 292
6.4 Numerical Integration 300
Checkup for Chapter 6 307
Review Problems 312

Chapter 7 Calculus of Several Variables 325
7.1 Functions of Several Variables 325
7.2 Partial Derivatives 329
7.3 Optimizing Functions of Two Variables 336
7.4 The Method of Least Squares 346
7.5 Constrained Optimization: The Method of Lagrange Multipliers 353
7.6 Double Integrals 362
Checkup for Chapter 7 371
Review Problems 375

,Chapter 1

Functions, Graphs, and Limits

9. 1
1.1 Functions f (t ) = (2t − 1)−3/2 =
√
,
( 2t − 1)3
f(x) = 3x + 5, 1
1. f (1) = √ = 1,
f (0) = 3(0) + 5 = 5 [ 2(1) − 1]3
f (−1) = 3(−1) + 5 = 2 f (5) = √ 1 = 1 =1 ,
f (2) = 3(2) + 5 = 11 [ 2(5) − 1]3 √
1 [ 9]3 27
1 1
f (13) = √ = √ = .
[ 2(13) − 1]3 [ 25]3 125
3. f(x) = 3x2 + 5x − 2, 11. f(x) = x − |x − 2|,

f (0) = 3(0)2 + 5(0) − 2 = −2, f (1) = 1 − |1 − 2 |= 1 − | − 1|= 1 − 1 = 0,
f (2) = 2 − |2 − 2|= 2 − |0|= 2,
f (−2) = 3(−2)2 + 5(−2) − 2 = 0,
f (3) = 3 − |3 − 2 |= 3 − |1|= 3 − 1 = 2.
f (1) = 3(1)2 + 5(1) − 2 = 6.
13. −2x + 4 if x ≤ 1
h(x) =
x2 + 1 if x > 1
1 h(3) = (3)2 + 1 = 10
5. g(x) = x + , h(1) = −2(1) + 4 = 2
x h(0) = −2(0) + 4 = 4
1
g(−1) = −1 + = −2, h(−3) = −2(−3) + 4 = 10
−1 x
1
15. g(x) .
g(1) = 1 + = 2, =
1 1 +2 x2
g(2) = 2 1 5. Since 1 + x /= 0 for any real number, the domain is
+ =
2 2 the set of all real numbers.
√
17. f (t ) = 1 − t.
, Since negative numbers do not have real square
7. h(t) = t 2 + 2t + 4,
roots, the domain is all real numbers such that
, √ 1 —t ≥ 0, or t ≤ 1. Therefore, the domain is not the
h(2) = 22 + 2(2) + 4 = 2 3,
, set of all real numbers.
2h(0) = 0
, √ x2 + 5
= + (−4) 19. g(x) = .
4 = 2,+ 2(−4) + 4 = 2 3
h(−4) 2
+ 2(0) x+2
1

,
,2 Chapter 1. Functions, Graphs, and Limits




Since denominators cannot be 0, the domain consists 35. f(x) = 4x − x2
of all real numbers such that x /= −2.
f(x + h) − f(x) 4(x + h) − (x + h)2 − (4x − x2)
√ =
21. f(x) = 2x + 6. h h
4x + 4h − (x2 + 2xh + h2) − 4x + x2
Since negative numbers do not have real square =
roots, the domain is all real numbers such that h
2x + 6 ≥ 0, or x ≥ −3. 4x + 4h − x2 − 2xh − h2 − 4x + x2
= h
t +2
23. f (t ) = √ . 4h − 2xh − h2 h(4 − 2x − h)
9 − 2t = =
Since negative numbers do not have real square roots h h
and denominators cannot be zero, the domain is the = 4 − 2x − h
set of all real numbers such that 9 − t 2 > 0, namely
37. f(x) x
−3 < t < 3. =
x+1
25. f(u) = 3u2 + 2u − 6 and g(x) = x + 2, so f(x + h) − f(x) x+h − x
= (x+h)+1 x+1
f(g(x)) = f(x + 2) = 3(x + 2)2 + 2(x + 2) − 6 h h
x+h − x (x + h + 1)(x + 1)
= 3x2 + 14x + 10. = x+h+1
· (x + h + 1)(x + 1)
x+1
h
27. f(u) = (u − 1)3 + 2u2 and g(x) = x + 1, so (x + h)(x + 1) − x(x + h + 1)
=
f(g(x)) = f(x + 1) h(x + 1)(x + h + 1)
x2 + hx + x + h − x2 − xh − x
= [(x + 1) − 1]3 + 2(x + 1)2 = h(x + 1)(x + h + 1)
= x3 + 2x2 + 4x + 2. h 1
= =
1 h(x + 1)(x + h + 1) (x + 1)(x + h + 1)
29. f(u) = and g(x) = x − 1, so √
u2
39. f (g(x)) = f (1√− 3x) = 1 − 3x
1 √
f(g(x)) = f(x − 1) = (x − 1)2 . g (f (x)) = g x = 1− 3 x
To solve √ √
1 − 3x = 1 − 3 x, square both sides, so
√ √
31. f(u) = u + 1 and g(x) = x2 − 1, so 1 − 3x = 1 − 6 x + 9x
f(g(x)) = f(x 2 − 1) √
—√ 3x = −6 x + 9x
6 x = 12x
= (x 2 − 1) + 1 √
, x = 2x
= x2 = |x|. squaring both sides again,
f(x) = 4 − 5x x = 4x2
33.
f(x + h) − f(x) 4 − 5(x + h) − (4 − 5x) 0 = 4x2 − x
=
h h 0 = x(4x − 1)
4 − 5x − 5h − 4 + 5x −5h
= = −5 1
h h x = 0, x=
4

,1.1 Functions 3




Since squaring both sides can introduce extraneous 51. f(x) = (x − 1)2 + 2(x − 1) + 3 can be rewritten as
solutions, need to check these values. g(h(x)) with g(u) = u2 + 2u + 3 and h(x) = x − 1.
, √
1 − 3(0) =? 1 − 3 0 1
53. f(x) =
1= 1 +1
x2
, , can be rewritten as g (h(x))
1 1
1− 3 = 1− 3
?
4 4 1
1 =? 1 − 3 with g(u) = u
2 2 2
and h(x) = x + 1.
1 1
/= − √
2 2 4
55. f (x) = 3
2−x +
Also check remaining value to see if is in domain 2−x
of f and g functions. Since f (0) and g(0) are both can be rewritten as g (h(x)) with
defined, f(g(x)) = g(f(x)) when x = 0. √ 4
g(u) = 3
u+
41. u
2 x+3 + 3 and h(x) = 2 − x.
x +3 =
x −2
=
f (g(x)) = f x+3
x
57. D(x) = − 0.02x + 29; C(x) = 1.43x2 + 18.3x +
x −2 x−2 −
1
15.6
+3
2x+3
2x + 3 = x−1 =x (a) R(x) = xD(x) = x(−0.02x + 29)
g (f(x)) = g x −1 2x+3 − 2 2
x−1 = −0.02x + 29x
Answer will be all real #’s for which f and g are P(x) = R(x) − C(x)
defined. So, f (g(x)) = g (f(x)) for all real #’s
except x = 1 and x = 2. = −0.02x2 + 29x − 1.43x2 + 18.3x + 15.6
2 = −1.45x2 + 10.7x − 15.6
43. f(x) = 2x − 3x + 1, (b) 0 when
P(x) >
f(x − 2) = 2(x − 2)2 − 3(x − 2) + 1
− 1.45x + 10.7x− 15.6 > 0
2

= 2x2 − 11x + 15. Using the quadratic formula, the zeros of P are
,
45. f(x) = (x + 1)5 − 3x2, −10.7 ± (10.7)2 − (4)(−1.45)(−15.6)
x=
f(x − 1) = [(x − 1) + 1]5 − 3(x − 1)2 2(−1.45)
= x − 3x + 6x − 3.
5 2 x = 2, 5.38
√ so, P(x) > 0 when 2 < x < 5.38.
47. f(x) = x,
√ 59. D(x) = −0.5x + 39; C(x) = 1.5x2 + 9.2x + 67
f(x + 3x − 1) =
2
x2 + 3x − 1.
x −1 (a) R(x) = xD(x) = x(−0.5x + 39)
49. f(x) = ,
x = −0.5x2 + 39x
P(x) = R(x) − C(x)
(x + 1) − 1
f(x + 1) =
x+1 = −0.5x2 + 39x − 1.5x2 + 9.2x + 67
x
= . = −2x2 + 29.8x − 67
x+1

,4 Chapter 1. Functions, Graphs, and Limits




(b) P(x) > 0 when (e) Need to find x when W(x) = 150.
− 2x + 29.8x− 67 > 0
2
600x
Using the quadratic formula, the zeros of P are =
150
, 300 − x
−29.8 ± (29.8)2 − (4)(−2)(−67) (150)(300 − x) = (1)(600x)
x= 300 − x = 4x
2(−2)
x ≈ 2.76, 12.14 x = 60
After 150 worker-hours, 60% of the households
so, P(x) > 0 when 2.76 < x < 12.14. have received a new telephone book.
150x
65. C(x)
61. C(q) = 0.01q2 + 0.9q + 2 =
200 − x
(a) C(10) = 0.01(10)2 + 0.9(10) + 2 (a) All real numbers except x = 200.
= $12 (b) All real numbers for which 0 ≤ x ≤ 100. If
(b) Cost of 10th unit x < 0 or x > 200 then C(x) < 0 but cost is
non-negative. x > 100 means more than 100%.
= Cost of 10 units − Cost of 9 units 150(50)
(c) C(50) = = 50 million dollars.
= C(10) − C(9) 200 − 50
150(100)
C(9) = 0.01(9)2 + 0.9(9) + 2 = $10.91 (d) C(100) = 200 − 100 = 150
= $12 − $10.91 = $1.09 C(100) −
C(50) = 100 million dollars.
150x
(e) = 37.5
600x 200 − x
63. W(x)
= 187.5x = 37.5(200),
300 − x 7, 500
x= = 40%.
(a) 300 —x /=0 187.5
x / 300
6
The domain= is all real numbers except 300. 67. P(t) = 20 −
(b) Typically, the domain would be restricted to t +1
theh first quadrant. That is, x ≥ 0. However, 6
since x is a percentage, the restriction should be (a) P (9) = 20 − or 19,400 people.
0 ≤ x ≤ 100. 9+1
6
(c) When x = 50, (b) P (8) = 20 −
600(50) +20
P (9) − P (8)8= 1 3 20 − 2
= 1



W (50) =
300 − 50 — − 3 15
5
This accounts for about 1 of 1,000 people, or

= 120 worker-hours = 15
(d) To distribute to all of the households, x 100
and
600(100) 67 people.
W (100) = (c) P(t) approaches 20, or 20,000 people.
Writing exercise–Answers will vary.
69. S(r) = C(R2 − r2)
= 1.76 × 105(1.22 × 10−4 − r2).
300 − 100
= 300 worker-hours

,1.1 Functions 5




(a) b b S(0)= (1.76 × 105)(1.44 × 10−4) = The b positive b root b leads b to b t b(0.04t b +
25.344cm/sec. b0.2) b = b0 bor bt b= b0. b(Disregard bt b < b0.)
bThe bnegative broot bproduces b imaginary
(b) b S(0.6 b× b10−2)
b numbers. b t b = b0 b now.
= b1.76 b× b105(1.44 b× b104 b− b0.62 b×
75. b (a) b c(p) b = b0.4p b + b1 band b p(t) b = b8 b+ b0.2tb2
b10−4)
c(t) b = b0.4(8 b+ b0.2tb2) b + b1 b= b0.08tb2 b + b4.2 b PPM.
= b1.76 b× b105(1.08 b× b104)
= 0.08(2) + 4.2 = 0.32 + 4.2 =
(b) b b c b (2) 2

= b19.008 bcm/sec. b4.52PPM.
√ b 2 b
(c) b 6.2 b= b0.08tb2 b+ b4.2, t b2 b = = b25, b or b t b = b5
71. b b s(A) b= b2 b 9
3

. A years. 0.08
√
(a) b b bs(8)= 2.9 3 b b8= 2.9 × 2 = 5.8
Since b the b number b of b species b should b be 4x2 b − b3
77. b To b find b the b domain b of b f(x)
b an = 2x2 x 3 , b Press
y b= b. + −
integer, byou bwould bexpect bto bfind
bapproximatelyb6 bspecies.
√bb
√3bb
Enter b (4x b ∧ b2 b − b3) b ÷ b(2x b ∧ b2 b + bx b − b3) b for b y1 b=
(b) b b s1 b = b2.9 A b and bs2 b = b2.9 b 3 b 2A Press graph .
s2 b= b2.9 b b b 2 b b √ √
b A b= 2b For b a b better b view b of b the b vertical b asymptotes, b press
b 2.9 b b A b b =√ √
2s1.
3 3 zoom

(c) b 100 b= b2.9 √ 3 3b A 3 and benter bZoom bln. bUse barrow bbuttons bto bmove
cross-hair bto bthe bleft-most bvertical basymptote.
100 √ 3bb bWhenbit b appears b cross-hair b is b on bthe b line, b zoom
=b A
2.9 b in b again b for ba bmore baccurate breading. bMove
100 b b 3 b√ bcross-hair bagain bto b be b on bthe b line.
= −b It b appears
3b
=b bthat bx 1.5 bis bnot
2.9
b3 bin bthe bdomain bof bf b. bZoom bout bonce bto bmove
bb b
100 b A bcross-hairbto bthe b rightmost b vertical b asymptote
b b3 bdomain
and b repeat
of f.bthe bprocedure b of b zoom b in b to b find
=A b that b x b = b1 bis b not b the
2.9
The b domain b consists b of b all b values b except b x b =
Need ban b area bof b approximately b41,002 b−1.5band bx b = b1.
b squarebmiles.
√
79. b For b f(x) b= b2 x b − b1 band b g(x) b= bx3 b− b1.2, b to b find
4, b374
73. b bQ(p) b= 2 and f b (g(2.3)) b, bwe bmust bfind bg b(2.3) bfirst band bthen binput
p
p(t) b = b0.04tb2 b + b0.2t b + that bans√wer binto bf b. bPress b by b = b .
b12
Input b2 b (x b − b1) b for by1 b= band bpress b enter b .
(a) b b Q(t)=
b 4, b374 Input b x b ∧ b3 b− b1.2 b for b y2 b =.
(0.04tb2 b + b0.2t b + b12)2 Use bthe bwindow bdimensions b[−15, b15]1 bby
4, b374 4, b374 b[−10, b10]1. b Use b the b value b function
(b) Q(10) b = =b b under b the
(4 b+ b2 b+ b12)2 324 calc bmenu, binput b2.3, band bpress b enter b .
Use b↑b b band↓ b b b barrows bto bbe bsure bthat
= by∧2 b b−b b bx b
= b13.5 bkg/week. b b b b3 b b b b1.2 bis bdisplayed bin bthe bupper bleft
(c) 30.375 b 4, b374 bcorner. bThe blower bright bcorner bdisplay bshould
= b
(0.04tb + b0.2t b + b12)2
2
bread by b = b10.967
Use bthe bvalue√function bagain band binput b10.967.
4, b374 Verify by1= 2 (x b − b1) b is b displayed b in b the b upper b left
(0.04tb2 + 0.2t + 12)2 = 144 = 122
30.375 = corner.
b±12. =
So b0.04t + b0.2t b + b12 b =
b2

, T
6 h Chapter 1. Functions, Graphs, and Limits
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