7 inverse functions -...

7 INVERSE FUNCTIONS:Exponential, Logarithmic, and Inverse Trigonometric Functions

7.1 Inverse Functions

1. (a) See Definition 1.

(b) It must pass the Horizontal Line Test.

3. f is not one-to-one because 2 6= 6, but f(2) = 2.0 = f(6).

5. No horizontal line intersects the graph of f more than once. Thus, by the Horizontal Line Test, f is one-to-one.

7. The horizontal line y = 0 (the x-axis) intersects the graph of f in more than one point. Thus, by the Horizontal Line Test,

f is not one-to-one.

9. The graph of f(x) = x2 − 2x is a parabola with axis of symmetry x = − b

2a= − −2

2(1)= 1. Pick any x-values equidistant

from 1 to find two equal function values. For example, f(0) = 0 and f(2) = 0, so f is not one-to-one.

11. g(x) = 1/x. x1 6= x2 ⇒ 1/x1 6= 1/x2 ⇒ g (x1) 6= g (x2), so g is one-to-one.

Geometric solution: The graph of g is the hyperbola shown in Figure 14 in Section 1.2. It passes the Horizontal Line Test,

so g is one-to-one.

13. h(x) = 1 + cosx is not one-to-one since h(a+ 2πn) = h(a) for any real number a and any integer n.

15. A football will attain every height h up to its maximum height twice: once on the way up, and again on the way down. Thus,

even if t1 does not equal t2, f(t1) may equal f(t2), so f is not 1-1.

17. Since f(2) = 9 and f is 1-1, we know that f−1(9) = 2. Remember, if the point (2, 9) is on the graph of f , then the point

(9, 2) is on the graph of f−1.

19. h(x) = x+√x ⇒ h0(x) = 1 + 1/(2

√x ) > 0 on (0,∞). So h is increasing and hence, 1-1. By inspection,

h(4) = 4 +√4 = 6, so h−1(6) = 4.

21. We solve C = 59 (F − 32) for F : 9

5C = F − 32 ⇒ F = 95C + 32. This gives us a formula for the inverse function, that

is, the Fahrenheit temperature F as a function of the Celsius temperature C. F ≥ −459.67 ⇒ 95C + 32 ≥ −459.67 ⇒

95C ≥ −491.67 ⇒ C ≥ −273.15, the domain of the inverse function.

23. y = f(x) = 3− 2x ⇒ 2x = 3− y ⇒ x =3− y

2. Interchange x and y: y =

3− x

2. So f−1(x) = 3− x

2.

25. f(x) =√10− 3x ⇒ y =

√10− 3x (y ≥ 0) ⇒ y2 = 10− 3x ⇒ 3x = 10− y2 ⇒ x = − 1

3y2 + 10

3 .

Interchange x and y: y = − 13x2 + 10

3. So f−1(x) = − 1

3x2 + 10

3. Note that the domain of f−1 is x ≥ 0.

245

246 ¤ CHAPTER 7 INVERSE FUNCTIONS

27. For f(x) = 1−√x1 +

√x

, the domain is x ≥ 0. f(0) = 1 and as x increases, y decreases. As x →∞,

1−√x1 +

√x· 1/

√x

1/√x=1/√x− 1

1/√x+ 1

→ −11= −1, so the range of f is −1 < y ≤ 1. Thus, the domain of f−1 is −1 < x ≤ 1.

y =1−√x1 +

√x

⇒ y(1 +√x ) = 1−√x ⇒ y + y

√x = 1−√x ⇒ √

x+ y√x = 1− y ⇒

√x(1 + y) = 1− y ⇒ √

x =1− y

1 + y⇒ x =

1− y

1 + y

2

. Interchange x and y: y =1− x

1 + x

2

. So

f−1(x) =1− x

1 + x

2

with −1 < x ≤ 1.

29. y = f(x) = x4 + 1 ⇒ y − 1 = x4 ⇒ x = 4√y − 1 (not ± since

x ≥ 0). Interchange x and y : y = 4√x− 1. So f−1(x) = 4

√x− 1. The

graph of y = 4√x− 1 is just the graph of y = 4

√x shifted right one unit.

From the graph, we see that f and f−1 are reflections about the line y = x.

31. Reflect the graph of f about the line y = x. The points (−1,−2), (1,−1),(2, 2), and (3, 3) on f are reflected to (−2,−1), (−1, 1), (2, 2), and (3, 3)

on f−1.

33. (a) x1 6= x2 ⇒ x31 6= x32 ⇒ f(x1) 6= f(x2), so f is one-to-one.

(b) f 0(x) = 3x2 and f(2) = 8 ⇒ f−1(8) = 2, so (f−1)0(8) = 1/f 0(f−1(8)) = 1/f 0(2) = 112 .

(c) y = x3 ⇒ x = y1/3. Interchanging x and y gives y = x1/3,

so f−1(x) = x1/3. Domain f−1 = range(f) = R.

Range(f−1) = domain(f) = R.

(d) f−1(x) = x1/3 ⇒ (f−1)0(x) = 13x−2/3 ⇒

(f−1)0(8) = 13

14= 1

12as in part (b).

(e)

35. (a) Since x ≥ 0, x1 6= x2 ⇒ x21 6= x22 ⇒ 9− x21 6= 9− x22 ⇒ f(x1) 6= f(x2), so f is 1-1.

(b) f 0(x) = −2x and f(1) = 8 ⇒ f−1(8) = 1, so (f−1)0(8) = 1

f 0(f−1(8))=

1

f 0(1)=

1

−2 = −1

2.

SECTION 7.1 INVERSE FUNCTIONS ¤ 247

(c) y = 9− x2 ⇒ x2 = 9− y ⇒ x =√9− y.

Interchange x and y: y =√9− x, so f−1(x) =

√9− x.

Domain(f−1) = range (f) = [0, 9].

Range(f−1) = domain (f) = [0, 3].

(d) (f−1)0(x) = −1/ 2√9− x ⇒ (f−1)0(8) = − 12

as in part (b).

(e)

37. f(0) = 4 ⇒ f−1(4) = 0, and f(x) = 2x3 + 3x2 + 7x+ 4 ⇒ f 0(x) = 6x2 + 6x+ 7 and f 0(0) = 7.

Thus, (f−1)0(4) = 1

f 0(f−1(4))=

1

f 0(0)=1

7.

39. f(0) = 3 ⇒ f−1(3) = 0, and f(x) = 3 + x2 + tan(πx/2) ⇒ f 0(x) = 2x+ π2sec2(πx/2) and

f 0(0) = π2 · 1 = π

2 . Thus, f−10(3) = 1/f 0 f−1(3) = 1/f 0(0) = 2/π.

41. f(4) = 5 ⇒ f−1(5) = 4. Thus, (f−1)0(5) = 1

f 0(f−1(5))=

1

f 0(4)=

1

2/3=3

2.

43. We see that the graph of y = f(x) =√x3 + x2 + x+ 1 is increasing, so f is 1-1.

Enter x = y3 + y2 + y + 1 and use your CAS to solve the equation for y.

Using Derive, we get two (irrelevant) solutions involving imaginary expressions,

as well as one which can be simplified to the following:

y = f−1(x) = − 3√46

3√D − 27x2 + 20− 3

√D + 27x2 − 20 + 3

√2

where D = 3√3√27x4 − 40x2 + 16.

Maple and Mathematica each give two complex expressions and one real expression, and the real expression is equivalent

to that given by Derive. For example, Maple’s expression simplifies to 16

M2/3 − 8− 2M1/3

2M1/3, where

M = 108x2 + 12√48− 120x2 + 81x4 − 80.

45. (a) If the point (x, y) is on the graph of y = f(x), then the point (x− c, y) is that point shifted c units to the left. Since f is

1-1, the point (y, x) is on the graph of y = f−1(x) and the point corresponding to (x− c, y) on the graph of f is

(y, x− c) on the graph of f−1. Thus, the curve’s reflection is shifted down the same number of units as the curve itself is

shifted to the left. So an expression for the inverse function is g−1(x) = f−1(x)− c.

(b) If we compress (or stretch) a curve horizontally, the curve’s reflection in the line y = x is compressed (or stretched)

vertically by the same factor. Using this geometric principle, we see that the inverse of h(x) = f(cx) can be expressed as

h−1(x) = (1/c) f−1(x).


7.2 Exponential Functions and Their Derivatives

1. (a) f(x) = ax, a > 0 (b) R

(c) (0,∞) (d) See Figures 6(c), 6(b), and 6(a), respectively.

3. All of these graphs approach 0 as x→−∞, all of them pass through the point

(0, 1), and all of them are increasing and approach∞ as x→∞. The larger the

base, the faster the function increases for x > 0, and the faster it approaches 0 as

x→−∞.

5. The functions with bases greater than 1 (3x and 10x) are increasing, while those

with bases less than 1 13

x and 110

x are decreasing. The graph of 13

x is the

reflection of that of 3x about the y-axis, and the graph of 110

x is the reflection of

that of 10x about the y-axis. The graph of 10x increases more quickly than that of

3x for x > 0, and approaches 0 faster as x→−∞.

7. We start with the graph of y = 4x (Figure 3) and then

shift 3 units downward. This shift doesn’t affect the

domain, but the range of y = 4x − 3 is (−3,∞) .There is a horizontal asymptote of y = −3.

y = 4x y = 4x − 3

9. We start with the graph of y = 2x (Figure 3),

reflect it about the y-axis, and then about the

x-axis (or just rotate 180◦ to handle both

reflections) to obtain the graph of y = −2−x.

In each graph, y = 0 is the horizontal

asymptote.y = 2x y = 2−x y = −2−x

SECTION 7.2 EXPONENTIAL FUNCTIONS AND THEIR DERIVATIVES ¤ 249

11. We start with the graph of y = ex (Figure 12 ) and reflect about the y-axis to get the graph of y = e−x. Then we compress the

graph vertically by a factor of 2 to obtain the graph of y = 12e−x and then reflect about the x-axis to get the graph of

y = − 12e−x. Finally, we shift the graph upward one unit to get the graph of y = 1− 1

2e−x.

13. (a) To find the equation of the graph that results from shifting the graph of y = ex 2 units downward, we subtract 2 from the

original function to get y = ex − 2.

(b) To find the equation of the graph that results from shifting the graph of y = ex 2 units to the right, we replace x with x− 2in the original function to get y = e(x−2).

(c) To find the equation of the graph that results from reflecting the graph of y = ex about the x-axis, we multiply the original

function by −1 to get y = −ex.

(d) To find the equation of the graph that results from reflecting the graph of y = ex about the y-axis, we replace x with −x in

the original function to get y = e−x.

(e) To find the equation of the graph that results from reflecting the graph of y = ex about the x-axis and then about the

y-axis, we first multiply the original function by −1 (to get y = −ex) and then replace x with −x in this equation to

get y = −e−x.

15. (a) The denominator 1 + ex is never equal to zero because ex > 0, so the domain of f(x) = 1/(1 + ex) is R.

(b) 1− ex = 0 ⇔ ex = 1 ⇔ x = 0, so the domain of f(x) = 1/(1− ex) is (−∞, 0) ∪ (0,∞).

17. Use y = Cax with the points (1, 6) and (3, 24). 6 = Ca1 C = 6a

and 24 = Ca3 ⇒ 24 =6

aa3 ⇒

4 = a2 ⇒ a = 2 [since a > 0] and C = 62= 3. The function is f(x) = 3 · 2x.

19. 2 ft = 24 in, f(24) = 242 in = 576 in = 48 ft. g(24) = 224 in = 224/(12 · 5280) mi ≈ 265 mi

21. The graph of g finally surpasses that of f at x ≈ 35.8.

23. limx→∞

(1.001)x =∞ by (3), since 1.001 > 1.


25. Divide numerator and denominator by e3x: limx→∞

e3x − e−3x

e3x + e−3x= lim

x→∞1− e−6x

1 + e−6x=1− 01 + 0

= 1

27. Let t = 3/(2− x). As x→ 2+, t→−∞. So limx→2+

e3/(2−x) = limt→−∞

et = 0 by (10).

29. Since −1 ≤ cosx ≤ 1 and e−2x > 0, we have −e−2x ≤ e−2x cosx ≤ e−2x. We know that limx→∞

(−e−2x) = 0 and

limx→∞

e−2x = 0, so by the Squeeze Theorem, limx→∞

(e−2x cosx) = 0.

31. By the Product Rule, f(x) = (x3 + 2x)ex ⇒

f 0(x) = (x3 + 2x)(ex)0 + ex(x3 + 2x)0 = (x3 + 2x)ex + ex(3x2 + 2)

= ex[(x3 + 2x) + (3x2 + 2)] = ex(x3 + 3x2 + 2x+ 2)

33. By (9), y = eax3 ⇒ y0 = eax

3 d

dx(ax3) = 3ax2eax

3.

35. f(u) = e1/u ⇒ f 0(u) = e1/u · d

du

1

u= e1/u

−1u2

=−1u2

e1/u

37. By (9), F (t) = et sin 2t ⇒ F 0(t) = et sin 2t(t sin 2t)0 = et sin 2t(t · 2 cos 2t+ sin 2t · 1) = et sin 2t(2t cos 2t+ sin 2t)

39. y =√1 + 2e3x ⇒ y0 =

1

2(1 + 2e3x)−1/2

d

dx(1 + 2e3x) =

1

2√1 + 2e3x

(2e3x · 3) = 3e3x√1 + 2e3x

41. y = eex ⇒ y0 = ee

x · d

dx(ex) = ee

x · ex or eex+x

43. By the Quotient Rule, y = aex + b

cex + d⇒

y0 =(cex + d)(aex)− (aex + b)(cex)

(cex + d)2=(acex + ad− acex − bc)ex

(cex + d)2=(ad− bc)ex

(cex + d)2.

45. y = cos 1− e2x

1 + e2x⇒

y0 = − sin 1− e2x

1 + e2x· d

dx

1− e2x

1 + e2x= − sin 1− e2x

1 + e2x· (1 + e2x)(−2e2x)− (1− e2x)(2e2x)

(1 + e2x)2

= − sin 1− e2x

1 + e2x· −2e

2x (1 + e2x) + (1− e2x)

(1 + e2x)2= − sin 1− e2x

1 + e2x· −2e

2x(2)

(1 + e2x)2=

4e2x

(1 + e2x)2· sin 1− e2x

1 + e2x

47. y = e2x cosπx ⇒ y0 = e2x(−π sinπx) + (cosπx)(2e2x) = e2x(2 cosπx− π sinπx).

At (0, 1), y0 = 1(2− 0) = 2, so an equation of the tangent line is y − 1 = 2(x− 0), or y = 2x+ 1.

49. d

dxex

2y =d

dx(x+ y) ⇒ ex

2y(x2y0 + y · 2x) = 1 + y0 ⇒ x2ex2yy0 + 2xyex

2y = 1 + y0 ⇒

x2ex2yy0 − y0 = 1− 2xyex2y ⇒ y0(x2ex

2y − 1) = 1− 2xyex2y ⇒ y0 =1− 2xyex2yx2ex2y − 1


51. y = ex + e−x/2 ⇒ y0 = ex − 12e−x/2 ⇒ y00 = ex + 1

4e−x/2, so

2y00 − y0 − y = 2 ex + 14e−x/2 − ex − 1

2e−x/2 − ex + e−x/2 = 0.

53. y = erx ⇒ y0 = rerx ⇒ y00 = r2erx, so if y = erx satisfies the differential equation y00 + 6y0 + 8y = 0,

then r2erx + 6rerx + 8erx = 0; that is, erx(r2 + 6r + 8) = 0. Since erx > 0 for all x, we must have r2 + 6r + 8 = 0,

or (r + 2)(r + 4) = 0, so r = −2 or −4.

55. f(x) = e2x ⇒ f 0(x) = 2e2x ⇒ f 00(x) = 2 · 2e2x = 22e2x ⇒f 000(x) = 22 · 2e2x = 23e2x ⇒ · · · ⇒ f (n)(x) = 2ne2x

57. (a) f(x) = ex + x is continuous on R and f(−1) = e−1 − 1 < 0 < 1 = f(0), so by the Intermediate Value Theorem,

ex + x = 0 has a root in (−1, 0).

(b) f(x) = ex + x ⇒ f 0(x) = ex + 1, so xn+1 = xn − exn + xnexn + 1

. Using x1 = −0.5, we get x2 ≈ −0.566311,

x3 ≈ −0.567143 ≈ x4, so the root is −0.567143 to six decimal places.

59. (a) limt→∞

p(t) = limt→∞

1

1 + ae−kt=

1

1 + a · 0 = 1, since k > 0 ⇒ −kt→ −∞ ⇒ e−kt → 0.

(b) p(t) = (1 + ae−kt)−1 ⇒ dp

dt= −(1 + ae−kt)−2(−kae−kt) = kae−kt

(1 + ae−kt)2

(c)From the graph of p(t) = (1 + 10e−0.5t)−1, it seems that p(t) = 0.8

(indicating that 80% of the population has heard the rumor) when

t ≈ 7.4 hours.

61. f(x) = x− ex ⇒ f 0(x) = 1− ex = 0 ⇔ ex = 1 ⇔ x = 0. Now f 0(x) > 0 for all x < 0 and f 0(x) < 0 for all

x > 0, so the absolute maximum value is f(0) = 0− 1 = −1.

63. f(x) = xe−x2/8, [−1, 4]. f 0(x) = x · e−x2/8 · (−x

4) + e−x

2/8 · 1 = e−x2/8(−x2

4+ 1). Since e−x

2/8 is never 0,

f 0(x) = 0 ⇒ −x2/4 + 1 = 0 ⇒ 1 = x2/4 ⇒ x2 = 4 ⇒ x = ±2, but −2 is not in the given interval, [−1, 4].

f(−1) = −e−1/8 ≈ −0.88, f(2) = 2e−1/2 ≈ 1.21, and f(4) = 4e−2 ≈ 0.54. So f(2) = 2e−1/2 is the absolute maximum

value and f(−1) = −e−1/8 is the absolute minimum value.

65. (a) f(x) = (1− x)e−x ⇒ f 0(x) = (1− x)(−e−x) + e−x(−1) = e−x(x− 2) > 0 ⇒ x > 2, so f is increasing on

(2,∞) and decreasing on (−∞, 2).

(b) f 00(x) = e−x(1) + (x− 2)(−e−x) = e−x(3− x) > 0 ⇔ x < 3, so f is CU on (−∞, 3) and CD on (3,∞).

(c) f 00 changes sign at x = 3, so there is an IP at 3,−2e−3 .


67. y = f(x) = e−1/(x+1) A. D = {x | x 6= −1} = (−∞,−1) ∪ (−1,∞) B. No x-intercept; y-intercept = f(0) = e−1

C. No symmetry D. limx→±∞

e−1/(x+1) = 1 since −1/(x+ 1)→ 0, so y = 1 is a HA. limx→−1+

e−1/(x+1) = 0 since

−1 /(x+ 1) → −∞, limx→−1−

e−1/(x+1) = ∞ since −1/(x+ 1) → ∞, so x = −1 is a VA.

E. f 0(x) = e−1/(x+1)/(x+ 1)2 ⇒ f 0(x) > 0 for all x except 1, so

f is increasing on (−∞,−1) and (−1,∞). F. No extreme values

G. f 00(x) = e−1/(x+1)

(x+ 1)4+

e−1/(x+1)(−2)(x+ 1)3

= −e−1/(x+1)(2x+ 1)(x+ 1)4

⇒

f 00(x) > 0 ⇔ 2x+ 1 < 0 ⇔ x < − 12

, so f is CU on (−∞,−1)and −1,− 1

2, and CD on − 1

2,∞ . f has an IP at − 1

2, e−2 .

H.

69. S(t) = Atpe−kt with A = 0.01, p = 4, and k = 0.07. We will find the

zeros of f 00 for f(t) = tpe−kt.

f 0(t) = tp(−ke−kt) + e−kt(ptp−1) = e−kt(−ktp + ptp−1)

f 00(t) = e−kt(−kptp−1 + p(p− 1)tp−2) + (−ktp + ptp−1)(−ke−kt)= tp−2e−kt[−kpt+ p(p− 1) + k2t2 − kpt]

= tp−2e−kt(k2t2 − 2kpt+ p2 − p)

Using the given values of p and k gives us f 00(t) = t2e−0.07t(0.0049t2 − 0.56t+ 12). So S00(t) = 0.01f 00(t) and its zeros

are t = 0 and the solutions of 0.0049t2 − 0.56t+ 12 = 0, which are t1 = 2007≈ 28.57 and t2 = 600

7≈ 85.71.

At t1 minutes, the rate of increase of the level of medication in the bloodstream is at its greatest and at t2 minutes, the rate of

decrease is the greatest.

71. f(x) = ex3−x → 0 as x→ −∞, and

f(x)→∞ as x→∞. From the graph,

it appears that f has a local minimum of

about f(0.58) = 0.68, and a local

maximum of about f(−0.58) = 1.47.

To find the exact values, we calculate

f 0(x) = 3x2 − 1 ex3−x, which is 0 when 3x2 − 1 = 0 ⇔ x = ± 1√

3. The negative root corresponds to the local

maximum f − 1√3= e(−1/

√3)3− (−1/√3) = e2

√3/9, and the positive root corresponds to the local minimum

f 1√3= e(1/

√3)3− (1/

√3) = e−2

√3/9. To estimate the inflection points, we calculate and graph

f 00(x) =d

dx3x2 − 1 ex

3−x = 3x2 − 1 ex3−x 3x2 − 1 + ex

3−x(6x) = ex3−x 9x4 − 6x2 + 6x+ 1 .

From the graph, it appears that f 00(x) changes sign (and thus f has inflection points) at x ≈ −0.15 and x ≈ −1.09. From the

graph of f , we see that these x-values correspond to inflection points at about (−0.15, 1.15) and (−1.09, 0.82).


73. Let u = −3x. Then du = −3 dx, so 5

0e−3x dx = − 1

3

−150

eu du = − 13 [e

u]−150 = − 13e−15 − e0 = 1

31− e−15 .

75. Let u = 1 + ex. Then du = ex dx, so ex√1 + ex dx =

√udu = 2

3u3/2 +C = 2

3 (1 + ex)3/2 + C.

77. (ex + e−x)2 dx = (e2x + 2 + e−2x) dx = 12e2x + 2x− 1

2e−2x +C

79. sinx ecos x dxu = cosx,

du = − sinx dx = eu (−du) = −eu +C = −ecos x + C

81. Let u =√x. Then du = 1

2√xdx, so e

√x

√xdx = 2 eu du = 2eu + C = 2e

√x +C.

83. Area = 1

0e3x − ex dx = 1

3e3x − ex

1

0= 1

3e3 − e − 1

3− 1 = 1

3e3 − e+ 2

3≈ 4.644

85. V =1

0π(ex)2 dx = π

1

0e2x dx = 1

2π e2x1

0= π

2e2 − 1

87. (a) erf(x) = 2√π

x

0

e−t2

dt ⇒x

0

e−t2

dt =

√π

2erf(x). By Property 5 of definite integrals in Section 5.2,

b

0e−t

2dt =

a

0e−t

2dt+

b

ae−t

2dt, so

b

a

e−t2

dt =b

0

e−t2

dt−a

0

e−t2

dt =

√π

2erf(b)−

√π

2erf(a) = 1

2

√π [erf(b)− erf(a)].

(b) y = ex2

erf(x) ⇒ y0 = 2xex2

erf(x) + ex2

erf 0(x) = 2xy + ex2 · 2√

πe−x

2

[by FTC1] = 2xy +2√π

.

89. We use Theorem 7.1.7. Note that f(0) = 3 + 0 + e0 = 4, so f−1(4) = 0. Also f 0(x) = 1 + ex. Therefore,

f−1 0(4) =

1

f 0(f−1(4))=

1

f 0(0)=

1

1 + e0=1

2.

91. From the graph, it appears that f is an odd function (f is undefined for x = 0).

To prove this, we must show that f(−x) = −f(x).

f(−x) = 1− e1/(−x)

1 + e1/(−x)=1− e(−1/x)

1 + e(−1/x)=1− 1

e1/x

1 +1

e1/x

· e1/x

e1/x=

e1/x − 1e1/x + 1

= −1− e1/x

1 + e1/x= −f(x)

so f is an odd function.

93. (a) Let f(x) = ex − 1− x. Now f(0) = e0 − 1 = 0, and for x ≥ 0, we have f 0(x) = ex − 1 ≥ 0. Now, since f(0) = 0 and

f is increasing on [0,∞), f(x) ≥ 0 for x ≥ 0 ⇒ ex − 1− x ≥ 0 ⇒ ex ≥ 1 + x.

(b) For 0 ≤ x ≤ 1, x2 ≤ x, so ex2 ≤ ex [since ex is increasing]. Hence [from (a)] 1 + x2 ≤ ex

2 ≤ ex.

So 43=

1

01 + x2 dx ≤ 1

0ex

2dx ≤ 1

0ex dx = e− 1 < e ⇒ 4

3≤ 1

0ex

2dx ≤ e.


95. (a) By Exercise 93(a), the result holds for n = 1. Suppose that ex ≥ 1 + x+x2

2!+ · · ·+ xk

k!for x ≥ 0.

Let f(x) = ex − 1− x− x2

2!− · · ·− xk

k!− xk+1

(k + 1)!. Then f 0(x) = ex − 1− x− · · ·− xk

k!≥ 0 by assumption. Hence

f(x) is increasing on (0,∞). So 0 ≤ x implies that 0 = f(0) ≤ f(x) = ex − 1− x− · · ·− xk

k!− xk+1

(k + 1)!, and hence

ex ≥ 1 + x+ · · ·+ xk

k!+

xk+1

(k + 1)!for x ≥ 0. Therefore, for x ≥ 0, ex ≥ 1 + x+

x2

2!+ · · ·+ xn

n!for every positive

integer n, by mathematical induction.

(b) Taking n = 4 and x = 1 in (a), we have e = e1 ≥ 1 + 12+ 1

6+ 1

24= 2.7083 > 2.7.

(c) ex ≥ 1 + x+ · · ·+ xk

k!+

xk+1

(k + 1)!⇒ ex

xk≥ 1

xk+

1

xk−1+ · · ·+ 1

k!+

x

(k + 1)!≥ x

(k + 1)!.

But limx→∞

x

(k + 1)!=∞, so lim

x→∞ex

xk=∞.

7.3 Logarithmic Functions

1. (a) It is defined as the inverse of the exponential function with base a, that is, loga x = y ⇔ ay = x.

(b) (0,∞) (c) R (d) See Figure 1.

3. (a) log5 125 = 3 since 53 = 125. (b) log31

27= −3 since 3−3 = 1

33=1

27.

5. (a) log5 125= log5 5

−2 = −2 by (2). (b) eln 15 = 15 by (8).

7. (a) log2 6− log2 15 + log2 20 = log2( 615 ) + log2 20 [by Law 2]

= log2(615· 20) [by Law 1]

= log2 8, and log2 8 = 3 since 23 = 8.

(b) log3 100− log3 18− log3 50 = log310018

− log3 50 = log3 10018·50

= log3(19), and log3

19= −2 since 3−2 = 1

9.

9. log2x3y

z2= log2(x

3y)− log2 z2 = log2 x3 + log2 y − log2 z2 = 3 log2 x+ log2 y − 2 log2 z

[assuming that the variables are positive]

11. ln(uv)10 = 10 ln(uv) = 10(lnu+ ln v) = 10 lnu+ 10 ln v

13. log10 a− log10 b+ log10 c = log10a

b+ log10 c = log10

a

b· c = log10

ac

b

15. ln 5 + 5 ln 3 = ln 5 + ln 35 [by Law 3]

= ln(5 · 35) [by Law 1]

= ln1215

SECTION 7.3 LOGARITHMIC FUNCTIONS ¤ 255

17. ln(1 + x2) + 12lnx− ln sinx = ln(1 + x2) + lnx1/2 − ln sinx = ln[(1 + x2)

√x ]− ln sinx = ln (1 + x2)

√x

sinx

19. (a) log12 e =ln e

ln 12=

1

ln 12≈ 0.402430

(b) log6 13.54 =ln 13.54

ln 6≈ 1.454240

(c) log2 π =lnπ

ln 2≈ 1.651496

21. To graph these functions, we use log1.5 x =lnx

ln 1.5and log50 x =

lnx

ln 50.

These graphs all approach−∞ as x→ 0+, and they all pass through the

point (1, 0). Also, they are all increasing, and all approach∞ as x→∞.

The functions with larger bases increase extremely slowly, and the ones with

smaller bases do so somewhat more quickly. The functions with large bases

approach the y-axis more closely as x→ 0+.

23. (a) Shift the graph of y = log10 x five units to the left to

obtain the graph of y = log10(x+5). Note the vertical

asymptote of x = −5.

y = log10 x y = log10(x+ 5)

(b) Reflect the graph of y = lnx about the x-axis to obtain

the graph of y = − lnx.

y = lnx y = − lnx

25. (a) 2 lnx = 1 ⇒ lnx = 12⇒ x = e1/2 =

√e

(b) e−x = 5 ⇒ −x = ln 5 ⇒ x = − ln 5

27. (a) 2x−5 = 3 ⇔ log2 3 = x− 5 ⇔ x = 5 + log2 3.

Or: 2x−5 = 3 ⇔ ln 2x−5 = ln3 ⇔ (x− 5) ln 2 = ln 3 ⇔ x− 5 = ln 3

ln 2⇔ x = 5 +

ln 3

ln 2

(b) lnx+ ln(x− 1) = ln(x(x− 1)) = 1 ⇔ x(x− 1) = e1 ⇔ x2 − x− e = 0. The quadratic formula (with a = 1,

b = −1, and c = −e) gives x = 121±√1 + 4e , but we reject the negative root since the natural logarithm is not

defined for x < 0. So x = 121 +

√1 + 4e .

29. 3xex + x2ex = 0 ⇔ xex(3 + x) = 0 ⇔ x = 0 or −3

31. ln(lnx) = 1 ⇔ eln(ln x) = e1 ⇔ lnx = e1 = e ⇔ eln x = ee ⇔ x = ee

33. e2x − ex − 6 = 0 ⇔ (ex − 3)(ex + 2) = 0 ⇔ ex = 3 or −2 ⇒ x = ln 3 since ex > 0.


35. (a) e2+5x = 100 ⇒ ln e2+5x = ln 100 ⇒ 2 + 5x = ln 100 ⇒ 5x = ln 100− 2 ⇒x = 1

5(ln 100− 2) ≈ 0.5210

(b) ln(ex − 2) = 3 ⇒ ex − 2 = e3 ⇒ ex = e3 + 2 ⇒ x = ln(e3 + 2) ≈ 3.0949

37. (a) ex < 10 ⇒ ln ex < ln 10 ⇒ x < ln 10 ⇒ x ∈ (−∞, ln 10)

(b) lnx > −1 ⇒ elnx > e−1 ⇒ x > e−1 ⇒ x ∈ (1/e,∞)

39. 3 ft = 36 in, so we need x such that log2 x = 36 ⇔ x = 236 = 68,719,476,736. In miles, this is

68,719,476,736 in · 1 ft12 in

· 1 mi5280 ft

≈ 1,084,587.7 mi.

41. If I is the intensity of the 1989 San Francisco earthquake, then log10(I/S) = 7.1 ⇒log10(16I/S) = log10 16 + log10(I/S) = log10 16 + 7.1 ≈ 8.3.

43. (a) n = 100 · 2t/3 ⇒ n

100= 2t/3 ⇒ log2

n

100=

t

3⇒ t = 3 log2

n

100. Using formula (7), we can write this

as t = f−1(n) = 3 · ln(n/100)ln 2

. This function tells us how long it will take to obtain n bacteria (given the number n).

(b) n = 50,000 ⇒ t = f−1(50,000) = 3 · ln50,000100

ln 2= 3

ln 500

ln 2≈ 26.9 hours

45. Let t = x2 − 9. Then as x→ 3+, t→ 0+, and limx→3+

ln(x2 − 9) = limt→0+

ln t = −∞ by (8).

47. limx→0

ln(cosx) = ln 1 = 0. [ln(cosx) is continuous at x = 0 since it is the composite of two continuous functions.]

49. limx→∞

[ln(1 + x2)− ln(1 + x)] = limx→∞

ln1 + x2

1 + x= ln lim

x→∞1 + x2

1 + x= ln lim

x→∞

1x+ x

1x+ 1

=∞, since the limit in

parentheses is∞.

51. f(x) = log10(x2 − 9). Df = x | x2 − 9 > 0 = x |x| > 3 = (−∞,−3) ∪ (3,∞)

53. (a) For f(x) =√3− e2x, we must have 3− e2x ≥ 0 ⇒ e2x ≤ 3 ⇒ 2x ≤ ln 3 ⇒ x ≤ 1

2ln 3.

Thus, the domain of f is (−∞, 12ln 3].

(b) y = f(x) =√3− e2x [note that y ≥ 0] ⇒ y2 = 3− e2x ⇒ e2x = 3− y2 ⇒ 2x = ln(3− y2) ⇒

x = 12ln(3− y2). Interchange x and y: y = 1

2ln(3− x2). So f−1(x) = 1

2ln(3− x2). For the domain of f−1, we must

have 3− x2 > 0 ⇒ x2 < 3 ⇒ |x| < √3 ⇒ −√3 < x <√3 ⇒ 0 ≤ x <

√3 since x ≥ 0. Note that the

domain of f−1, [0,√3 ), equals the range of f .

55. y = ln(x+ 3) ⇒ ey = eln(x+3) = x+ 3 ⇒ x = ey − 3.

Interchange x and y: the inverse function is y = ex − 3.

57. y = f(x) = ex3 ⇒ ln y = x3 ⇒ x = 3

√ln y. Interchange x and y: y = 3

√lnx. So f−1(x) = 3

√lnx.

SECTION 7.4 DERIVATIVES OF LOGARITHMIC FUNCTIONS ¤ 257

59. y = log10 1 +1

x⇒ 10y = 1 +

1

x⇒ 1

x= 10y − 1 ⇒ x =

1

10y − 1 .

Interchange x and y: y =1

10x − 1 is the inverse function.

61. f(x) = e3x − ex ⇒ f 0(x) = 3e3x − ex. Thus, f 0(x) > 0 ⇔ 3e3x > ex ⇔ 3e3x

ex>

ex

ex⇔ 3e2x > 1 ⇔

e2x > 13⇔ 2x > ln 1

3= − ln 3 ⇔ x > − 1

2ln 3, so f is increasing on − 1

2ln 3,∞ .

63. y = f(x) = (x2 − 2)e−x ⇒ y0 = (x2 − 2)(−e−x) + e−x(2x) = e−x(−x2 + 2x+ 2) ⇒y00 = e−x(−2x+ 2) + (−x2 + 2x+ 2)(−e−x) = e−x(x2 − 4x) = xe−x(x− 4).y00 > 0 ⇒ x < 0 or x > 4 ⇒ f is CU on (−∞, 0) and (4,∞).

65. (a) We have to show that −f(x) = f(−x).

−f(x) = − ln x+√x2 + 1 = ln x+

√x2 + 1

−1= ln

1

x+√x2 + 1

= ln1

x+√x2 + 1

· x−√x2 + 1

x−√x2 + 1 = lnx−√x2 + 1x2 − x2 − 1 = ln

√x2 + 1− x = f(−x)

Thus, f is an odd function.

(b) Let y = ln x+√x2 + 1 . Then ey = x+

√x2 + 1 ⇔ (ey − x)2 = x2 + 1 ⇔ e2y − 2xey + x2 = x2 + 1 ⇔

2xey = e2y − 1 ⇔ x =e2y − 12ey

= 12(ey − e−y). Thus, the inverse function is f−1(x) = 1

2(ex − e−x).

67. x1/ lnx = 2 ⇒ ln(x1/ lnx) = ln(2) ⇒ 1

lnx· lnx = ln 2 ⇒ 1 = ln 2, a contradiction, so the given equation has no

solution. The function f(x) = x1/ lnx = (eln x)1/ ln x = e1 = e for all x > 0, so the function f(x) = x1/ lnx is the constant

function f(x) = e.

69. (a) Let ε > 0 be given. We need N such that |ax − 0| < ε when x < N . But ax < ε ⇔ x < loga ε. Let N = loga ε.

Then x < N ⇒ x < loga ε ⇒ |ax − 0| = ax < ε, so limx→−∞

ax = 0.

(b) Let M > 0 be given. We need N such that ax > M when x > N . But ax > M ⇔ x > logaM . Let N = logaM .

Then x > N ⇒ x > logaM ⇒ ax > M , so limx→∞

ax =∞.

71. ln(x2 − 2x− 2) ≤ 0 ⇒ 0 < x2 − 2x− 2 ≤ 1. Now x2 − 2x− 2 ≤ 1 gives x2 − 2x− 3 ≤ 0 and hence

(x− 3)(x+ 1) ≤ 0. So −1 ≤ x ≤ 3. Now 0 < x2 − 2x− 2 ⇒ x < 1−√3 or x > 1 +√3. Therefore,

ln(x2 − 2x− 2) ≤ 0 ⇔ −1 ≤ x < 1−√3 or 1 +√3 < x ≤ 3.

7.4 Derivatives of Logarithmic Functions

1. The differentiation formula for logarithmic functions, d

dx(loga x) =

1

x ln a, is simplest when a = e because ln e = 1.

3. f(x) = sin(lnx) ⇒ f 0(x) = cos(lnx) · d

dxlnx = cos(lnx) · 1

x=cos(lnx)

x


5. f(x) = log2(1− 3x) ⇒ f 0(x) =1

(1− 3x) ln 2d

dx(1− 3x) = −3

(1− 3x) ln 2 or 3

(3x− 1) ln 2

7. f(x) = 5√lnx = (lnx)1/5 ⇒ f 0(x) = 1

5(lnx)−4/5

d

dx(lnx) =

1

5(lnx)4/5· 1x=

1

5x 5 (lnx)4

9. f(x) = sinx ln(5x) ⇒ f 0(x) = sinx · 15x· ddx(5x)+ ln(5x) · cosx = sinx · 5

5x+cosx ln(5x) =

sinx

x+cosx ln(5x)

11. F (t) = ln (2t+ 1)3

(3t− 1)4 = ln(2t+ 1)3 − ln(3t − 1)4 = 3 ln(2t+ 1)− 4 ln(3t− 1) ⇒

F 0(t) = 3 · 1

2t+ 1· 2− 4 · 1

3t− 1 · 3 =6

2t+ 1− 12

3t− 1 , or combined, −6(t+ 3)(2t+ 1)(3t− 1) .

13. g(x) = ln x√x2 − 1 = lnx + ln(x2 − 1)1/2 = lnx + 1

2ln(x2 − 1) ⇒

g0(x) =1

x+1

2· 1

x2 − 1 · 2x =1

x+

x

x2 − 1 =x2 − 1 + x · xx(x2 − 1) =

2x2 − 1x(x2 − 1)

15. f(u) = lnu

1 + ln(2u)⇒

f 0(u) =[1 + ln(2u)] · 1

u − lnu · 12u · 2

[1 + ln(2u)]2=

1u [1 + ln(2u)− lnu][1 + ln(2u)]2

=1 + (ln 2 + lnu)− lnu

u[1 + ln(2u)]2=

1 + ln 2

u[1 + ln(2u)]2

17. h(t) = t3 − 3t ⇒ h0(t) = 3t2 − 3t ln 3

19. y = ln 2− x− 5x2 ⇒ y0 =1

2− x− 5x2 · (−1− 10x) =−10x− 12− x− 5x2 or 10x+ 1

5x2 + x− 2

21. y = ln(e−x + xe−x) = ln(e−x(1 + x)) = ln(e−x) + ln(1 + x) = −x+ ln(1 + x) ⇒

y0 = −1 + 1

1 + x=−1− x+ 1

1 + x= − x

1 + x

23. y = 2x log10√x = 2x log10 x

1/2 = 2x · 12 log10 x = x log10 x ⇒ y0 = x · 1

x ln 10+ log10 x · 1 =

1

ln 10+ log10 x

Note: 1

ln 10=ln e

ln 10= log10 e, so the answer could be written as 1

ln 10+ log10 x = log10 e+ log10 x = log10 ex.

25. Using Formula 7 and the Chain Rule, y = 5−1/x ⇒ y0 = 5−1/x(ln 5) −1 · −x−2 = 5−1/x(ln 5)/x2

27. y = x2 ln(2x) ⇒ y0 = x2 · 12x· 2 + ln(2x) · (2x) = x + 2x ln(2x) ⇒

y00 = 1 + 2x · 12x· 2 + ln(2x) · 2 = 1 + 2 + 2 ln(2x) = 3 + 2 ln(2x)

29. y = ln x+√1 + x2 ⇒

y0 =1

x+√1 + x2

d

dxx+

√1 + x2 =

1

x+√1 + x2

1 + 12(1 + x2)−1/2(2x)

=1

x+√1 + x2

1 +x√1 + x2

=1

x+√1 + x2

·√1 + x2 + x√1 + x2

=1√1 + x2

⇒

y00 = − 12 (1 + x2)−3/2(2x) =

−x(1 + x2)3/2


31. f(x) = x

1− ln(x− 1) ⇒

f 0(x) =[1− ln(x− 1)] · 1− x · −1

x− 1[1− ln(x− 1)]2 =

(x− 1)[1− ln(x− 1)] + x

x− 1[1− ln(x− 1)]2 =

x− 1− (x− 1) ln(x− 1) + x

(x− 1)[1− ln(x− 1)]2

=2x− 1− (x− 1) ln(x− 1)(x− 1)[1− ln(x− 1)]2

Dom(f) = {x | x− 1 > 0 and 1− ln(x− 1) 6= 0} = {x | x > 1 and ln(x− 1) 6= 1}= x | x > 1 and x− 1 6= e1 = {x | x > 1 and x 6= 1 + e} = (1, 1 + e) ∪ (1 + e,∞)

33. f(x) = ln(x2 − 2x) ⇒ f 0(x) =1

x2 − 2x (2x− 2) =2(x− 1)x(x− 2) .

Dom(f) = {x | x(x− 2) > 0} = (−∞, 0) ∪ (2,∞).

35. f(x) = lnx

1 + x2⇒ f 0(x) =

(1 + x2)1

x− (lnx)(2x)

(1 + x2)2, so f 0(1) = 2(1)− 0(2)

22=2

4=1

2.

37. y = ln xex2

= lnx+ ln ex2

= lnx+ x2 ⇒ y0 =1

x+ 2x. At (1, 1), the slope of the tangent line is

y0(1) = 1 + 2 = 3, and an equation of the tangent line is y − 1 = 3(x− 1), or y = 3x− 2.

39. f(x) = sinx+ lnx ⇒ f 0(x) = cosx+ 1/x.

This is reasonable, because the graph shows that f increases when f 0 is

positive, and f 0(x) = 0 when f has a horizontal tangent.

41. y = (2x+ 1)5(x4 − 3)6 ⇒ ln y = ln (2x+ 1)5(x4 − 3)6 ⇒ ln y = 5 ln(2x+ 1) + 6 ln(x4 − 3) ⇒1

yy0 = 5 · 1

2x+ 1· 2 + 6 · 1

x4 − 3 · 4x3 ⇒

y0 = y10

2x+ 1+

24x3

x4 − 3 = (2x+ 1)5(x4 − 3)6 10

2x+ 1+

24x3

x4 − 3 .

[The answer could be simplified to y0 = 2(2x+ 1)4(x4 − 3)5(29x4 + 12x3 − 15), but this is unnecessary.]

43. y = sin2 x tan4 x

(x2 + 1)2⇒ ln y = ln(sin2 x tan4 x) − ln(x2 + 1)2 ⇒

ln y = ln(sinx)2 + ln(tanx)4 − ln(x2 + 1)2 ⇒ ln y = 2 ln |sinx|+ 4 ln |tanx|− 2 ln(x2 + 1) ⇒1

yy0 = 2 · 1

sinx· cosx+ 4 · 1

tanx· sec2 x− 2 · 1

x2 + 1· 2x ⇒ y0 =

sin2 x tan4 x

(x2 + 1)22 cotx+

4 sec2 x

tanx− 4x

x2 + 1

45. y = xx ⇒ ln y = lnxx ⇒ ln y = x lnx ⇒ y0/y = x(1/x) + (lnx) · 1 ⇒ y0 = y(1 + lnx) ⇒y0 = xx(1 + lnx)


47. y = x sin x ⇒ ln y = lnx sinx ⇒ ln y = sinx lnx ⇒ y0

y= (sinx) · 1

x+ (lnx)(cosx) ⇒

y0 = ysinx

x+ lnx cosx ⇒ y0 = x sin x

sinx

x+ lnx cosx

49. y = (cosx)x ⇒ ln y = ln(cosx)x ⇒ ln y = x ln cosx ⇒ 1

yy0 = x · 1

cosx· (− sinx) + ln cosx · 1 ⇒

y0 = y ln cosx− x sinx

cosx⇒ y0 = (cosx)x(ln cosx− x tanx)

51. y = (tanx)1/x ⇒ ln y = ln(tanx)1/x ⇒ ln y =1

xln tanx ⇒

1

yy0 =

1

x· 1

tanx· sec2 x+ ln tanx · − 1

x2⇒ y0 = y

sec2 x

x tanx− ln tanx

x2⇒

y0 = (tanx)1/xsec2 x

x tanx− ln tanx

x2or y0 = (tanx)1/x · 1

xcscx secx− ln tanx

x

53. y = ln(x2 + y2) ⇒ y0 =1

x2 + y2d

dx(x2 + y2) ⇒ y0 =

2x+ 2yy0

x2 + y2⇒ x2y0 + y2y0 = 2x+ 2yy0 ⇒

x2y0 + y2y0 − 2yy0 = 2x ⇒ (x2 + y2 − 2y)y0 = 2x ⇒ y0 =2x

x2 + y2 − 2y

55. f(x) = ln(x− 1) ⇒ f 0(x) =1

(x− 1) = (x− 1)−1 ⇒ f 00(x) = −(x− 1)−2 ⇒ f 000 (x) = 2(x− 1)−3 ⇒

f (4)(x) = −2 · 3(x− 1)−4 ⇒ · · · ⇒ f (n)(x) = (−1)n−1 · 2 · 3 · 4 · · · · · (n− 1)(x− 1)−n = (−1)n−1 (n− 1)!(x− 1)n

57. From the graph, it appears that the curves y = (x− 4)2 and y = lnx intersect

just to the left of x = 3 and to the right of x = 5, at about x = 5.3. Let

f(x) = lnx− (x− 4)2. Then f 0(x) = 1/x− 2(x− 4), so Newton’s Method

says that xn+1 = xn − f(xn)/f0(xn) = xn − lnxn − (xn − 4)2

1/xn − 2(xn − 4) . Taking

x0 = 3, we get x1 ≈ 2.957738, x2 ≈ 2.958516 ≈ x3, so the first root is 2.958516, to six decimal places. Taking x0 = 5, we

get x1 ≈ 5.290755, x2 ≈ 5.290718 ≈ x3, so the second (and final) root is 5.290718, to six decimal places.

59. f(x) = lnx√x

⇒ f 0(x) =

√x (1/x)− (lnx)[1/(2√x )]

x=2− lnx2x3/2

⇒

f 00(x) =2x3/2(−1/x)− (2− lnx)(3x1/2)

4x3=3 lnx− 84x5/2

> 0 ⇔ lnx > 83⇔ x > e8/3, so f is CU on (e8/3,∞)

and CD on (0, e8/3). The inflection point is e8/3, 83e−4/3 .

61. y = f(x) = ln(sinx)

A. D = {x in R | sinx > 0} =∞

n=−∞(2nπ, (2n+ 1)π) = · · · ∪ (−4π,−3π) ∪ (−2π,−π) ∪ (0, π) ∪ (2π, 3π) ∪ · · ·

B. No y-intercept; x-intercepts: f(x) = 0 ⇔ ln(sinx) = 0 ⇔ sinx = e0 = 1 ⇔ x = 2nπ + π2

for each


integer n. C. f is periodic with period 2π. D. limx→(2nπ)+

f(x) = −∞ and limx→[(2n+1)π]−

f(x) = −∞, so the lines

x = nπ are VAs for all integers n. E. f 0(x) = cosx

sinx= cotx, so f 0(x) > 0 when 2nπ < x < 2nπ + π

2for each

integer n, and f 0(x) < 0 when 2nπ + π2< x < (2n+ 1)π. Thus, f is increasing on 2nπ, 2nπ + π

2and

decreasing on 2nπ + π2 , (2n+ 1)π for each integer n.

F. Local maximum values f 2nπ + π2= 0, no local minimum.

G. f 00(x) = − csc2 x < 0, so f is CD on (2nπ, (2n+ 1)π) for

each integer n. No IP

H.

63. y = f(x) = ln(1 + x2) A. D = R B. Both intercepts are 0. C. f(−x) = f(x), so the curve is symmetric about the

y-axis. D. limx→±∞

ln(1 + x2) =∞, no asymptotes. E. f 0(x) = 2x

1 + x2> 0 ⇔

x > 0, so f is increasing on (0,∞) and decreasing on (−∞, 0) .

F. f(0) = 0 is a local and absolute minimum.

G. f 00(x) = 2(1 + x2)− 2x(2x)(1 + x2)2

=2(1− x2)

(1 + x2)2> 0 ⇔

|x| < 1, so f is CU on (−1, 1), CD on (−∞,−1) and (1,∞). IP

(1, ln 2) and (−1, ln 2).

H.

65. We use the CAS to calculate f 0(x) = 2 + sinx+ x cosx

2x+ x sinxand

f 00(x) =2x2 sinx+ 4 sinx− cos2 x+ x2 + 5

x2(cos2 x− 4 sinx− 5) . From the graphs, it

seems that f 0 > 0 (and so f is increasing) on approximately the intervals

(0, 2.7), (4.5, 8.2) and (10.9, 14.3). It seems that f 00 changes sign

(indicating inflection points) at x ≈ 3.8, 5.7, 10.0 and 12.0.

Looking back at the graph of f(x) = ln(2x+ x sinx), this implies that the inflection points have approximate coordinates

(3.8, 1.7), (5.7, 2.1), (10.0, 2.7), and (12.0, 2.9).

67. (a) Using a calculator or CAS, we obtain the model Q = abt with a ≈ 100.0124369 and b ≈ 0.000045145933.

(b) Use Q0(t) = abt ln b (from Formula 7) with the values of a and b from part (a) to get Q0(0.04) ≈ −670.63 μA.

The result of Example 2 in Section 2.1 was −670 μA.

69.4

2

3

xdx = 3

4

2

1

xdx = 3 ln |x| 4

2= 3(ln 4− ln 2) = 3 ln 4

2= 3 ln 2

71.2

1

dt

8− 3t = −13ln |8− 3t|

2

1

= −13ln 2− −1

3ln 5 =

1

3(ln 5− ln 2) = 1

3ln5

2

Or: Let u = 8 − 3t. Then du = −3 dt, so2

1

dt

8− 3t =2

5

− 13du

u= −1

3ln |u|

2

5

= −13ln 2− −1

3ln 5 =

1

3(ln 5− ln 2) = 1

3ln5

2.


73.e

1

x2 + x+ 1

xdx =

e

1

x+ 1 +1

xdx = 1

2x2 + x+ lnx

e

1= 1

2e2 + e+ 1 − 1

2+ 1 + 0

= 12e2 + e− 1

2

75. Let u = lnx. Then du = dx

x⇒ (lnx)2

xdx = u2 du =

1

3u3 + C =

1

3(lnx)3 +C.

77. sin 2x

1 + cos2 xdx = 2

sinx cosx

1 + cos2 xdx = 2I. Let u = cosx. Then du = − sinxdx, so

2I = −2 udu

1 + u2= −2 · 12 ln(1 + u2) + C = − ln(1 + u2) +C = − ln(1 + cos2 x) + C.

Or: Let u = 1 + cos2 x.

79.2

1

10t dt =10t

ln 10

2

1

=102

ln 10− 101

ln 10=100− 10ln 10

=90

ln 10

81. (a) d

dx(ln |sinx|+ C) =

1

sinxcosx = cotx

(b) Let u = sinx. Then du = cosxdx, so cotxdx =cosx

sinxdx =

du

u= ln |u|+C = ln |sinx|+C.

83. The cross-sectional area is π 1/√x+ 1

2= π/(x + 1). Therefore, the volume is

1

0

π

x+ 1dx = π[ln(x+ 1)]10 = π(ln 2− ln 1) = π ln 2.

85. W =V2V1

P dV =1000

600

C

VdV = C

1000

600

1

VdV = C ln |V | 1000

600= C(ln 1000− ln 600) = C ln 1000

600= C ln 5

3.

Initially, PV = C, where P = 150 kPa and V = 600 cm3, so C = (150)(600) = 90,000. Thus,

W = 90, 000 ln 53≈ 45, 974 kPa · cm3 = 45, 974(103 Pa)(10−6 m3) = 45, 974 Pa·m3 = 45, 974 N·m [Pa = N/m2]

= 45, 974 J

87. f(x) = 2x+ lnx ⇒ f 0(x) = 2 + 1/x. If g = f−1, then f(1) = 2 ⇒ g(2) = 1, so

g0(2) = 1/f 0(g(2)) = 1/f 0(1) = 13 .

89. The curve and the line will determine a region when they intersect at two or

more points. So we solve the equation x/(x2 + 1) = mx ⇒ x = 0 or

mx2 +m− 1 = 0 ⇒ x = 0 or x =± −4(m)(m− 1)

2m= ± 1

m− 1.

Note that if m = 1, this has only the solution x = 0, and no region is

determined. But if 1/m− 1 > 0 ⇔ 1/m > 1 ⇔ 0 < m < 1, then there are two solutions. [Another way of seeing this

SECTION 7.2* THE NATURAL LOGARITHMIC FUNCTION ¤ 263

is to observe that the slope of the tangent to y = x/(x2 + 1) at the origin is y0 = 1 and therefore we must have 0 < m < 1.]

Note that we cannot just integrate between the positive and negative roots, since the curve and the line cross at the origin.

Since mx and x/(x2 + 1) are both odd functions, the total area is twice the area between the curves on the interval

0, 1/m− 1 . So the total area enclosed is

2

√1/m−1

0

x

x2 + 1−mx dx= 2 1

2ln(x2 + 1)− 1

2mx2

√1/m−1

0

= ln1

m− 1 + 1 −m

1

m− 1 − (ln 1− 0)

= ln1

m+m− 1 = m− lnm− 1

91. If f(x) = ln (1 + x), then f 0(x) =1

1 + x, so f 0(0) = 1.

Thus, limx→0

ln(1 + x)

x= lim

x→0

f(x)

x= lim

x→0

f(x)− f(0)

x− 0 = f 0(0) = 1.

7.2* The Natural Logarithmic Function

1. ln r2

3√s= ln r2 − ln 3√s = 2 ln r − (ln 3 + ln s1/2) = 2 ln r − ln 3− 1

2 ln s

3. ln(uv)10 = 10 ln(uv) = 10(lnu+ ln v) = 10 lnu+ 10 ln v

5. ln 5 + 5 ln 3 = ln 5 + ln 35 [by Law 3]

= ln(5 · 35) [by Law 1]

= ln1215

7. ln(1 + x2) + 12lnx− ln sinx = ln(1 + x2) + lnx1/2 − ln sinx = ln[(1 + x2)

√x ]− ln sinx = ln (1 + x2)

√x

sinx

9. Reflect the graph of y = lnx about the x-axis to obtain

the graph of y = − lnx.

y = lnx y = − lnx

11.

y = lnx y = ln(x+ 3)

13. Let t = x2 − 9. Then as x→ 3+, t→ 0+, and limx→3+

ln(x2 − 9) = limt→0+

ln t = −∞ by (4).


15. f(x) =√x lnx ⇒ f 0(x) =

1

2√xlnx+

√x

1

x=lnx+ 2

2√x

17. f(x) = sin(lnx) ⇒ f 0(x) = cos(lnx) · d

dxlnx = cos(lnx) · 1

x=cos(lnx)

x

19. f(x) = 5√lnx = (lnx)1/5 ⇒ f 0(x) = 1

5 (lnx)−4/5 d

dx(lnx) =

1

5(lnx)4/5· 1x=

1

5x 5 (lnx)4

21. f(x) = sinx ln(5x) ⇒ f 0(x) = sinx · 15x· ddx(5x)+ ln(5x) · cosx = sinx · 5

5x+cosx ln(5x) =

sinx

x+cosx ln(5x)

23. g(x) = ln a− x

a+ x= ln(a− x)− ln(a+ x) ⇒

g0(x) =1

a− x(−1)− 1

a+ x=−(a+ x)− (a− x)

(a− x)(a+ x)=

−2aa2 − x2

25. F (t) = ln (2t+ 1)3

(3t− 1)4 = ln(2t+ 1)3 − ln(3t − 1)4 = 3 ln(2t+ 1)− 4 ln(3t− 1) ⇒

F 0(t) = 3 · 1

2t+ 1· 2− 4 · 1

3t− 1 · 3 =6

2t+ 1− 12

3t− 1 , or combined, −6(t+ 3)(2t+ 1)(3t− 1) .

27. g(x) = ln x√x2 − 1 = lnx + ln(x2 − 1)1/2 = lnx + 1

2 ln(x2 − 1) ⇒

g0(x) =1

x+1

2· 1

x2 − 1 · 2x =1

x+

x

x2 − 1 =x2 − 1 + x · xx(x2 − 1) =

2x2 − 1x(x2 − 1)

29. f(u) = lnu

1 + ln(2u)⇒

f 0(u) =[1 + ln(2u)] · 1

u− lnu · 1

2u· 2

[1 + ln(2u)]2=

1u[1 + ln(2u)− lnu][1 + ln(2u)]2

=1 + (ln 2 + lnu)− lnu

u[1 + ln(2u)]2=

1 + ln 2

u[1 + ln(2u)]2

31. y = ln 2− x− 5x2 ⇒ y0 =1

2− x− 5x2 · (−1− 10x) =−10x− 12− x− 5x2 or 10x+ 1

5x2 + x− 2

33. y = tan [ln(ax+ b)] ⇒ y0 = sec2[ln(ax+ b)] · 1

ax+ b· a = sec2[ln(ax+ b)]

a

ax+ b

35. y = x2 ln(2x) ⇒ y0 = x2 · 12x· 2 + ln(2x) · (2x) = x + 2x ln(2x) ⇒

y00 = 1 + 2x · 12x· 2 + ln(2x) · 2 = 1 + 2 + 2 ln(2x) = 3 + 2 ln(2x)

37. f(x) = x

1− ln(x− 1) ⇒

f 0(x) =[1− ln(x− 1)] · 1− x · −1

x− 1[1− ln(x− 1)]2 =

(x− 1)[1− ln(x− 1)] + x

x− 1[1− ln(x− 1)]2 =

x− 1− (x− 1) ln(x− 1) + x

(x− 1)[1− ln(x− 1)]2

=2x− 1− (x− 1) ln(x− 1)(x− 1)[1− ln(x− 1)]2

Dom(f) = {x | x− 1 > 0 and 1− ln(x− 1) 6= 0} = {x | x > 1 and ln(x− 1) 6= 1}= x | x > 1 and x− 1 6= e1 = {x | x > 1 and x 6= 1 + e} = (1, 1 + e) ∪ (1 + e,∞)


39. f(x) =√1− lnx is defined ⇔ x > 0 [so that lnx is defined] and 1 − lnx ≥ 0 ⇔

x > 0 and lnx ≤ 1 ⇔ 0 < x ≤ e, so the domain of f is (0, e]. Now

f 0(x) =1

2(1− lnx)−1/2 · d

dx(1− lnx) = 1

2√1− lnx · − 1

x=

−12x√1− lnx .

41. f(x) = lnx

1 + x2⇒ f 0(x) =

(1 + x2)1

x− (lnx)(2x)

(1 + x2)2, so f 0(1) = 2(1)− 0(2)

22=2

4=1

2.

43. f(x) = sinx+ lnx ⇒ f 0(x) = cosx+ 1/x.

This is reasonable, because the graph shows that f increases when f 0 is

positive, and f 0(x) = 0 when f has a horizontal tangent.

45. y = sin(2 lnx) ⇒ y0 = cos(2 lnx) · 2x

. At (1, 0), y0 = cos 0 · 21= 2, so an equation of the tangent line is

y − 0 = 2 · (x− 1), or y = 2x− 2.

47. y = ln(x2 + y2) ⇒ y0 =1

x2 + y2d

dx(x2 + y2) ⇒ y0 =

2x+ 2yy0

x2 + y2⇒ x2y0 + y2y0 = 2x+ 2yy0 ⇒

x2y0 + y2y0 − 2yy0 = 2x ⇒ (x2 + y2 − 2y)y0 = 2x ⇒ y0 =2x

x2 + y2 − 2y

49. f(x) = ln(x− 1) ⇒ f 0(x) =1

(x− 1) = (x− 1)−1 ⇒ f 00(x) = −(x− 1)−2 ⇒ f 000 (x) = 2(x− 1)−3 ⇒

f (4)(x) = −2 · 3(x− 1)−4 ⇒ · · · ⇒ f (n)(x) = (−1)n−1 · 2 · 3 · 4 · · · · · (n− 1)(x− 1)−n = (−1)n−1 (n− 1)!(x− 1)n

51. From the graph, it appears that the curves y = (x− 4)2 and y = lnx intersect

just to the left of x = 3 and to the right of x = 5, at about x = 5.3. Let

f(x) = lnx− (x− 4)2. Then f 0(x) = 1/x− 2(x− 4), so Newton’s Method

says that xn+1 = xn − f(xn)/f0(xn) = xn − lnxn − (xn − 4)2

1/xn − 2(xn − 4) . Taking

x0 = 3, we get x1 ≈ 2.957738, x2 ≈ 2.958516 ≈ x3, so the first root is 2.958516, to six decimal places. Taking x0 = 5, we

get x1 ≈ 5.290755, x2 ≈ 5.290718 ≈ x3, so the second (and final) root is 5.290718, to six decimal places.

53. y = f(x) = ln(sinx)

A. D = {x in R | sinx > 0} =∞

n=−∞(2nπ, (2n+ 1)π) = · · · ∪ (−4π,−3π) ∪ (−2π,−π) ∪ (0, π) ∪ (2π, 3π) ∪ · · ·

B. No y-intercept; x-intercepts: f(x) = 0 ⇔ ln(sinx) = 0 ⇔ sinx = e0 = 1 ⇔ x = 2nπ + π2

for each


integer n. C. f is periodic with period 2π. D. limx→(2nπ)+

f(x) = −∞ and limx→[(2n+1)π]−

f(x) = −∞, so the lines

x = nπ are VAs for all integers n. E. f 0(x) = cosx

sinx= cotx, so f 0(x) > 0 when 2nπ < x < 2nπ + π

2for each

integer n, and f 0(x) < 0 when 2nπ + π2 < x < (2n+ 1)π. Thus, f is increasing on 2nπ, 2nπ + π

2and

decreasing on 2nπ + π2, (2n+ 1)π for each integer n.

F. Local maximum values f 2nπ + π2= 0, no local minimum.

G. f 00(x) = − csc2 x < 0, so f is CD on (2nπ, (2n+ 1)π) for

each integer n. No IP

H.

55. y = f(x) = ln(1 + x2) A. D = R B. Both intercepts are 0. C. f(−x) = f(x), so the curve is symmetric about the

y-axis. D. limx→±∞

ln(1 + x2) =∞, no asymptotes. E. f 0(x) = 2x

1 + x2> 0 ⇔

x > 0, so f is increasing on (0,∞) and decreasing on (−∞, 0) .

F. f(0) = 0 is a local and absolute minimum.

G. f 00(x) = 2(1 + x2)− 2x(2x)(1 + x2)2

=2(1− x2)

(1 + x2)2> 0 ⇔

|x| < 1, so f is CU on (−1, 1), CD on (−∞,−1) and (1,∞). IP

(1, ln 2) and (−1, ln 2).

H.

57. We use the CAS to calculate f 0(x) = 2 + sinx+ x cosx

2x+ x sinxand

f 00(x) =2x2 sinx+ 4 sinx− cos2 x+ x2 + 5

x2(cos2 x− 4 sinx− 5) . From the graphs, it

seems that f 0 > 0 (and so f is increasing) on approximately the intervals

(0, 2.7), (4.5, 8.2) and (10.9, 14.3). It seems that f 00 changes sign

(indicating inflection points) at x ≈ 3.8, 5.7, 10.0 and 12.0.

Looking back at the graph of f(x) = ln(2x+ x sinx), this implies that the inflection points have approximate coordinates

(3.8, 1.7), (5.7, 2.1), (10.0, 2.7), and (12.0, 2.9).

59. y = (2x+ 1)5(x4 − 3)6 ⇒ ln y = ln (2x+ 1)5(x4 − 3)6 ⇒ ln y = 5 ln(2x+ 1) + 6 ln(x4 − 3) ⇒1

yy0 = 5 · 1

2x+ 1· 2 + 6 · 1

x4 − 3 · 4x3 ⇒

y0 = y10

2x+ 1+

24x3

x4 − 3 = (2x+ 1)5(x4 − 3)6 10

2x+ 1+

24x3

x4 − 3 .

[The answer could be simplified to y0 = 2(2x+ 1)4(x4 − 3)5(29x4 + 12x3 − 15), but this is unnecessary.]


61. y = sin2 x tan4 x

(x2 + 1)2⇒ ln y = ln(sin2 x tan4 x) − ln(x2 + 1)2 ⇒

ln y = ln(sinx)2 + ln(tanx)4 − ln(x2 + 1)2 ⇒ ln y = 2 ln |sinx|+ 4 ln |tanx|− 2 ln(x2 + 1) ⇒1

yy0 = 2 · 1

sinx· cosx+ 4 · 1

tanx· sec2 x− 2 · 1

x2 + 1· 2x ⇒ y0 =

sin2 x tan4 x

(x2 + 1)22 cotx+

4 sec2 x

tanx− 4x

x2 + 1

63.4

2

3

xdx = 3

4

2

1

xdx = 3 ln |x| 4

2= 3(ln 4− ln 2) = 3 ln 4

2= 3 ln 2

65.2

1

dt

8− 3t = −13ln |8− 3t|

2

1

= −13ln 2− −1

3ln 5 =

1

3(ln 5− ln 2) = 1

3ln5

2

Or: Let u = 8 − 3t. Then du = −3 dt, so2

1

dt

8− 3t =2

5

− 13 du

u= −1

3ln |u|

2

5

= −13ln 2− −1

3ln 5 =

1

3(ln 5− ln 2) = 1

3ln5

2.

67.e

1

x2 + x+ 1

xdx =

e

1

x+ 1 +1

xdx = 1

2x2 + x+ lnx

e

1= 1

2e2 + e+ 1 − 1

2+ 1 + 0

= 12e

2 + e− 12

69. Let u = lnx. Then du = dx

x⇒ (lnx)2

xdx = u2 du =

1

3u3 + C =

1

3(lnx)3 +C.

71. sin 2x

1 + cos2 xdx = 2

sinx cosx

1 + cos2 xdx = 2I. Let u = cosx. Then du = − sinxdx, so

2I = −2 udu

1 + u2= −2 · 1

2ln(1 + u2) + C = − ln(1 + u2) +C = − ln(1 + cos2 x) + C.

Or: Let u = 1 + cos2 x.

73. (a) d

dx(ln |sinx|+ C) =

1

sinxcosx = cotx

(b) Let u = sinx. Then du = cosxdx, so cotxdx =cosx

sinxdx =

du

u= ln |u|+C = ln |sinx|+C.

75. The cross-sectional area is π 1/√x+ 1

2= π/(x + 1). Therefore, the volume is

1

0

π

x+ 1dx = π[ln(x+ 1)]10 = π(ln 2− ln 1) = π ln 2.

77. W =V2V1

P dV =1000

600

C

VdV = C

1000

600

1

VdV = C ln |V | 1000

600= C(ln 1000− ln 600) = C ln 1000

600= C ln 5

3.

Initially, PV = C, where P = 150 kPa and V = 600 cm3, so C = (150)(600) = 90,000. Thus,

W = 90, 000 ln 53≈ 45, 974 kPa · cm3 = 45, 974(103 Pa)(10−6 m3) = 45, 974 Pa·m3 = 45, 974 N·m [Pa = N/m2]

= 45, 974 J

79. f(x) = 2x+ lnx ⇒ f 0(x) = 2 + 1/x. If g = f−1, then f(1) = 2 ⇒ g(2) = 1, so

g0(2) = 1/f 0(g(2)) = 1/f 0(1) = 13

.


81. (a) We interpret ln 1.5 as the area under the curve y = 1/x from x = 1 to

x = 1.5. The area of the rectangle BCDE is 12· 23= 1

3. The area of the

trapezoid ABCD is 12 · 12 1 + 2

3= 5

12 . Thus, by comparing areas, we

observe that 13< ln 1.5 < 5

12.

(b) With f(t) = 1/t, n = 10, and ∆t = 0.05, we have

ln 1.5 =1.5

1(1/t) dt ≈ (0.05)[f(1.025) + f(1.075) + · · ·+ f(1.475)]

= (0.05) 11.025 +

11.075 + · · ·+ 1

1.475≈ 0.4054

83. The area of Ri is 1

i+ 1and so 1

2+1

3+ · · ·+ 1

n<

n

1

1

tdt = lnn.

The area of Si is 1i

and so 1 + 1

2+ · · ·+ 1

n− 1 >n

1

1

tdt = lnn.

85. The curve and the line will determine a region when they intersect at two or

more points. So we solve the equation x/(x2 + 1) = mx ⇒ x = 0 or

mx2 +m− 1 = 0 ⇒ x = 0 or x =± −4(m)(m− 1)

2m= ± 1

m− 1.

Note that if m = 1, this has only the solution x = 0, and no region is

determined. But if 1/m− 1 > 0 ⇔ 1/m > 1 ⇔ 0 < m < 1, then there are two solutions. [Another way of seeing this

is to observe that the slope of the tangent to y = x/(x2 + 1) at the origin is y0 = 1 and therefore we must have 0 < m < 1.]

Note that we cannot just integrate between the positive and negative roots, since the curve and the line cross at the origin.

Since mx and x/(x2 + 1) are both odd functions, the total area is twice the area between the curves on the interval

0, 1/m− 1 . So the total area enclosed is

2

√1/m−1

0

x

x2 + 1−mx dx= 2 1

2ln(x2 + 1)− 1

2mx2

√1/m−1

0

= ln1

m− 1 + 1 −m

1

m− 1 − (ln 1− 0)

= ln1

m+m− 1 = m− lnm− 1

SECTION 7.3* THE NATURAL EXPONENTIAL FUNCTION ¤ 269

87. If f(x) = ln (1 + x), then f 0(x) =1

1 + x, so f 0(0) = 1.

Thus, limx→0

ln(1 + x)

x= lim

x→0

f(x)

x= lim

x→0

f(x)− f(0)

x− 0 = f 0(0) = 1.

7.3* The Natural Exponential Function

1. The function value at x = 0 is 1 and the slope at x = 0 is 1.

3. (a) e−2 ln 5 = eln 5−2 (4)= 5−2 =

1

52=1

25(b) ln ln ee

10 (5)= ln(e10)

(5)= 10

5. (a) 2 lnx = 1 ⇒ lnx = 12 ⇒ x = e1/2 =

√e (b) e−x = 5 ⇒ −x = ln 5 ⇒ x = − ln 5

7. (a) e3x+1 = k ⇔ 3x+ 1 = ln k ⇔ x = 13(lnk − 1)

(b) lnx+ ln(x− 1) = ln(x(x− 1)) = 1 ⇔ x(x− 1) = e1 ⇔ x2 − x− e = 0. The quadratic formula (with a = 1,

b = −1, and c = −e) gives x = 121±√1 + 4e , but we reject the negative root since the natural logarithm is not

defined for x < 0. So x = 121 +

√1 + 4e .

9. 3xex + x2ex = 0 ⇔ xex(3 + x) = 0 ⇔ x = 0 or −3

11. e2x − ex − 6 = 0 ⇔ (ex − 3)(ex + 2) = 0 ⇔ ex = 3 or −2 ⇒ x = ln 3 since ex > 0.

13. (a) e2+5x = 100 ⇒ ln e2+5x = ln 100 ⇒ 2 + 5x = ln 100 ⇒ 5x = ln 100− 2 ⇒x = 1

5(ln 100− 2) ≈ 0.5210

(b) ln(ex − 2) = 3 ⇒ ex − 2 = e3 ⇒ ex = e3 + 2 ⇒ x = ln(e3 + 2) ≈ 3.0949

15. (a) ex < 10 ⇒ ln ex < ln 10 ⇒ x < ln 10 ⇒ x ∈ (−∞, ln 10)

(b) lnx > −1 ⇒ elnx > e−1 ⇒ x > e−1 ⇒ x ∈ (1/e,∞)

17.

y = ex y = e−x


19. We start with the graph of y = ex (Figure 2) and reflect about the y-axis to get the graph of y = e−x. Then we compress the

graph vertically by a factor of 2 to obtain the graph of y = 12e−x and then reflect about the x-axis to get the graph of

y = − 12e−x. Finally, we shift the graph upward one unit to get the graph of y = 1− 1

2e−x.

21. (a) For f(x) =√3− e2x, we must have 3− e2x ≥ 0 ⇒ e2x ≤ 3 ⇒ 2x ≤ ln 3 ⇒ x ≤ 1

2 ln 3.

Thus, the domain of f is (−∞, 12 ln 3].

(b) y = f(x) =√3− e2x [note that y ≥ 0] ⇒ y2 = 3− e2x ⇒ e2x = 3− y2 ⇒ 2x = ln(3− y2) ⇒

x = 12 ln(3− y2). Interchange x and y: y = 1

2 ln(3− x2). So f−1(x) = 12 ln(3− x2). For the domain of f−1, we must

have 3− x2 > 0 ⇒ x2 < 3 ⇒ |x| < √3 ⇒ −√3 < x <√3 ⇒ 0 ≤ x <

√3 since x ≥ 0. Note that the

domain of f−1, [0,√3 ), equals the range of f .

23. y = ln(x+ 3) ⇒ ey = eln(x+3) = x+ 3 ⇒ x = ey − 3.

Interchange x and y: the inverse function is y = ex − 3.

25. y = f(x) = ex3 ⇒ ln y = x3 ⇒ x = 3

√ln y. Interchange x and y: y = 3

√lnx. So f−1(x) = 3

√lnx.

27. Divide numerator and denominator by e3x: limx→∞

e3x − e−3x

e3x + e−3x= lim

x→∞1− e−6x

1 + e−6x=1− 01 + 0

= 1

29. Let t = 3/(2− x). As x→ 2+, t→−∞. So limx→2+

e3/(2−x) = limt→−∞

et = 0 by (6).

31. Since −1 ≤ cosx ≤ 1 and e−2x > 0, we have −e−2x ≤ e−2x cosx ≤ e−2x. We know that limx→∞

(−e−2x) = 0 and

limx→∞

e−2x = 0, so by the Squeeze Theorem, limx→∞

(e−2x cosx) = 0.

33. By the Product Rule, f(x) = (x3 + 2x)ex ⇒

f 0(x) = (x3 + 2x)(ex)0 + ex(x3 + 2x)0 = (x3 + 2x)ex + ex(3x2 + 2)

= ex[(x3 + 2x) + (3x2 + 2)] = ex(x3 + 3x2 + 2x+ 2)

35. By (9), y = eax3 ⇒ y0 = eax

3 d

dx(ax3) = 3ax2eax

3.

37. f(u) = e1/u ⇒ f 0(u) = e1/u · d

du

1

u= e1/u

−1u2

=−1u2

e1/u

39. By (9), F (t) = et sin 2t ⇒ F 0(t) = et sin 2t(t sin 2t)0 = et sin 2t(t · 2 cos 2t+ sin 2t · 1) = et sin 2t(2t cos 2t+ sin 2t)


41. y =√1 + 2e3x ⇒ y0 =

1

2(1 + 2e3x)−1/2

d

dx(1 + 2e3x) =

1

2√1 + 2e3x

(2e3x · 3) = 3e3x√1 + 2e3x

43. y = eex ⇒ y0 = ee

x · d

dx(ex) = ee

x · ex or eex+x

45. By the Quotient Rule, y = aex + b

cex + d⇒

y0 =(cex + d)(aex)− (aex + b)(cex)

(cex + d)2=(acex + ad− acex − bc)ex

(cex + d)2=(ad− bc)ex

(cex + d)2.

47. y = cos 1− e2x

1 + e2x⇒

y0 = − sin 1− e2x

1 + e2x· d

dx

1− e2x

1 + e2x= − sin 1− e2x

1 + e2x· (1 + e2x)(−2e2x)− (1− e2x)(2e2x)

(1 + e2x)2

= − sin 1− e2x

1 + e2x· −2e

2x (1 + e2x) + (1− e2x)

(1 + e2x)2= − sin 1− e2x

1 + e2x· −2e

2x(2)

(1 + e2x)2=

4e2x

(1 + e2x)2· sin 1− e2x

1 + e2x

49. y = e2x cosπx ⇒ y0 = e2x(−π sinπx) + (cosπx)(2e2x) = e2x(2 cosπx− π sinπx).

At (0, 1), y0 = 1(2− 0) = 2, so an equation of the tangent line is y − 1 = 2(x− 0), or y = 2x+ 1.

51. d

dxex

2y =d

dx(x+ y) ⇒ ex

2y(x2y0 + y · 2x) = 1 + y0 ⇒ x2ex2yy0 + 2xyex

2y = 1 + y0 ⇒

x2ex2yy0 − y0 = 1− 2xyex2y ⇒ y0(x2ex

2y − 1) = 1− 2xyex2y ⇒ y0 =1− 2xyex2yx2ex2y − 1

53. y = ex + e−x/2 ⇒ y0 = ex − 12e−x/2 ⇒ y00 = ex + 1

4e−x/2, so

2y00 − y0 − y = 2 ex + 14e−x/2 − ex − 1

2e−x/2 − ex + e−x/2 = 0.

55. y = erx ⇒ y0 = rerx ⇒ y00 = r2erx, so if y = erx satisfies the differential equation y00 + 6y0 + 8y = 0,

then r2erx + 6rerx + 8erx = 0; that is, erx(r2 + 6r + 8) = 0. Since erx > 0 for all x, we must have r2 + 6r + 8 = 0,

or (r + 2)(r + 4) = 0, so r = −2 or −4.

57. f(x) = e2x ⇒ f 0(x) = 2e2x ⇒ f 00(x) = 2 · 2e2x = 22e2x ⇒f 000(x) = 22 · 2e2x = 23e2x ⇒ · · · ⇒ f (n)(x) = 2ne2x

59. (a) f(x) = ex + x is continuous on R and f(−1) = e−1 − 1 < 0 < 1 = f(0), so by the Intermediate Value Theorem,

ex + x = 0 has a root in (−1, 0).

(b) f(x) = ex + x ⇒ f 0(x) = ex + 1, so xn+1 = xn − exn + xnexn + 1

. Using x1 = −0.5, we get x2 ≈ −0.566311,

x3 ≈ −0.567143 ≈ x4, so the root is −0.567143 to six decimal places.

61. (a) limt→∞

p(t) = limt→∞

1

1 + ae−kt=

1

1 + a · 0 = 1, since k > 0 ⇒ −kt→ −∞ ⇒ e−kt → 0.

(b) p(t) = (1 + ae−kt)−1 ⇒ dp

dt= −(1 + ae−kt)−2(−kae−kt) = kae−kt

(1 + ae−kt)2


(c)From the graph of p(t) = (1 + 10e−0.5t)−1, it seems that p(t) = 0.8

(indicating that 80% of the population has heard the rumor) when

t ≈ 7.4 hours.

63. f(x) = x− ex ⇒ f 0(x) = 1− ex = 0 ⇔ ex = 1 ⇔ x = 0. Now f 0(x) > 0 for all x < 0 and f 0(x) < 0 for all

x > 0, so the absolute maximum value is f(0) = 0− 1 = −1.

65. f(x) = xe−x2/8, [−1, 4]. f 0(x) = x · e−x2/8 · (−x

4) + e−x

2/8 · 1 = e−x2/8(−x2

4+ 1). Since e−x

2/8 is never 0,

f 0(x) = 0 ⇒ −x2/4 + 1 = 0 ⇒ 1 = x2/4 ⇒ x2 = 4 ⇒ x = ±2, but −2 is not in the given interval, [−1, 4].f(−1) = −e−1/8 ≈ −0.88, f(2) = 2e−1/2 ≈ 1.21, and f(4) = 4e−2 ≈ 0.54. So f(2) = 2e−1/2 is the absolute maximum

value and f(−1) = −e−1/8 is the absolute minimum value.

67. (a) f(x) = (1− x)e−x ⇒ f 0(x) = (1− x)(−e−x) + e−x(−1) = e−x(x− 2) > 0 ⇒ x > 2, so f is increasing on

(2,∞) and decreasing on (−∞, 2).

(b) f 00(x) = e−x(1) + (x− 2)(−e−x) = e−x(3− x) > 0 ⇔ x < 3, so f is CU on (−∞, 3) and CD on (3,∞).

(c) f 00 changes sign at x = 3, so there is an IP at 3,−2e−3 .

69. y = f(x) = e−1/(x+1) A. D = {x | x 6= −1} = (−∞,−1) ∪ (−1,∞) B. No x-intercept; y-intercept = f(0) = e−1

C. No symmetry D. limx→±∞

e−1/(x+1) = 1 since −1/(x+ 1)→ 0, so y = 1 is a HA. limx→−1+

e−1/(x+1) = 0 since

−1 /(x+ 1) → −∞, limx→−1−

e−1/(x+1) = ∞ since −1/(x+ 1) → ∞, so x = −1 is a VA.

E. f 0(x) = e−1/(x+1)/(x+ 1)2 ⇒ f 0(x) > 0 for all x except 1, so

f is increasing on (−∞,−1) and (−1,∞). F. No extreme values

G. f 00(x) = e−1/(x+1)

(x+ 1)4+

e−1/(x+1)(−2)(x+ 1)3

= −e−1/(x+1)(2x+ 1)(x+ 1)4

⇒

f 00(x) > 0 ⇔ 2x+ 1 < 0 ⇔ x < − 12

, so f is CU on (−∞,−1)and −1,− 1

2, and CD on − 1

2,∞ . f has an IP at − 1

2, e−2 .

H.

71. S(t) = Atpe−kt with A = 0.01, p = 4, and k = 0.07. We will find the

zeros of f 00 for f(t) = tpe−kt.

f 0(t) = tp(−ke−kt) + e−kt(ptp−1) = e−kt(−ktp + ptp−1)

f 00(t) = e−kt(−kptp−1 + p(p− 1)tp−2) + (−ktp + ptp−1)(−ke−kt)= tp−2e−kt[−kpt+ p(p− 1) + k2t2 − kpt]

= tp−2e−kt(k2t2 − 2kpt+ p2 − p)

Using the given values of p and k gives us f 00(t) = t2e−0.07t(0.0049t2 − 0.56t+ 12). So S00(t) = 0.01f 00(t) and its zeros


are t = 0 and the solutions of 0.0049t2 − 0.56t+ 12 = 0, which are t1 = 2007 ≈ 28.57 and t2 = 600

7 ≈ 85.71.

At t1 minutes, the rate of increase of the level of medication in the bloodstream is at its greatest and at t2 minutes, the rate of

decrease is the greatest.

73. f(x) = ex3−x → 0 as x→ −∞, and

f(x)→∞ as x→∞. From the graph,

it appears that f has a local minimum of

about f(0.58) = 0.68, and a local

maximum of about f(−0.58) = 1.47.

To find the exact values, we calculate

f 0(x) = 3x2 − 1 ex3−x, which is 0 when 3x2 − 1 = 0 ⇔ x = ± 1√

3. The negative root corresponds to the local

maximum f − 1√3= e(−1/

√3)3− (−1/√3) = e2

√3/9, and the positive root corresponds to the local minimum

f 1√3= e(1/

√3)3− (1/

√3) = e−2

√3/9. To estimate the inflection points, we calculate and graph

f 00(x) =d

dx3x2 − 1 ex

3−x = 3x2 − 1 ex3−x 3x2 − 1 + ex

3−x(6x) = ex3−x 9x4 − 6x2 + 6x+ 1 .

From the graph, it appears that f 00(x) changes sign (and thus f has inflection points) at x ≈ −0.15 and x ≈ −1.09. From the

graph of f , we see that these x-values correspond to inflection points at about (−0.15, 1.15) and (−1.09, 0.82).

75. Let u = −3x. Then du = −3 dx, so 5

0e−3x dx = − 1

3

−150

eu du = − 13[eu]−150 = − 1

3e−15 − e0 = 1

31− e−15 .

77. Let u = 1 + ex. Then du = ex dx, so ex√1 + ex dx =

√udu = 2

3u3/2 +C = 2

3(1 + ex)3/2 + C.

79. (ex + e−x)2 dx = (e2x + 2 + e−2x) dx = 12e2x + 2x− 1

2e−2x +C

81. sinx ecos x dxu = cosx,

du = − sinx dx = eu (−du) = −eu +C = −ecos x + C

83. Let u =√x. Then du = 1

2√xdx, so e

√x

√xdx = 2 eu du = 2eu + C = 2e

√x +C.

85. Area = 1

0e3x − ex dx = 1

3e3x − ex

1

0= 1

3e3 − e − 1

3− 1 = 1

3e3 − e+ 2

3≈ 4.644

87. V =1

0π(ex)2 dx = π

1

0e2x dx = 1

2π e2x

1

0= π

2e2 − 1

89. (a) erf(x) = 2√π

x

0

e−t2

dt ⇒x

0

e−t2

dt =

√π

2erf(x). By Property 5 of definite integrals in Section 5.2,

b

0e−t

2

dt =a

0e−t

2

dt+b

ae−t

2

dt, so

b

a

e−t2

dt =b

0

e−t2

dt−a

0

e−t2

dt =

√π

2erf(b)−

√π

2erf(a) = 1

2

√π [erf(b)− erf(a)].

(b) y = ex2

erf(x) ⇒ y0 = 2xex2

erf(x) + ex2

erf 0(x) = 2xy + ex2 · 2√

πe−x

2

[by FTC1] = 2xy +2√π

.


91. We use Theorem 7.1.7. Note that f(0) = 3 + 0 + e0 = 4, so f−1(4) = 0. Also f 0(x) = 1 + ex. Therefore,

f−1 0(4) =

1

f 0(f−1(4))=

1

f 0(0)=

1

1 + e0=1

2.

93. From the graph, it appears that f is an odd function (f is undefined for x = 0).

To prove this, we must show that f(−x) = −f(x).

f(−x) = 1− e1/(−x)

1 + e1/(−x)=1− e(−1/x)

1 + e(−1/x)=1− 1

e1/x

1 +1

e1/x

· e1/x

e1/x=

e1/x − 1e1/x + 1

= −1− e1/x

1 + e1/x= −f(x)

so f is an odd function.

95. Using the second law of logarithms and Equation 5, we have ln(ex/ey) = ln ex − ln ey = x− y = ln(ex− y). Since ln is a

one-to-one function, it follows that ex/ey = ex− y .

97. (a) Let f(x) = ex − 1− x. Now f(0) = e0 − 1 = 0, and for x ≥ 0, we have f 0(x) = ex − 1 ≥ 0. Now, since f(0) = 0 and

f is increasing on [0,∞), f(x) ≥ 0 for x ≥ 0 ⇒ ex − 1− x ≥ 0 ⇒ ex ≥ 1 + x.

(b) For 0 ≤ x ≤ 1, x2 ≤ x, so ex2 ≤ ex [since ex is increasing]. Hence [from (a)] 1 + x2 ≤ ex

2 ≤ ex.

So 43=

1

01 + x2 dx ≤ 1

0ex

2

dx ≤ 1

0ex dx = e− 1 < e ⇒ 4

3≤ 1

0ex

2

dx ≤ e.

99. (a) By Exercise 97(a), the result holds for n = 1. Suppose that ex ≥ 1 + x+x2

2!+ · · ·+ xk

k!for x ≥ 0.

Let f(x) = ex − 1− x− x2

2!− · · ·− xk

k!− xk+1

(k + 1)!. Then f 0(x) = ex − 1− x− · · ·− xk

k!≥ 0 by assumption. Hence

f(x) is increasing on (0,∞). So 0 ≤ x implies that 0 = f(0) ≤ f(x) = ex − 1− x− · · ·− xk

k!− xk+1

(k + 1)!, and hence

ex ≥ 1 + x+ · · ·+ xk

k!+

xk+1

(k + 1)!for x ≥ 0. Therefore, for x ≥ 0, ex ≥ 1 + x+

x2

2!+ · · ·+ xn

n!for every positive

integer n, by mathematical induction.

(b) Taking n = 4 and x = 1 in (a), we have e = e1 ≥ 1 + 12+ 1

6+ 1

24= 2.7083 > 2.7.

(c) ex ≥ 1 + x+ · · ·+ xk

k!+

xk+1

(k + 1)!⇒ ex

xk≥ 1

xk+

1

xk−1+ · · ·+ 1

k!+

x

(k + 1)!≥ x

(k + 1)!.

But limx→∞

x

(k + 1)!=∞, so lim

x→∞ex

xk=∞.

SECTION 7.4* GENERAL LOGARITHMIC AND EXPONENTIAL FUNCTIONS ¤ 275

7.4* General Logarithmic and Exponential Functions

1. (a) ax = ex ln a

(b) The domain of f(x) = ax is R.

(c) The range of f(x) = ax [a 6= 1] is (0,∞).(d) (i) See Figure 1. (ii) See Figure 3. (iii) See Figure 2.

3. 5√7 = (eln 5)

√7 = e

√7 ln 5 5. (cosx)x = (eln cosx)x = ex ln(cos x).

7. (a) log5 125 = 3 since 53 = 125. (b) log31

27= −3 since 3−3 = 1

33=1

27.

9. (a) log2 6− log2 15 + log2 20 = log2( 615 ) + log2 20 [by Law 2]

= log2(615· 20) [by Law 1]

= log2 8, and log2 8 = 3 since 23 = 8.

(b) log3 100− log3 18− log3 50 = log310018

− log3 50 = log3 10018·50

= log3(19 ), and log3

19= −2 since 3−2 = 1

9 .

11. All of these graphs approach 0 as x→−∞, all of them pass through the point

(0, 1), and all of them are increasing and approach∞ as x→∞. The larger the

base, the faster the function increases for x > 0, and the faster it approaches 0 as

x→−∞.

13. (a) log12 e =ln e

ln 12=

1

ln 12≈ 0.402430

(b) log6 13.54 =ln 13.54

ln 6≈ 1.454240

(c) log2 π =lnπ

ln 2≈ 1.651496

15. To graph these functions, we use log1.5 x =lnx

ln 1.5and log50 x =

lnx

ln 50.

These graphs all approach−∞ as x→ 0+, and they all pass through the

point (1, 0). Also, they are all increasing, and all approach∞ as x→∞.

The functions with larger bases increase extremely slowly, and the ones with

smaller bases do so somewhat more quickly. The functions with large bases

approach the y-axis more closely as x→ 0+.

17. Use y = Cax with the points (1, 6) and (3, 24). 6 = Ca1 C = 6a

and 24 = Ca3 ⇒ 24 =6

aa3 ⇒

4 = a2 ⇒ a = 2 [since a > 0] and C = 62 = 3. The function is f(x) = 3 · 2x.

19. (a) 2 ft = 24 in, f(24) = 242 in = 576 in = 48 ft. g(24) = 224 in = 224/(12 · 5280) mi ≈ 265 mi

(b) 3 ft = 36 in, so we need x such that log2 x = 36 ⇔ x = 236 = 68,719,476,736. In miles, this is

68,719,476,736 in · 1 ft12 in

· 1 mi5280 ft

≈ 1,084,587.7 mi.


21. limx→∞

(1.001)x =∞ by Figure 1, since 1.001 > 1.

23. limt→∞

2−t2

= limu→−∞

2u [where u = −t2] = 0

25. h(t) = t3 − 3t ⇒ h0(t) = 3t2 − 3t ln 3

27. Using Formula 4 and the Chain Rule, y = 5−1/x ⇒ y0 = 5−1/x(ln 5) −1 · −x−2 = 5−1/x(ln 5)/x2

29. f(u) = (2u + 2−u)10 ⇒

f 0(u) = 10(2u + 2−u)9d

du(2u + 2−u) = 10(2u + 2−u)9 2u ln 2 + 2−u ln 2 · (−1)

= 10 ln 2(2u + 2−u)9(2u − 2−u)

31. f(x) = log2(1− 3x) ⇒ f 0(x) =1

(1− 3x) ln 2d

dx(1− 3x) = −3

(1− 3x) ln 2 or 3

(3x− 1) ln 2

33. y = 2x log10√x = 2x log10 x

1/2 = 2x · 12log10 x = x log10 x ⇒ y0 = x · 1

x ln 10+ log10 x · 1 =

1

ln 10+ log10 x

Note: 1

ln 10=ln e

ln 10= log10 e, so the answer could be written as 1

ln 10+ log10 x = log10 e+ log10 x = log10 ex.

35. y = xx ⇒ ln y = lnxx ⇒ ln y = x lnx ⇒ y0/y = x(1/x) + (lnx) · 1 ⇒ y0 = y(1 + lnx) ⇒y0 = xx(1 + lnx)

37. y = x sin x ⇒ ln y = lnx sinx ⇒ ln y = sinx lnx ⇒ y0

y= (sinx) · 1

x+ (lnx)(cosx) ⇒

y0 = ysinx

x+ lnx cosx ⇒ y0 = x sin x

sinx

x+ lnx cosx

39. y = (cosx)x ⇒ ln y = ln(cosx)x ⇒ ln y = x ln cosx ⇒ 1

yy0 = x · 1

cosx· (− sinx) + ln cosx · 1 ⇒

y0 = y ln cosx− x sinx

cosx⇒ y0 = (cosx)x(ln cosx− x tanx)

41. y = (tanx)1/x ⇒ ln y = ln(tanx)1/x ⇒ ln y =1

xln tanx ⇒

1

yy0 =

1

x· 1

tanx· sec2 x+ ln tanx · − 1

x2⇒ y0 = y

sec2 x

x tanx− ln tanx

x2⇒

y0 = (tanx)1/xsec2 x

x tanx− ln tanx

x2or y0 = (tanx)1/x · 1

xcscx secx− ln tanx

x

43. y = 10x ⇒ y0 = 10x ln 10, so at (1, 10), the slope of the tangent line is 101 ln 10 = 10 ln 10, and its equation is

y − 10 = 10 ln 10(x− 1), or y = (10 ln 10)x+ 10(1− ln 10).

45.2

1

10t dt =10t

ln 10

2

1

=102

ln 10− 101

ln 10=100− 10ln 10

=90

ln 10

SECTION 7.4* GENERAL LOGARITHMIC AND EXPONENTIAL FUNCTIONS ¤ 277

47. log10 x

xdx =

(lnx)/(ln 10)

xdx =

1

ln 10

lnx

xdx. Now put u = lnx, so du = 1

xdx, and the expression becomes

1

ln 10udu =

1

ln 1012u2 + C1 =

1

2 ln 10(lnx)2 + C.

Or: The substitution u = log10 x gives du = dx

x ln 10and we get log10 x

xdx = 1

2ln 10(log10 x)

2 + C.

49. Let u = sin θ. Then du = cos θ dθ and 3sin θ cos θ dθ = 3udu =3u

ln 3+ C =

1

ln 33sin θ +C.

51. A =0

−1(2x − 5x) dx+

1

0

(5x − 2x) dx = 2x

ln 2− 5x

ln 5

0

−1+

5x

ln 5− 2x

ln 2

1

0

=1

ln 2− 1

ln 5− 1/2

ln 2− 1/5

ln 5+

5

ln 5− 2

ln 2− 1

ln 5− 1

ln 2

=16

5 ln 5− 1

2 ln 2

53. We see that the graphs of y = 2x and y = 1 + 3−x intersect at x ≈ 0.6. We

let f(x) = 2x − 1− 3−x and calculate f 0(x) = 2x ln 2 + 3−x ln 3, and

using the formula xn+1 = xn − f(xn) /f0(xn) (Newton’s Method), we get

x1 = 0.6, x2 ≈ x3 ≈ 0.600967. So, correct to six decimal places, the root

occurs at x = 0.600967.

55. y = log10 1 +1

x⇒ 10y = 1 +

1

x⇒ 1

x= 10y − 1 ⇒ x =

1

10y − 1 .

Interchange x and y: y =1

10x − 1 is the inverse function.

57. If I is the intensity of the 1989 San Francisco earthquake, then log10(I/S) = 7.1 ⇒log10(16I/S) = log10 16 + log10(I/S) = log10 16 + 7.1 ≈ 8.3.

59. We find I with the loudness formula from Exercise 58, substituting I0 = 10−12 and L = 50:

50 = 10 log10I

10−12⇔ 5 = log10

I

10−12⇔ 105 =

I

10−12⇔ I = 10−7 watt/m2. Now we differentiate L with

respect to I: L = 10 log10I

I0⇒ dL

dI= 10

1

(I/I0) ln 10

1

I0=

10

ln 10

1

I. Substituting I = 10−7, we get

L0 (50) =10

ln 10

1

10−7=108

ln 10≈ 4.34× 107 dB

watt/m2.

61. (a) Using a calculator or CAS, we obtain the model Q = abt with a ≈ 100.0124369 and b ≈ 0.000045145933.

(b) Use Q0(t) = abt ln b (from Formula 4) with the values of a and b from part (a) to get Q0(0.04) ≈ −670.63 μA.

The result of Example 2 in Section 2.1 was −670 μA.

63. Using Definition 1 and the second law of exponents for ex, we have ax− y = e(x− y) ln a = ex ln a− y ln a =ex ln a

ey ln a=

ax

ay.


65. Let loga x = r and loga y = s. Then ar = x and as = y.

(a) xy = aras = ar+ s ⇒ loga (xy) = r + s = loga x+ loga y

(b) x

y=

ar

as= ar− s ⇒ loga

x

y= r − s = loga x− loga y

(c) xy = (ar)y = ary ⇒ loga (xy) = ry = y loga x

7.5 Exponential Growth and Decay

1. The relative growth rate is 1P

dP

dt= 0.7944, so dP

dt= 0.7944P and, by Theorem 2, P (t) = P (0)e0.7944t = 2e0.7944t.

Thus, P (6) = 2e0.7944(6) ≈ 234.99 or about 235 members.

3. (a) By Theorem 2, P (t) = P (0)ekt = 100ekt. Now P (1) = 100ek(1) = 420 ⇒ ek = 420100

⇒ k = ln 4.2.

So P (t) = 100e(ln 4.2)t = 100(4.2)t.

(b) P (3) = 100(4.2)3 = 7408.8 ≈ 7409 bacteria

(c) dP/dt = kP ⇒ P 0(3) = k · P (3) = (ln 4.2) 100(4.2)3 [from part (a)] ≈ 10,632 bacteria/hour

(d) P (t) = 100(4.2)t = 10,000 ⇒ (4.2)t = 100 ⇒ t = (ln 100)/(ln 4.2) ≈ 3.2 hours

5. (a) Let the population (in millions) in the year t be P (t). Since the initial time is the year 1750, we substitute t− 1750 for t in

Theorem 2, so the exponential model gives P (t) = P (1750)ek(t−1750). Then P (1800) = 980 = 790ek(1800−1750) ⇒980790 = ek(50) ⇒ ln 980

790 = 50k ⇒ k = 150 ln

980790 ≈ 0.0043104. So with this model, we have

P (1900) = 790ek(1900−1750) ≈ 1508 million, and P (1950) = 790ek(1950−1750) ≈ 1871 million. Both of these

estimates are much too low.

(b) In this case, the exponential model gives P (t) = P (1850)ek(t−1850) ⇒ P (1900) = 1650 = 1260ek(1900−1850) ⇒ln 1650

1260= k(50) ⇒ k = 1

50ln 1650

1260≈ 0.005393. So with this model, we estimate

P (1950) = 1260ek(1950−1850) ≈ 2161 million. This is still too low, but closer than the estimate of P (1950) in part (a).

(c) The exponential model gives P (t) = P (1900)ek(t−1900) ⇒ P (1950) = 2560 = 1650ek(1950−1900) ⇒ln 2560

1650 = k(50) ⇒ k = 150 ln

25601650 ≈ 0.008785. With this model, we estimate

P (2000) = 1650ek(2000−1900) ≈ 3972 million. This is much too low. The discrepancy is explained by the fact that the

world birth rate (average yearly number of births per person) is about the same as always, whereas the mortality rate

(especially the infant mortality rate) is much lower, owing mostly to advances in medical science and to the wars in the first

part of the twentieth century. The exponential model assumes, among other things, that the birth and mortality rates will

remain constant.

7. (a) If y = [N2O5] then by Theorem 2, dydt= −0.0005y ⇒ y(t) = y(0)e−0.0005t = Ce−0.0005t.

(b) y(t) = Ce−0.0005t = 0.9C ⇒ e−0.0005t = 0.9 ⇒ −0.0005t = ln 0.9 ⇒ t = −2000 ln 0.9 ≈ 211 s

SECTION 7.5 EXPONENTIAL GROWTH AND DECAY ¤ 279

9. (a) If y(t) is the mass (in mg) remaining after t years, then y(t) = y(0)ekt = 100ekt.

y(30) = 100e30k = 12 (100) ⇒ e30k = 1

2 ⇒ k = −(ln 2)/30 ⇒ y(t) = 100e−(ln 2)t/30 = 100 · 2−t/30

(b) y(100) = 100 · 2−100/30 ≈ 9.92 mg

(c) 100e−(ln 2)t/30 = 1 ⇒ −(ln 2)t/30 = ln 1100 ⇒ t = −30 ln 0.01ln 2 ≈ 199.3 years

11. Let y(t) be the level of radioactivity. Thus, y(t) = y(0)e−kt and k is determined by using the half-life:

y(5730) = 12y(0) ⇒ y(0)e−k(5730) = 1

2y(0) ⇒ e−5730k = 12 ⇒ −5730k = ln 1

2 ⇒ k = − ln12

5730=ln 2

5730.

If 74% of the 14C remains, then we know that y(t) = 0.74y(0) ⇒ 0.74 = e−t(ln 2)/5730 ⇒ ln 0.74 = − t ln 2

5730⇒

t = −5730(ln 0.74)ln 2

≈ 2489 ≈ 2500 years.

13. (a) Using Newton’s Law of Cooling, dTdt

= k(T − Ts), we have dT

dt= k(T − 75). Now let y = T − 75, so

y(0) = T (0)− 75 = 185− 75 = 110, so y is a solution of the initial-value problem dy/dt = ky with y(0) = 110 and by

Theorem 2 we have y(t) = y(0)ekt = 110ekt.

y(30) = 110e30k = 150− 75 ⇒ e30k = 75110

= 1522

⇒ k = 130ln 15

22, so y(t) = 110e

130t ln( 1522 ) and

y(45) = 110e4530

ln( 1522 ) ≈ 62◦F. Thus, T (45) ≈ 62 + 75 = 137◦F.

(b) T (t) = 100 ⇒ y(t) = 25. y(t) = 110e130t ln( 1522 ) = 25 ⇒ e

130t ln( 1522 ) = 25

110⇒ 1

30t ln 15

22= ln 25

110⇒

t =30 ln 25

110

ln 1522

≈ 116 min.

15. dT

dt= k(T − 20). Letting y = T − 20, we get dy

dt= ky, so y(t) = y(0)ekt. y(0) = T (0)− 20 = 5− 20 = −15, so

y(25) = y(0)e25k = −15e25k, and y(25) = T (25)− 20 = 10− 20 = −10, so −15e25k = −10 ⇒ e25k = 23

. Thus,

25k = ln 23

and k = 125 ln

23

, so y(t) = y(0)ekt = −15e(1/25) ln(2/3)t. More simply, e25k = 23 ⇒ ek = 2

3

1/25 ⇒

ekt = 23

t/25 ⇒ y(t) = −15 · 23

t/25.

(a) T (50) = 20 + y(50) = 20− 15 · 23

50/25= 20− 15 · 2

3

2= 20− 20

3 = 13.3̄◦C

(b) 15 = T (t) = 20 + y(t) = 20− 15 · 23

t/25 ⇒ 15 · 23

t/25= 5 ⇒ 2

3

t/25= 1

3 ⇒

(t/25) ln 23= ln 1

3⇒ t = 25 ln 1

3ln 2

3≈ 67.74 min.

17. (a) Let P (h) be the pressure at altitude h. Then dP/dh = kP ⇒ P (h) = P (0)ekh = 101.3ekh.

P (1000) = 101.3e1000k = 87.14 ⇒ 1000k = ln 87.14101.3

⇒ k = 11000

ln 87.14101.3

⇒

P (h) = 101.3 e1

1000h ln( 87.14101.3 ), so P (3000) = 101.3e3 ln(

87.14101.3 ) ≈ 64.5 kPa.

(b) P (6187) = 101.3 e61871000

ln( 87.14101.3 ) ≈ 39.9 kPa


19. (a) Using A = A0 1 +r

n

nt

with A0 = 3000, r = 0.05, and t = 5, we have:

(i) Annually: n = 1; A = 3000 1 + 0.051

1·5= $3828.84

(ii) Semiannually: n = 2; A = 3000 1 + 0.052

2·5= $3840.25

(iii) Monthly: n = 12; A = 3000 1 + 0.0512

12·5= $3850.08

(iv) Weekly: n = 52; A = 3000 1 + 0.0552

52·5= $3851.61

(v) Daily: n = 365; A = 3000 1 + 0.05365

365·5= $3852.01

(vi) Continuously: A = 3000e(0.05)5 = $3852.08

(b) dA/dt = 0.05A and A(0) = 3000.

7.6 Inverse Trigonometric Functions

1. (a) sin−1√32

= π3

since sin π3=√32

and π3

is in −π2, π2

.

(b) cos−1(−1) = π since cosπ = −1 and π is in [0, π].

3. (a) arctan 1 = π4

since tan π4= 1 and π

4is in −π

2, π2

.

(b) sin−1 1√2= π

4 since sin π4 =

1√2

and π4 is in −π

2 ,π2

.

5. (a) In general, tan(arctanx) = x for any real number x. Thus, tan(arctan 10) = 10.

(b) sin−1 sin 7π3

= sin−1 sin π3= sin−1

√32= π

3since sin π

3=√32

and π3

is in −π2, π2

.

[Recall that 7π3= π

3+ 2π and the sine function is periodic with period 2π.]

7. Let θ = sin−1 23

.

Then tan sin−1 23

= tan θ =2√5

.

9. Let θ = tan−1√2. Then

sin 2 tan−1√2 = sin(2θ) = 2 sin θ cos θ

= 2

√2√3

1√3

=2√2

3

11. Let y = sin−1 x. Then −π2 ≤ y ≤ π

2 ⇒ cos y ≥ 0, so cos(sin−1 x) = cos y = 1− sin2 y = √1− x2.

13. Let y = tan−1 x. Then tan y = x, so from the triangle we see that

sin(tan−1 x) = sin y =x√1 + x2

.

SECTION 7.6 INVERSE TRIGONOMETRIC FUNCTIONS ¤ 281

15. The graph of sin−1 x is the reflection of the graph of

sinx about the line y = x.

17. Let y = cos−1 x. Then cos y = x and 0 ≤ y ≤ π ⇒ − sin y dy

dx= 1 ⇒

dy

dx= − 1

sin y= − 1

1− cos2 y = −1√1− x2

. [Note that sin y ≥ 0 for 0 ≤ y ≤ π.]

19. Let y = cot−1 x. Then cot y = x ⇒ − csc2 y dy

dx= 1 ⇒ dy

dx= − 1

csc2 y= − 1

1 + cot2 y= − 1

1 + x2.

21. Let y = csc−1 x. Then csc y = x ⇒ − csc y cot y dy

dx= 1 ⇒

dy

dx= − 1

csc y cot y= − 1

csc y csc2 y − 1 = −1

x√x2 − 1 . Note that cot y ≥ 0 on the domain of csc−1 x.

23. y = tan−1√x ⇒ y0 =

1

1 +√x

2 ·d

dx

√x =

1

1 + x12x−1/2 =

1

2√x (1 + x)

25. y = sin−1(2x + 1) ⇒y0 =

1

1− (2x+ 1)2 ·d

dx(2x+ 1) =

1

1− (4x2 + 4x+ 1) · 2 =2√−4x2 − 4x =

1√−x2 − x

27. G(x) =√1− x2 arccosx ⇒ G0(x) =

√1− x2 · −1√

1− x2+ arccosx · 1

2(1− x2)−1/2(−2x) = −1− x arccosx√

1− x2

29. y = cos−1(e2x) ⇒ y0 = − 1

1− (e2x)2 ·d

dx(e2x) = − 2e2x√

1− e4x

31. y = arctan(cos θ) ⇒ y0 =1

1 + (cos θ)2(− sin θ) = − sin θ

1 + cos2 θ

33. h(t) = cot−1(t) + cot−1(1/t) ⇒

h0(t) = − 1

1 + t2− 1

1 + (1/t)2· ddt

1

t= − 1

1 + t2− t2

t2 + 1· − 1

t2= − 1

1 + t2+

1

t2 + 1= 0.

Note that this makes sense because h(t) = π

2for t > 0 and h(t) = 3π

2for t < 0.

35. y = arccos b+ a cosx

a+ b cosx⇒

y0 = − 1

1− b+ a cosx

a+ b cosx

2

(a+ b cosx)(−a sinx)− (b+ a cosx)(−b sinx)(a+ b cosx)2

=1√

a2 + b2 cos2 x− b2 − a2 cos2 x

(a2 − b2) sinx

|a+ b cosx| =1√

a2 − b2√1− cos2 x

(a2 − b2) sinx

|a+ b cosx| =

√a2 − b2

|a+ b cosx|sinx

|sinx|

But 0 ≤ x ≤ π, so |sinx| = sinx. Also a > b > 0 ⇒ b cosx ≥ −b > −a, so a+ b cosx > 0. Thus y0 =√a2 − b2

a+ b cosx.


37. g(x) = cos−1(3 − 2x) ⇒ g0(x) = − 1

1− (3− 2x)2(−2) = 2

1− (3− 2x)2 .

Domain(g) = {x | −1 ≤ 3− 2x ≤ 1} = {x | −4 ≤ −2x ≤ −2} = {x | 2 ≥ x ≥ 1} = [1, 2].

Domain(g0) = x | 1− (3− 2x)2 > 0 = x | (3− 2x)2 < 1 = {x | |3− 2x| < 1}= {x | −1 < 3− 2x < 1} = {x | −4 < −2x < −2} = {x | 2 > x > 1} = (1, 2)

39. g(x) = x sin−1x

4+√16− x2 ⇒ g0(x) = sin−1

x

4+

x

4 1− (x/4)2 −x√

16− x2= sin−1

x

4⇒

g0(2) = sin−1 12= π

6

41. f(x) =√1− x2 arcsinx ⇒ f 0(x) =

√1− x2 · 1√

1− x2+ arcsinx · 1

21− x2

−1/2(−2x) = 1− x arcsinx√

1− x2

Note that f 0 = 0 where the graph of f has a horizontal tangent. Also note

that f 0 is negative when f is decreasing and f 0 is positive when f is

increasing.

43. limx→−1+

sin−1 x = sin−1(−1) = −π2

45. Let t = ex. As x→∞, t→∞. limx→∞

arctan(ex) = limt→∞

arctan t = π2

by (8).

47. From the figure, tanα = 5

xand tanβ = 2

3− x. Since

α+ β + θ = 180◦ = π, θ = π − tan−1 5

x− tan−1 2

3− x⇒

dθ

dx= − 1

1 +5

x

2 − 5

x2− 1

1 +2

3− x

2

2

(3− x)2

=x2

x2 + 25· 5x2− (3− x)2

(3− x)2 + 4· 2

(3− x)2.

Now dθ

dx= 0 ⇒ 5

x2 + 25=

2

x2 − 6x+ 13 ⇒ 2x2 + 50 = 5x2 − 30x+ 65 ⇒

3x2 − 30x+ 15 = 0 ⇒ x2 − 10x+ 5 = 0 ⇒ x = 5± 2√5. We reject the root with the + sign, since it is larger

than 3. dθ/dx > 0 for x < 5− 2√5 and dθ/dx < 0 for x > 5− 2√5, so θ is maximized when

|AP | = x = 5− 2√5 ≈ 0.53.

49. dx

dt= 2 ft/s, sin θ = x

10⇒ θ = sin−1

x

10, dθdx

=1/10

1− (x/10)2 ,

dθ

dt=

dθ

dx

dx

dt=

1/10

1− (x/10)2 (2) rad/s, dθ

dt x=6

=2/10

1− (6/10)2 rad/s = 14

rad/s

SECTION 7.6 INVERSE TRIGONOMETRIC FUNCTIONS ¤ 283

51. y = f(x) = sin−1(x/(x+ 1)) A. D = {x | −1 ≤ x/(x+ 1) ≤ 1}. For x > −1 we have −x− 1 ≤ x ≤ x+ 1 ⇔2x ≥ −1 ⇔ x ≥ − 1

2, so D = − 1

2,∞ . B. Intercepts are 0 C. No symmetry

D. limx→∞

sin−1x

x+ 1= lim

x→∞sin−1

1

1 + 1/x= sin−1 1 = π

2, so y = π

2is a HA.

E. f 0(x) = 1

1− [x/(x+ 1)]2(x+ 1)− x

(x+ 1)2=

1

(x+ 1)√2x+ 1

> 0,

so f is increasing on − 12,∞ . F. No local maximum or minimum,

f − 12= sin−1 (−1) = −π

2is an absolute minimum

G. f 00(x) = −√2x+ 1 + (x+ 1)/

√2x+ 1

(x+ 1)2(2x+ 1)

= − 3x+ 2

(x+ 1)2(2x+ 1)3/2< 0 on D, so f is CD on − 1

2,∞ .

H.

53. y = f(x) = x− tan−1 x A. D = R B. Intercepts are 0 C. f(−x) = −f(x), so the curve is symmetric about the

origin. D. limx→∞

(x− tan−1 x) =∞ and limx→−∞

(x− tan−1 x) = −∞, no HA.

But f(x)− x− π2= − tan−1 x+ π

2 → 0 as x→∞, and

f(x)− x+ π2= − tan−1 x− π

2 → 0 as x→−∞, so y = x± π2 are

slant asymptotes. E. f 0(x) = 1− 1

x2 + 1=

x2

x2 + 1> 0, so f is

increasing on R. F. No extrema

G. f 00(x) = (1 + x2)(2x)− x2(2x)

(1 + x2)2=

2x

(1 + x2)2> 0 ⇔ x > 0, so

f is CU on (0,∞), CD on (−∞, 0). IP at (0, 0).

H.

55. f(x) = arctan(cos(3 arcsinx)). We use a CAS to compute f 0 and f 00, and to graph f , f 0, and f 00:

From the graph of f 0, it appears that the only maximum occurs at x = 0 and there are minima at x = ±0.87. From the graph

of f 00, it appears that there are inflection points at x = ±0.52.

57. f(x) = 2 + x2

1 + x2=1 + (1 + x2)

1 + x2=

1

1 + x2+ 1 ⇒ F (x) = tan−1 x+ x+C

59.√3/2

1/2

6√1− t2

dt = 6

√3/2

1/2

1√1− t2

dt = 6 sin−1 t√3/2

1/2= 6 sin−1

√32

− sin−1 12

= 6 π3− π

6= 6 π

6= π


61. Let u = 4x. Then du = 4 dx, so√3/4

0

dx

1 + 16x2=1

4

√3

0

1

1 + u2du = 1

4tan−1 u

√3

0= 1

4tan−1

√3− tan−1 0 = 1

4π3− 0 = π

12.

63. Let u = 1 + x2. Then du = 2xdx, so

1 + x

1 + x2dx =

1

1 + x2dx+

x

1 + x2dx = tan−1 x+

12du

u= tan−1 x+ 1

2ln|u|+C

= tan−1 x+ 12 ln 1 + x2 + C = tan−1 x+ 1

2 ln(1 + x2) +C [since 1 + x2 > 0].

65. Let u = sin−1 x. Then du = 1√1− x2

dx, so dx√1− x2 sin−1 x

=1

udu = ln |u|+C = ln sin−1 x +C.

67. Let u = t3. Then du = 3t2 dt and t2√1− t6

dt =13du√

1− u2= 1

3sin−1 u+C = 1

3sin−1(t3) + C.

69. Let u =√x. Then du = dx

2√x

and dx√x (1 + x)

=2 du

1 + u2= 2 tan−1 u+ C = 2 tan−1

√x+ C.

71. Let u = x/a. Then du = dx/a, so dx√a2 − x2

=dx

a 1− (x/a)2 =du√1− u2

= sin−1 u+C = sin−1x

a+C.

73. The integral represents the area below the curve y = sin−1 x on the interval

x ∈ [0, 1]. The bounding curves are y = sin−1 x ⇔ x = sin y, y = 0 and

x = 1. We see that y ranges between sin−1 0 = 0 and sin−1 1 = π2

. So we have to

integrate the function x = 1− sin y between y = 0 and y = π2 :

1

0sin−1 xdx = π/2

0(1− sin y) dy = π

2 + cosπ2− (0 + cos 0) = π

2 − 1.

75. (a) arctan 12+ arctan 1

3= arctan

12+ 1

3

1− 12· 13

= arctan 1 =π

4

(b) 2 arctan 13+ arctan 1

7= arctan 1

3+ arctan 1

3+ arctan 1

7= arctan

13+ 1

3

1− 13· 13

+ arctan 17

= arctan 34 + arctan

17 = arctan

34+ 1

7

1− 34 · 17

= arctan 1 =π

4

77. Let f(x) = 2 sin−1 x− cos−1(1− 2x2). Then f 0(x) = 2√1− x2

− 4x

1− (1− 2x2)2=

2√1− x2

− 4x

2x√1− x2

= 0

[since x ≥ 0]. Thus f 0(x) = 0 for all x ∈ [0, 1). Thus f(x) = C. To find C let x = 0. Thus

2 sin−1(0)− cos−1(1) = 0 = C. Therefore we see that f(x) = 2 sin−1 x− cos−1(1− 2x2) = 0 ⇒

2 sin−1 x = cos−1(1− 2x2).

SECTION 7.7 HYPERBOLIC FUNCTIONS ¤ 285

79. y = sec−1 x ⇒ sec y = x ⇒ sec y tan ydy

dx= 1 ⇒ dy

dx=

1

sec y tan y. Now tan2 y = sec2 y − 1 = x2 − 1, so

tan y = ±√x2 − 1. For y ∈ 0, π2 , x ≥ 1, so sec y = x = |x| and tan y ≥ 0 ⇒ dy

dx=

1

x√x2 − 1 =

1

|x|√x2 − 1 .

For y ∈ π2, π , x ≤ −1, so |x| = −x and tan y = −√x2 − 1 ⇒

dy

dx=

1

sec y tan y=

1

x −√x2 − 1 =1

(−x)√x2 − 1 =1

|x|√x2 − 1

7.7 Hyperbolic Functions

1. (a) sinh 0 = 12(e0 − e0) = 0 (b) cosh 0 = 1

2(e0 + e0) = 1

2(1 + 1) = 1

3. (a) sinh(ln 2) = eln 2 − e−ln 2

2=

eln 2 − (eln 2)−12

=2− 2−12

=2− 1

2

2=3

4

(b) sinh 2 = 12(e2 − e−2) ≈ 3.62686

5. (a) sech 0 = 1

cosh 0=1

1= 1 (b) cosh−1 1 = 0 because cosh 0 = 1.

7. sinh(−x) = 12[e−x − e−(−x)] = 1

2(e−x − ex) = − 1

2(e−x − ex) = − sinhx

9. coshx+ sinhx = 12(ex + e−x) + 1

2(ex − e−x) = 1

2(2ex) = ex

11. sinhx cosh y + coshx sinh y = 12(ex − e−x) 1

2(ey + e−y) + 1

2(ex + e−x) 1

2(ey − e−y)

= 14[(ex+y + ex−y − e−x+y − e−x−y) + (ex+y − ex−y + e−x+y − e−x−y)]

= 14 (2e

x+y − 2e−x−y) = 12 [e

x+y − e−(x+y)] = sinh(x+ y)

13. Divide both sides of the identity cosh2 x− sinh2 x = 1 by sinh2 x:

cosh2 x

sinh2 x− sinh2 x

sinh2 x=

1

sinh2 x⇔ coth2 x− 1 = csch2 x.

15. Putting y = x in the result from Exercise 11, we have

sinh 2x = sinh(x+ x) = sinhx coshx+ coshx sinhx = 2 sinhx coshx.

17. tanh(lnx) = sinh(lnx)

cosh(lnx)=(eln x − e− lnx)/2

(eln x + e− lnx)/2=

x− (eln x)−1x+ (eln x)−1

=x− x−1

x+ x−1=

x− 1/xx+ 1/x

=(x2 − 1)/x(x2 + 1)/x

=x2 − 1x2 + 1

19. By Exercise 9, (coshx+ sinhx)n = (ex)n = enx = coshnx+ sinhnx.

21. sechx = 1

coshx⇒ sechx =

1

5/3=3

5.

cosh2 x− sinh2 x = 1 ⇒ sinh2 x = cosh2 x− 1 = 53

2 − 1 = 169 ⇒ sinhx = 4

3 [because x > 0].

cschx =1

sinhx⇒ cschx =

1

4/3=3

4.

tanhx =sinhx

coshx⇒ tanhx =

4/3

5/3=4

5.

cothx =1

tanhx⇒ cothx =

1

4/5=5

4.


23. (a) limx→∞

tanhx = limx→∞

ex − e−x

ex + e−x· e−x

e−x= lim

x→∞1− e−2x

1 + e−2x=1− 01 + 0

= 1

(b) limx→−∞

tanhx = limx→−∞

ex − e−x

ex + e−x· e

x

ex= lim

x→−∞e2x − 1e2x + 1

=0− 10 + 1

= −1

(c) limx→∞

sinhx = limx→∞

ex − e−x

2=∞

(d) limx→−∞

sinhx = limx→−∞

ex − e−x

2= −∞

(e) limx→∞

sechx = limx→∞

2

ex + e−x= 0

(f ) limx→∞

cothx = limx→∞

ex + e−x

ex − e−x· e−x

e−x= lim

x→∞1 + e−2x

1− e−2x=1 + 0

1− 0 = 1 [Or: Use part (a)]

(g) limx→0+

cothx = limx→0+

coshx

sinhx=∞, since sinhx→ 0 through positive values and coshx→ 1.

(h) limx→0−

cothx = limx→0−

coshx

sinhx= −∞, since sinhx→ 0 through negative values and coshx→ 1.

(i) limx→−∞

cschx = limx→−∞

2

ex − e−x= 0

25. Let y = sinh−1 x. Then sinh y = x and, by Example 1(a), cosh2 y − sinh2 y = 1 ⇒ [with cosh y > 0]

cosh y = 1 + sinh2 y =√1 + x2. So by Exercise 9, ey = sinh y + cosh y = x+

√1 + x2 ⇒ y = ln x+

√1 + x2 .

27. (a) Let y = tanh−1 x. Then x = tanh y = sinh y

cosh y=(ey − e−y)/2(ey + e−y)/2

· ey

ey=

e2y − 1e2y + 1

⇒ xe2y + x = e2y − 1 ⇒

1+ x = e2y − xe2y ⇒ 1+ x = e2y(1− x) ⇒ e2y =1 + x

1− x⇒ 2y = ln

1 + x

1− x⇒ y = 1

2 ln1 + x

1− x.

(b) Let y = tanh−1 x. Then x = tanh y, so from Exercise 18 we have

e2y =1 + tanh y

1− tanh y =1 + x

1− x⇒ 2y = ln

1 + x

1− x⇒ y = 1

2 ln1 + x

1− x.

29. (a) Let y = cosh−1 x. Then cosh y = x and y ≥ 0 ⇒ sinh ydy

dx= 1 ⇒

dy

dx=

1

sinh y=

1

cosh2 y − 1=

1√x2 − 1 [since sinh y ≥ 0 for y ≥ 0]. Or: Use Formula 4.

(b) Let y = tanh−1 x. Then tanh y = x ⇒ sech2 ydy

dx= 1 ⇒ dy

dx=

1

sech2y=

1

1− tanh2 y =1

1− x2.

Or: Use Formula 5.

(c) Let y = csch−1 x. Then csch y = x ⇒ − csch y coth y dy

dx= 1 ⇒ dy

dx= − 1

csch y coth y. By Exercise 13,

coth y = ± csch2 y + 1 = ±√x2 + 1. If x > 0, then coth y > 0, so coth y =√x2 + 1. If x < 0, then coth y < 0,

so coth y = −√x2 + 1. In either case we have dy

dx= − 1

csch y coth y= − 1

|x|√x2 + 1 .

SECTION 7.7 HYPERBOLIC FUNCTIONS ¤ 287

(d) Let y = sech−1 x. Then sech y = x ⇒ − sech y tanh y dy

dx= 1 ⇒

dy

dx= − 1

sech y tanh y= − 1

sech y 1− sech2y= − 1

x√1− x2

. [Note that y > 0 and so tanh y > 0.]

(e) Let y = coth−1 x. Then coth y = x ⇒ − csch2 y dy

dx= 1 ⇒ dy

dx= − 1

csch2 y=

1

1− coth2 y =1

1− x2

by Exercise 13.

31. f(x) = x sinhx− coshx ⇒ f 0(x) = x (sinhx)0 + sinhx · 1− sinhx = x coshx

33. h(x) = ln(coshx) ⇒ h0(x) =1

coshx(coshx)0 =

sinhx

coshx= tanhx

35. y = ecosh 3x ⇒ y0 = ecosh 3x · sinh 3x · 3 = 3ecosh 3x sinh 3x

37. f(t) = sech2(et) = [sech(et)]2 ⇒f 0(t) = 2[sech(et)] [sech(et)]0 = 2 sech(et) − sech(et) tanh(et) · et = −2et sech2(et) tanh(et)

39. y = arctan(tanhx) ⇒ y0 =1

1 + (tanhx)2(tanhx)0 =

sech2 x

1 + tanh2 x

41. G(x) = 1− coshx1 + coshx

⇒

G0(x) =(1 + coshx)(− sinhx)− (1− coshx)(sinhx)

(1 + coshx)2=− sinhx− sinhx coshx− sinhx+ sinhx coshx

(1 + coshx)2

=−2 sinhx

(1 + coshx)2

43. y = tanh−1√x ⇒ y0 =

1

1−√x

2 ·1

2x−1/2 =

1

2√x (1− x)

45. y = x sinh−1(x/3) − √9 + x2 ⇒

y0 = sinh−1x

3+ x

1/3

1 + (x/3)2− 2x

2√9 + x2

= sinh−1x

3+

x√9 + x2

− x√9 + x2

= sinh−1x

3

47. y = coth−1√x2 + 1 ⇒ y0 =

1

1− (x2 + 1)2x

2√x2 + 1

= − 1

x√x2 + 1

49. As the depth d of the water gets large, the fraction 2πdL

gets large, and from Figure 3 or Exercise 23(a), tanh 2πd

L

approaches 1. Thus, v = gL

2πtanh

2πd

L≈ gL

2π(1) =

gL

2π.

51. (a) y = 20 cosh(x/20)− 15 ⇒ y0 = 20 sinh(x/20) · 120= sinh(x/20). Since the right pole is positioned at x = 7,

we have y0(7) = sinh 720≈ 0.3572.

(b) If α is the angle between the tangent line and the x-axis, then tanα = slope of the line = sinh 720 , so

α = tan−1 sinh 720

≈ 0.343 rad ≈ 19.66◦. Thus, the angle between the line and the pole is θ = 90◦ − α ≈ 70.34◦.


53. (a) y = A sinhmx + B coshmx ⇒ y0 = mA coshmx +mB sinhmx ⇒y00 = m2A sinhmx+m2B coshmx = m2(A sinhmx+B coshmx) = m2y

(b) From part (a), a solution of y00 = 9y is y(x) = A sinh 3x+B cosh 3x. So −4 = y(0) = A sinh 0 +B cosh 0 = B, so

B = −4. Now y0(x) = 3A cosh 3x− 12 sinh 3x ⇒ 6 = y0(0) = 3A ⇒ A = 2, so y = 2 sinh 3x− 4 cosh 3x.

55. The tangent to y = coshx has slope 1 when y0 = sinhx = 1 ⇒ x = sinh−1 1 = ln 1 +√2 , by Equation 3.

Since sinhx = 1 and y = coshx = 1 + sinh2 x, we have coshx =√2. The point is ln 1 +

√2 ,

√2 .

57. Let u = coshx. Then du = sinhxdx, so sinhx cosh2 xdx = u2 du = 13u

3 + C = 13 cosh

3 x+C.


2√x

and sinh√x√

xdx = sinhu · 2 du = 2 coshu+C = 2 cosh

√x+C.

61. coshx

cosh2 x− 1 dx =coshx

sinh2 xdx =

coshx

sinhx· 1

sinhxdx = cothx cschxdx = − cschx+ C

63. Let t = 3u. Then dt = 3 du and6

4

1√t2 − 9 dt =

2

4/3

1√9u2 − 9 3 du =

2

4/3

du√u2 − 1 = cosh−1 u

2

4/3= cosh−1 2− cosh−1 4

3or

= cosh−1 u2

4/3= ln u+

√u2 − 1

2

4/3= ln 2 +

√3 − ln 4 +

√7

3= ln

6 + 3√3

4 +√7

65. Let u = ex. Then du = ex dx and ex

1− e2xdx =

du

1− u2= tanh−1 u+ C = tanh−1(ex) + C

or 12ln

1 + ex

1− ex+ C .

67. (a) From the graphs, we estimate

that the two curves y = cosh 2x

and y = 1 + sinhx intersect at

x = 0 and at x = a ≈ 0.481.

(b) We have found the two roots of the equation cosh 2x = 1 + sinhx to be x = 0 and x = a ≈ 0.481. Note from the first

graph that 1 + sinhx > cosh 2x on the interval (0, a), so the area between the two curves is

A=a

0(1 + sinhx− cosh 2x) dx = x+ coshx− 1

2sinh 2x

a

0

= [a+ cosh a− 12sinh 2a]− [0 + cosh 0− 1

2sinh 0] ≈ 0.0402

SECTION 7.8 INDETERMINATE FORMS AND L’HOSPITAL’S RULE ¤ 289

69. If aex + be−x = α cosh(x+ β) [or α sinh(x+ β)], then

aex + be−x = α2ex+β ± e−x−β = α

2exeβ ± e−xe−β = α

2eβ ex ± α

2e−β e−x. Comparing coefficients of ex

and e−x, we have a = α2eβ (1) and b = ±α

2e−β (2). We need to find α and β. Dividing equation (1) by equation (2)

gives us ab= ±e2β ⇒ ( ) 2β = ln ±a

b⇒ β = 1

2ln ±a

b. Solving equations (1) and (2) for eβ gives us

eβ =2a

αand eβ = ± α

2b, so 2a

α= ± α

2b⇒ α2 = ±4ab ⇒ α = 2

√±ab.

( ) If ab> 0, we use the + sign and obtain a cosh function, whereas if a

b< 0, we use the − sign and obtain a sinh

function.

In summary, if a and b have the same sign, we have aex + be−x = 2√ab cosh x+ 1

2 lnab

, whereas, if a and b have the

opposite sign, then aex + be−x = 2√−ab sinh x+ 1

2ln −a

b.

7.8 Indeterminate Forms and L'Hospital's Rule

Note: The use of l’Hospital’s Rule is indicated by an H above the equal sign: H=

1. (a) limx→a

f(x)

g(x)is an indeterminate form of type 0

0.

(b) limx→a

f(x)

p(x)= 0 because the numerator approaches 0 while the denominator becomes large.

(c) limx→a

h(x)

p(x)= 0 because the numerator approaches a finite number while the denominator becomes large.

(d) If limx→a

p(x) =∞ and f(x)→ 0 through positive values, then limx→a

p(x)

f(x)=∞. [For example, take a = 0, p(x) = 1/x2,

and f(x) = x2.] If f(x)→ 0 through negative values, then limx→a

p(x)

f(x)= −∞. [For example, take a = 0, p(x) = 1/x2,

and f(x) = −x2.] If f(x)→ 0 through both positive and negative values, then the limit might not exist. [For example,

take a = 0, p(x) = 1/x2, and f(x) = x.]

(e) limx→a

p(x)

q(x)is an indeterminate form of type ∞∞ .

3. (a) When x is near a, f(x) is near 0 and p(x) is large, so f(x)− p(x) is large negative. Thus, limx→a

[f(x)− p(x)] = −∞.

(b) limx→a

[ p(x)− q(x)] is an indeterminate form of type∞−∞.

(c) When x is near a, p(x) and q(x) are both large, so p(x) + q(x) is large. Thus, limx→a

[p(x) + q(x)] =∞.

5. This limit has the form 00

. We can simply factor and simplify to evaluate the limit.

limx→1

x2 − 1x2 − x

= limx→1

(x+ 1)(x− 1)x(x− 1) = lim

x→1

x+ 1

x=1 + 1

1= 2


. limx→1

x9 − 1x5 − 1

H= lim

x→1

9x8

5x4=9

5limx→1

x4 =9

5(1) =

9

5


9. This limit has the form 00 . lim

x→(π/2)+

cosx

1− sinxH= lim

x→(π/2)+

− sinx− cosx = lim

x→(π/2)+tanx = −∞.


. limt→0

et − 1t3

H= lim

t→0

et

3t2=∞ since et → 1 and 3t2 → 0+ as t→ 0.


x→0

tan px

tan qxH= lim

x→0

p sec2 px

q sec2 qx=

p(1)2

q(1)2=

p

q

15. This limit has the form ∞∞ . lim

x→∞lnx√x

H= lim

x→∞1/x

12x−1/2 = lim

x→∞2√x= 0

17. limx→0+

[(lnx)/x] = −∞ since lnx→ −∞ as x→ 0+ and dividing by small values of x just increases the magnitude of the

quotient (lnx)/x. L’Hospital’s Rule does not apply.

19. This limit has the form ∞∞ . lim

x→∞ex

x3H= lim

x→∞ex

3x2H= lim

x→∞ex

6xH= lim

x→∞ex

6=∞


. limx→0

ex − 1− x

x2H= lim

x→0

ex − 12x

H= lim

x→0

ex

2=1

2


. limx→0

tanhx

tanxH= lim

x→0

sech 2x

sec2 x=sech2 0

sec2 0=1

1= 1


t→0

5t − 3tt

H= lim

t→0

5t ln 5− 3t ln 31

= ln 5− ln 3 = ln 53


. limx→0

sin−1 xx

H= lim

x→0

1/√1− x2

1= lim

x→0

1√1− x2

=1

1= 1


x→0

1− cosxx2

H= lim

x→0

sinx

2xH= lim

x→0

cosx

2=1

2

31. limx→0

x+ sinx

x+ cosx=0 + 0

0 + 1=0

1= 0. L’Hospital’s Rule does not apply.


x→1

1− x+ lnx

1 + cosπxH= lim

x→1

−1 + 1/x−π sinπx

H= lim

x→1

−1/x2−π2 cosπx =

−1−π2 (−1) = −

1

π2


. limx→1

xa − ax+ a− 1(x− 1)2

H= lim

x→1

axa−1 − a

2(x− 1)H= lim

x→1

a(a− 1)xa−22

=a(a− 1)

2


. limx→0

cosx− 1 + 12x

2

x4H= lim

x→0

− sinx+ x

4x3H= lim

x→0

− cosx+ 112x2

H= lim

x→0

sinx

24xH= lim

x→0

cosx

24=1

24

39. This limit has the form ∞ · 0.

limx→∞

x sin(π/x) = limx→∞

sin(π/x)

1/xH= lim

x→∞cos(π/x)(−π/x2)

−1/x2 = π limx→∞

cos(π/x) = π(1) = π

41. This limit has the form ∞ · 0. We’ll change it to the form 00

.

limx→0

cot 2x sin 6x = limx→0

sin 6x

tan 2xH= lim

x→0

6 cos 6x

2 sec2 2x=6(1)

2(1)2= 3

43. This limit has the form∞ · 0. limx→∞

x3e−x2= lim

x→∞x3

ex2H= lim

x→∞3x2

2xex2= lim

x→∞3x

2ex2H= lim

x→∞3

4xex2= 0


45. This limit has the form 0 · (−∞).

limx→1+

lnx tan(πx/2) = limx→1+

lnx

cot(πx/2)H= lim

x→1+

1/x

(−π/2) csc2(πx/2) =1

(−π/2)(1)2 = −2

π

47. This limit has the form∞−∞.

limx→1

x

x− 1 −1

lnx= lim

x→1

x lnx− (x− 1)(x− 1) lnx

H= lim

x→1

x(1/x) + lnx− 1(x− 1)(1/x) + lnx = lim

x→1

lnx

1− (1/x) + lnxH= lim

x→1

1/x

1/x2 + 1/x· x

2

x2= lim

x→1

x

1 + x=

1

1 + 1=1

2

49. We will multiply and divide by the conjugate of the expression to change the form of the expression.

limx→∞

√x2 + x− x = lim

x→∞

√x2 + x− x

1·√x2 + x+ x√x2 + x+ x

= limx→∞

x2 + x − x2√x2 + x+ x

= limx→∞

x√x2 + x+ x

= limx→∞

1

1 + 1/x+ 1=

1√1 + 1

=1

2

As an alternate solution, write√x2 + x− x as

√x2 + x−√x2, factor out

√x2, rewrite as ( 1 + 1/x− 1)/(1/x), and

apply l’Hospital’s Rule.

51. The limit has the form∞−∞ and we will change the form to a product by factoring out x.

limx→∞

(x− lnx) = limx→∞

x 1− lnx

x=∞ since lim

x→∞lnx

xH= lim

x→∞1/x

1= 0.

53. y = xx2 ⇒ ln y = x2 lnx, so lim

x→0+ln y = lim

x→0+x2 lnx = lim

x→0+

lnx

1/x2H= lim

x→0+

1/x

−2/x3 = limx→0+

−12x2 = 0 ⇒

limx→0+

xx2= lim

x→0+eln y = e0 = 1.

55. y = (1− 2x)1/x ⇒ ln y =1

xln(1− 2x), so lim

x→0ln y = lim

x→0

ln(1− 2x)x

H= lim

x→0

−2/(1− 2x)1

= −2 ⇒

limx→0

(1− 2x)1/x = limx→0

eln y = e−2.

57. y = 1 +3

x+5

x2

x

⇒ ln y = x ln 1 +3

x+5

x2⇒

limx→∞

ln y = limx→∞

ln 1 +3

x+5

x2

1/xH= lim

x→∞

− 3

x2− 10

x31 +

3

x+5

x2

−1/x2 = limx→∞

3 +10

x

1 +3

x+5

x2

= 3,

so limx→∞

1 +3

x+5

x2

x

= limx→∞

eln y = e3.

59. y = x1/x ⇒ ln y = (1/x) lnx ⇒ limx→∞

ln y = limx→∞

lnx

xH= lim

x→∞1/x

1= 0 ⇒

limx→∞

x1/x = limx→∞

eln y = e0 = 1

61. y = (4x+ 1)cot x ⇒ ln y = cotx ln(4x+ 1), so limx→0+

ln y = limx→0+

ln(4x+ 1)

tanxH= lim

x→0+

4

4x+ 1sec2 x

= 4 ⇒

limx→0+

(4x+ 1)cotx = limx→0+

eln y = e4.


63. y = (cosx)1/x2 ⇒ ln y =

1

x2ln cosx ⇒

limx→0+

ln y = limx→0+

ln cosx

x2H= lim

x→0+

− tanx2x

H= lim

x→0+

− sec2 x2

= −12

⇒

limx→0+

(cosx)1/x2

= limx→0+

eln y = e−1/2 = 1/√e

65. From the graph, if x = 500, y ≈ 7.36. The limit has the form 1∞.

Now y = 1 +2

x

x

⇒ ln y = x ln 1 +2

x⇒

limx→∞

ln y = limx→∞

ln(1 + 2/x)

1/xH= lim

x→∞

1

1 + 2/x− 2

x2

−1/x2

= 2 limx→∞

1

1 + 2/x= 2(1) = 2 ⇒

limx→∞

1 +2

x

x

= limx→∞

eln y = e2 [≈ 7.39]

67. From the graph, it appears that limx→0

f(x)

g(x)= lim

x→0

f 0(x)g0(x)

= 0.25.

We calculate limx→0

f(x)

g(x)= lim

x→0

ex − 1x3 + 4x

H= lim

x→0

ex

3x2 + 4=1

4.

69. y = f(x) = xe−x A. D = R B. Intercepts are 0 C. No symmetry

D. limx→∞

xe−x = limx→∞

x

exH= lim

x→∞1

ex= 0, so y = 0 is a HA. lim

x→−∞xe−x = −∞

E. f 0(x) = e−x − xe−x = e−x(1− x) > 0 ⇔ x < 1, so f is increasing

on (−∞, 1) and decreasing on (1,∞) . F. Absolute and local maximum

value f(1) = 1/e. G. f 00(x) = e−x (x− 2) > 0 ⇔ x > 2, so f is CU

on (2,∞) and CD on (−∞, 2). IP at 2, 2/e2

H.

71. y = f(x) = xe−x2

A. D = R B. Intercepts are 0 C. f(−x) = −f(x), so the curve is symmetric

about the origin. D. limx→±∞

xe−x2= lim

x→±∞x

ex2H= lim

x→±∞1

2xex2= 0, so y = 0 is a HA.

E. f 0(x) = e−x2 − 2x2e−x2 = e−x

2(1− 2x2) > 0 ⇔ x2 < 1

2⇔ |x| < 1√

2, so f is increasing on − 1√

2, 1√

2

and decreasing on −∞,− 1√2

and 1√2,∞ . F. Local maximum value f 1√

2= 1/

√2e, local minimum

value f − 1√2= −1/√2e G. f 00(x) = −2xe−x2(1− 2x2)− 4xe−x2 = 2xe−x2(2x2 − 3) > 0 ⇔


x > 32 or − 3

2 < x < 0, so f is CU on 32 ,∞ and

− 32 , 0 and CD on −∞,− 3

2and 0, 3

2.

IP are (0, 0) and ± 32,± 3

2e−3/2 .

H.

73. y = f(x) = x− ln(1 + x) A. D = {x | x > −1} = (−1,∞) B. Intercepts are 0 C. No symmetry

D. limx→−1+

[x− ln(1 + x)] =∞, so x = −1 is a VA. limx→∞

[x− ln(1 + x)] = limx→∞

x 1− ln(1 + x)

x=∞,

since limx→∞

ln(1 + x)

xH= lim

x→∞1/(1 + x)

1= 0.

E. f 0(x) = 1− 1

1 + x=

x

1 + x> 0 ⇔ x > 0 since x+ 1 > 0.

So f is increasing on (0,∞) and decreasing on (−1, 0).F. f(0) = 0 is an absolute minimum.

G. f 00(x) = 1/(1 + x)2 > 0, so f is CU on (−1,∞).

H.

75. (a) f(x) = x−x

(b) y = f(x) = x−x. We note that ln f(x) = lnx−x = −x lnx = − lnx1/x

, so

limx→0+

ln f(x)H= lim

x→0+− 1/x

−x−2 = limx→0+

x = 0. Thus limx→0+

f(x) = limx→0+

eln f(x) = e0 = 1.

(c) From the graph, it appears that there is a local and absolute maximum of about f(0.37) ≈ 1.44. To find the exact value, we

differentiate: f(x) = x−x = e−x lnx ⇒ f 0(x) = e−x lnx −x 1

x+ lnx(−1) = −x−x(1 + lnx). This is 0 only

when 1 + lnx = 0 ⇔ x = e−1. Also f 0(x) changes from positive to negative at e−1. So the maximum value is

f(1/e) = (1/e)−1/e = e1/e.

(d) We differentiate again to get

f 00(x) = −x−x(1/x) + (1 + lnx)2(x−x) = x−x[(1 + lnx)2 − 1/x]

From the graph of f 00(x), it seems that f 00(x) changes from negative to

positive at x = 1, so we estimate that f has an IP at x = 1.


77. (a) f(x) = x1/x

(b) Recall that ab = eb ln a. limx→0+

x1/x = limx→0+

e(1/x) ln x. As x→ 0+, lnxx→ −∞, so x1/x = e(1/x) lnx → 0. This

indicates that there is a hole at (0, 0). As x→∞, we have the indeterminate form∞0. limx→∞

x1/x = limx→∞

e(1/x) ln x,

but limx→∞

lnx

xH= lim

x→∞1/x

1= 0, so lim

x→∞x1/x = e0 = 1. This indicates that y = 1 is a HA.

(c) Estimated maximum: (2.72, 1.45). No estimated minimum. We use logarithmic differentiation to find any critical

numbers. y = x1/x ⇒ ln y =1

xlnx ⇒ y0

y=1

x· 1x+ (lnx) − 1

x2⇒ y0 = x1/x

1− lnxx2

= 0 ⇒

lnx = 1 ⇒ x = e. For 0 < x < e, y0 > 0 and for x > e, y0 < 0, so f(e) = e1/e is a local maximum value. This

point is approximately (2.7183, 1.4447), which agrees with our estimate.

(d) From the graph, we see that f 00(x) = 0 at x ≈ 0.58 and x ≈ 4.37. Since f 00

changes sign at these values, they are x-coordinates of inflection points.

79. If c < 0, then limx→−∞

f(x) = limx→−∞

xe−cx = limx→−∞

x

ecxH= lim

x→−∞1

cecx= 0, and lim

x→∞f(x) =∞.

If c > 0, then limx→−∞

f(x) = −∞, and limx→∞

f(x)H= lim

x→∞1

cecx= 0.

If c = 0, then f(x) = x, so limx→±∞

f(x) = ±∞, respectively.

So we see that c = 0 is a transitional value. We now exclude the case c = 0, since we know how the function behaves

in that case. To find the maxima and minima of f , we differentiate: f(x) = xe−cx ⇒f 0(x) = x(−ce−cx) + e−cx = (1− cx)e−cx. This is 0 when 1− cx = 0 ⇔ x = 1/c. If c < 0 then this

represents a minimum value of f(1/c) = 1/(ce), since f 0(x) changes from negative to positive at x = 1/c;

and if c > 0, it represents a maximum value. As |c| increases, the maximum or

minimum point gets closer to the origin. To find the inflection points, we

differentiate again: f 0(x) = e−cx(1− cx) ⇒f 00(x) = e−cx(−c) + (1− cx)(−ce−cx) = (cx− 2)ce−cx. This changes sign

when cx− 2 = 0 ⇔ x = 2/c. So as |c| increases, the points of inflection get

closer to the origin.


81. limx→∞

x√x2 + 1

H= lim

x→∞1

12(x2 + 1)−1/2(2x)

= limx→∞

√x2 + 1

x. Repeated applications of l’Hospital’s Rule result in the

original limit or the limit of the reciprocal of the function. Another method is to try dividing the numerator and denominator

by x: limx→∞

x√x2 + 1

= limx→∞

x/x

x2/x2 + 1/x2= lim

x→∞1

1 + 1/x2=1

1= 1

83. limE→0+

P (E) = limE→0+

eE + e−E

eE − e−E− 1

E

= limE→0+

E eE + e−E − 1 eE − e−E

(eE − e−E)E= lim

E→0+

EeE +Ee−E − eE + e−E

EeE −Ee−Eform is 0

0

H= lim

E→0+

EeE + eE · 1 +E −e−E + e−E · 1− eE + −e−EEeE + eE · 1− [E(−e−E) + e−E · 1]

= limE→0+

EeE −Ee−E

EeE + eE +Ee−E − e−E= lim

E→0+

eE − e−E

eE +eE

E+ e−E − e−E

E

[divide by E]

=0

2 + L, where L = lim

E→0+

eE − e−E

Eform is 0

0

H= lim

E→0+

eE + e−E

1=1 + 1

1= 2

Thus, limE→0+

P (E) =0

2 + 2= 0.

85. First we will find limn→∞

1 +r

n

nt

, which is of the form 1∞. y = 1 +r

n

nt

⇒ ln y = nt ln 1 +r

n, so

limn→∞

ln y = limn→∞

nt ln 1 +r

n= t lim

n→∞ln(1 + r/n)

1/nH= t lim

n→∞−r/n2

(1 + r/n)(−1/n2) = t limn→∞

r

1 + i/n= tr ⇒

limn→∞

y = ert. Thus, as n→∞, A = A0 1 +r

n

nt

→ A0ert.

87. Both numerator and denominator approach 0 as x→ 0, so we use l’Hospital’s Rule (and FTC1):

limx→0

S(x)

x3= lim

x→0

x

0sin(πt2/2)dt

x3H= lim

x→0

sin(πx2/2)

3x2H= lim

x→0

πx cos(πx2/2)

6x=

π

6· cos 0 = π

6

89. Since f(2) = 0, the given limit has the form 00 .

limx→0

f(2 + 3x) + f(2 + 5x)

xH= lim

x→0

f 0(2 + 3x) · 3 + f 0(2 + 5x) · 51

= f 0(2) · 3 + f 0(2) · 5 = 8f 0(2) = 8 · 7 = 56

91. Since limh→0

[f(x+ h)− f(x− h)] = f(x)− f(x) = 0 (f is differentiable and hence continuous) and limh→0

2h = 0, we use

l’Hospital’s Rule:

limh→0

f(x+ h)− f(x− h)

2hH= lim

h→0

f 0(x+ h)(1)− f 0(x− h)(−1)2

=f 0(x) + f 0(x)

2=2f 0(x)2

= f 0(x)

f(x+ h)− f(x− h)

2his the slope of the secant line between

(x− h, f(x− h)) and (x+ h, f(x+ h)). As h→ 0, this line gets closer

to the tangent line and its slope approaches f 0(x).


93. limx→∞

ex

xnH= lim

x→∞ex

nxn−1H= lim

x→∞ex

n(n− 1)xn−2H= · · · H

= limx→∞

ex

n!=∞

95. limx→0+

xα lnx = limx→0+

lnx

x−αH= lim

x→0+

1/x

−αx−α− 1 = limx→0+

xα

−α = 0 since α > 0.

97. Let the radius of the circle be r. We see that A(θ) is the area of the whole figure (a sector of the circle with radius 1), minus

the area of4OPR. But the area of the sector of the circle is 12r2θ (see Reference Page 1), and the area of the triangle is

12r |PQ| = 1

2r(r sin θ) = 1

2r2 sin θ. So we have A(θ) = 1

2r2θ − 1

2r2 sin θ = 1

2r2(θ − sin θ). Now by elementary

trigonometry, B(θ) = 12 |QR| |PQ| = 1

2 (r − |OQ|) |PQ| = 12 (r − r cos θ)(r sin θ) = 1

2r2(1− cos θ) sin θ.

So the limit we want is

limθ→0+

A(θ)

B(θ)= lim

θ→0+

12r

2(θ − sin θ)12r2(1− cos θ) sin θ

H= lim

θ→0+

1− cos θ(1− cos θ) cos θ + sin θ (sin θ)

= limθ→0+

1− cos θcos θ − cos2 θ + sin2 θ

H= lim

θ→0+

sin θ

− sin θ − 2 cos θ (− sin θ) + 2 sin θ (cos θ)

= limθ→0+

sin θ

− sin θ + 4 sin θ cos θ = limθ→0+

1

−1 + 4 cos θ =1

−1 + 4 cos 0 =1

3

99. (a) We show that limx→0

f(x)

xn= 0 for every integer n ≥ 0. Let y = 1

x2. Then

limx→0

f(x)

x2n= lim

x→0

e−1/x2

(x2)n= lim

y→∞yn

eyH= lim

y→∞nyn−1

eyH= · · · H

= limy→∞

n!

ey= 0 ⇒

limx→0

f(x)

xn= lim

x→0xn

f(x)

x2n= lim

x→0xn lim

x→0

f(x)

x2n= 0. Thus, f 0(0) = lim

x→0

f(x)− f(0)

x− 0 = limx→0

f(x)

x= 0.

(b) Using the Chain Rule and the Quotient Rule we see that f (n)(x) exists for x 6= 0. In fact, we prove by induction that for

each n ≥ 0, there is a polynomial pn and a non-negative integer kn with f (n)(x) = pn(x)f(x)/xkn for x 6= 0. This is

true for n = 0; suppose it is true for the nth derivative. Then f 0(x) = f(x)(2/x3), so

f (n+1)(x) = xkn [p0n(x) f(x) + pn(x) f0(x)]− knx

kn−1pn(x) f(x) x−2kn

= xknp0n(x) + pn(x)(2/x3)− knx

kn−1pn(x) f(x)x−2kn

= xkn+3p0n(x) + 2pn(x)− knxkn+2 pn(x) f(x)x

−(2kn+3)

which has the desired form.

Now we show by induction that f (n) (0) = 0 for all n. By part (a), f 0(0) = 0. Suppose that f (n)(0) = 0. Then

f (n+1)(0) = limx→0

f (n)(x)− f (n)(0)

x− 0 = limx→0

f (n)(x)

x= lim

x→0

pn(x) f(x)/xkn

x= lim

x→0

pn(x) f(x)

xkn+1

= limx→0

pn(x) limx→0

f(x)

xkn+1= pn(0) · 0 = 0

CHAPTER 7 REVIEW ¤ 297

7 Review

1. (a) A function f is called a one-to-one function if it never takes on the same value twice; that is, if f(x1) 6= f(x2) whenever

x1 6= x2. (Or, f is 1-1 if each output corresponds to only one input.)

Use the Horizontal Line Test: A function is one-to-one if and only if no horizontal line intersects its graph more thanonce.

(b) If f is a one-to-one function with domain A and range B, then its inverse function f−1 has domain B and range A and is

defined by

f−1(y) = x ⇔ f(x) = y

for any y in B. The graph of f−1 is obtained by reflecting the graph of f about the line y = x.

(c) (f−1)0(a) = 1

f 0(f−1(a))

2. (a) The function f(x) = ex has domain R and range (0,∞).

(b) The function f(x) = lnx has domain (0,∞) and range R.

(c) The graphs are reflections of one another about the line y = x. See Figure 7.3.3 or Figure 7.3*.1.

(d) loga x =lnx

ln a

3. (a) The inverse sine function f(x) = sin−1 x is defined as follows:

sin−1 x = y ⇔ sin y = x and −π

2≤ y ≤ π

2

Its domain is −1 ≤ x ≤ 1 and its range is −π

2≤ y ≤ π

2.

(b) The inverse cosine function f(x) = cos−1 x is defined as follows:

cos−1 x = y ⇔ cos y = x and 0 ≤ y ≤ π

Its domain is −1 ≤ x ≤ 1 and its range is 0 ≤ y ≤ π.

(c) The inverse tangent function f(x) = tan−1 x is defined as follows:

tan−1 x = y ⇔ tan y = x and −π

2< y <

π

2

Its domain is R and its range is −π

2< y <

π

2.

4. sinhx = ex − e−x

2, coshx = ex + e−x

2, tanhx = sinhx

coshx=

ex − e−x

ex + e−x

5. (a) y = ex ⇒ y0 = ex (b) y = ax ⇒ y0 = ax lna

(c) y = lnx ⇒ y0 = 1/x (d) y = loga x ⇒ y0 = 1/(x ln a)

(e) y = sin−1 x ⇒ y0 = 1/√1− x2 (f ) y = cos−1 x ⇒ y0 = −1/√1− x2

(g) y = tan−1 x ⇒ y0 = 1/(1 + x2) (h) y = sinhx ⇒ y0 = coshx


(i) y = coshx ⇒ y0 = sinhx ( j) y = tanhx ⇒ y0 = sech2 x

(k) y = sinh−1 x ⇒ y0 = 1/√1 + x2 (l) y = cosh−1 x ⇒ y0 = 1/

√x2 − 1

(m) y = tanh−1 x ⇒ y0 = 1/(1− x2)

6. (a) See Definition 7.2.7 (or 7.2*.5).

(b) e = limx→0

(1 + x)1/x

(c) The differentiation formula for y = ax [y0 = ax ln a] is simplest when a = e because ln e = 1.

(d) The differentiation formula for y = loga x [y0 = 1/(x ln a)] is simplest when a = e because ln e = 1.

7. (a) dy

dt= ky

(b) The equation in part (a) is an appropriate model for population growth, assuming that there is enough room and nutrition to

support the growth.

(c) If y(0) = y0, then the solution is y(t) = y0ekt.

8. (a) See l’Hospital’s Rule and the three notes that follow it in Section 7.8.

(b) Write fg as f

1/gor g

1/f.

(c) Convert the difference into a quotient using a common denominator, rationalizing, factoring, or some other method.

(d) Convert the power to a product by taking the natural logarithm of both sides of y = fg or by writing fg as eg ln f .

1. True. If f is one-to-one, with domain R, then f−1(f(6)) = 6 by the first cancellation equation [see (4) in Section 7.1].

3. False. For example, cos π2 = cos −π2

, so cosx is not 1-1.

5. True, since lnx is an increasing function on (0,∞).

7. True. We can divide by ex since ex 6= 0 for every x.

9. False. Let x = e. Then (lnx)6 = (ln e)6 = 16 = 1, but 6 lnx = 6 ln e = 6 · 1 = 6 6= 1 = (lnx)6.

11. False. ln 10 is a constant, so its derivative is 0.

13. False. The “−1” is not an exponent; it is an indication of an inverse function.

15. True. See Figure 2 in Section 7.7.

17. True. 16

2(1/x) dx = lnx

16

2= ln 16− ln 2 = ln 16

2= ln 8 = ln 23 = 3 ln 2


1. No. f is not 1-1 because the graph of f fails the Horizontal Line Test.

3. (a) f−1(3) = 7 since f(7) = 3. (b) (f−1)0(3) = 1

f 0(f−1(3))=

1

f 0(7)=1

8

5.

y = 5x − 1

7. Reflect the graph of y = lnx about the x-axis to obtain the graph

of y = − lnx.

y = lnx y = − lnx

9.

y = 2arctanx

11. (a) e2 ln 3 = (eln 3)2 = 32 = 9 (b) log10 25 + log10 4 = log10(25 · 4) = log10 100 = log10 102 = 2

13. lnx = 13⇔ loge x =

13⇒ x = e1/3

15. eex= 17 ⇒ ln ee

x= ln 17 ⇒ ex = ln 17 ⇒ ln ex = ln(ln 17) ⇒ x = ln ln 17

17. ln(x+ 1) + ln(x− 1) = 1 ⇒ ln [(x+ 1)(x− 1)] = 1 ⇒ ln(x2 − 1) = ln e ⇒ x2 − 1 = e ⇒x2 = e+ 1 ⇒ x =

√e+ 1 since ln(x− 1) is defined only when x > 1.

19. tan−1 x = 1 ⇒ tan tan−1 x = tan 1 ⇒ x = tan 1 (≈ 1.5574)

21. f(t) = t2 ln t ⇒ f 0(t) = t2 · 1t+ (ln t)(2t) = t+ 2t ln t or t(1 + 2 ln t)

23. h(θ) = etan 2θ ⇒ h0(θ) = etan 2θ · sec2 2θ · 2 = 2 sec2(2θ) etan 2θ

25. y = ln |sec 5x+ tan 5x| ⇒

y0 =1

sec 5x+ tan 5x(sec 5x tan 5x · 5 + sec2 5x · 5) = 5 sec 5x (tan 5x+ sec 5x)

sec 5x+ tan 5x= 5 sec 5x

27. y = ecx(c sinx− cosx) ⇒ y0 = cecx(c sinx− cosx) + ecx(c cosx+ sinx) = (c2 + 1)ecx sinx

29. y = ln(sec2 x) = 2 ln |secx| ⇒ y0 = (2/ secx)(secx tanx) = 2 tanx


31. y = e1/x

x2⇒ y0 =

x2(e1/x)0 − e1/x x20

(x2)2=

x2(e1/x)(−1/x2)− e1/x(2x)

x4=−e1/x(1 + 2x)

x4

33. y = 3x ln x ⇒ y0 = 3x ln x(ln 3)d

dx(x lnx) = 3x lnx(ln 3) x · 1

x+ lnx · 1 = 3x ln x(ln 3)(1 + lnx)

35. H(v) = v tan−1 v ⇒ H0(v) = v · 1

1 + v2+ tan−1 v · 1 = v

1 + v2+ tan−1 v

37. y = x sinh(x2) ⇒ y0 = x cosh(x2) · 2x+ sinh(x2) · 1 = 2x2 cosh(x2) + sinh(x2)

39. y = ln sinx− 12 sin

2 x ⇒ y0 =1

sinx· cosx− 1

2 · 2 sinx · cosx = cotx− sinx cosx

41. y = ln 1

x+

1

lnx= lnx−1 + (lnx)−1 = − lnx+ (lnx)−1 ⇒ y0 = −1 · 1

x+ (−1)(lnx)−2 · 1

x= − 1

x− 1

x (lnx)2

43. y = ln(cosh 3x) ⇒ y0 = (1/ cosh 3x)(sinh 3x)(3) = 3 tanh 3x

45. y = cosh−1(sinhx) ⇒ y0 = (coshx)/ sinh2 x− 1

47. y = cos e√tan 3x ⇒

y0 = − sin e√tan 3x · e

√tan 3x

0= − sin e

√tan 3x e

√tan 3x · 1

2(tan 3x)−1/2 · sec2(3x) · 3

=−3 sin e

√tan 3x e

√tan 3x sec2(3x)

2√tan 3x

49. f(x) = eg(x) ⇒ f 0(x) = eg(x)g0(x)

51. f(x) = ln |g(x)| ⇒ f 0(x) =1

g(x)g0(x) =

g0(x)g(x)

53. f(x) = 2x ⇒ f 0(x) = 2x ln 2 ⇒ f 00(x) = 2x(ln 2)2 ⇒ · · · ⇒ f (n)(x) = 2x(ln 2)n

55. We first show it is true for n = 1: f 0(x) = ex + xex = (x+ 1)ex. We now assume it is true for n = k:

f (k)(x) = (x+ k)ex. With this assumption, we must show it is true for n = k + 1:

f (k+1)(x) =d

dxf (k)(x) =

d

dx[(x+ k)ex] = ex + (x+ k)ex = [x+ (k + 1)] ex.

Therefore, f (n)(x) = (x+ n)ex by mathematical induction.

57. y = (2+ x)e−x ⇒ y0 = (2+ x)(−e−x) + e−x · 1 = e−x[−(2 + x) + 1] = e−x(−x− 1). At (0, 2), y0 = 1(−1) = −1,

so an equation of the tangent line is y − 2 = −1(x− 0), or y = −x+ 2.

59. y = [ln(x+ 4)]2 ⇒ y0 = 2[ln(x+ 4)]1 · 1

x+ 4· 1 = 2 ln(x+ 4)

x+ 4and y0 = 0 ⇔ ln(x+ 4) = 0 ⇔

x+ 4 = e0 ⇒ x+ 4 = 1 ⇔ x = −3, so the tangent is horizontal at the point (−3, 0).


61. (a) The line x− 4y = 1 has slope 14 . A tangent to y = ex has slope 1

4 when y0 = ex = 14 ⇒ x = ln 1

4 = − ln 4.

Since y = ex, the y-coordinate is 14

and the point of tangency is − ln 4, 14

. Thus, an equation of the tangent line

is y − 14= 1

4(x+ ln 4) or y = 1

4x+ 1

4(ln 4 + 1).

(b) The slope of the tangent at the point (a, ea) is d

dxex

x= a

= ea. Thus, an equation of the tangent line is

y − ea = ea(x− a). We substitute x = 0, y = 0 into this equation, since we want the line to pass through the origin:

0− ea = ea(0− a) ⇔ −ea = ea(−a) ⇔ a = 1. So an equation of the tangent line at the point (a, ea) = (1, e)

is y − e = e(x− 1) or y = ex.

63. limx→∞

e−3x = 0 since −3x→−∞ as x→∞ and limt→−∞

et = 0.

65. Let t = 2/(x− 3). As x→ 3−, t→ −∞. limx→3−

e2/(x−3) = limt→−∞

et = 0

67. Let t = sinhx. As x→ 0+, t→ 0+. limx→0+

ln(sinhx) = limt→0+

ln t = −∞

69. limx→∞

(1 + 2x)/2x

(1− 2x)/2x = limx→∞

1/2x + 1

1/2x − 1 =0 + 1

0− 1 = −1


x→0

tanπx

ln(1 + x)H= lim

x→0

π sec2 πx

1/(1 + x)=

π · 121/1

= π


x→0

e4x − 1− 4xx2

H= lim

x→0

4e4x − 42x

H= lim

x→0

16e4x

2= lim

x→08e4x = 8 · 1 = 8

75. This limit has the form∞ · 0. limx→∞

x3e−x = limx→∞

x3

exH= lim

x→∞3x2

exH= lim

x→∞6x

exH= lim

x→∞6

ex= 0

77. This limit has the form∞−∞.

limx→1+

x

x− 1 −1

lnx= lim

x→1+

x lnx− x+ 1

(x− 1) lnxH= lim

x→1+

x · (1/x) + lnx− 1(x− 1) · (1/x) + lnx = lim

x→1+

lnx

1− 1/x+ lnxH= lim

x→1+

1/x

1/x2 + 1/x=

1

1 + 1=1

2

79. y = f(x) = tan−1(1/x) A. D = {x | x 6= 0} B. No intercept C. f (−x) = −f(x), so the curve is symmetric

about the origin. D. limx→±∞

tan−1(1/x) = tan−1 0 = 0, so y = 0 is a HA. limx→0+

tan−1(1/x) = π2

and

limx→0−

tan−1(1/x) = −π2

since 1x→ ±∞ as x→ 0±.

E. f 0(x) = 1

1 + (1/x)2−1/x2 =

−1x2 + 1

⇒ f 0(x) < 0, so f is

decreasing on (−∞, 0) and (0,∞) . F. No maximum nor minimum

H.

G. f 00(x) = 2x

(x2 + 1)2> 0 ⇔ x > 0, so f is CU on (0,∞) and CD on (−∞, 0).


81. y = f(x) = x lnx A. D = (0,∞) B. No y-intercept; x-intercept 1. C. No symmetry D. No asymptote

[Note that the graph approaches the point (0, 0) as x→ 0+.]

E. f 0(x) = x(1/x) + (lnx)(1) = 1 + lnx, so f 0(x)→−∞ as x→ 0+ and

f 0(x)→∞ as x→∞. f 0(x) = 0 ⇔ lnx = −1 ⇔ x = e−1 = 1/e.

f 0(x) > 0 for x > 1/e, so f is decreasing on (0, 1/e) and increasing on

(1/e,∞). F. Local minimum: f(1/e) = −1/e. No local maximum.

G. f 00(x) = 1/x, so f 00(x) > 0 for x > 0. The graph is CU on (0,∞) and

there is no IP.

H.

83. y = f(x) = xe−2x A. D = R B. y-intercept: f(0) = 0; x-intercept: f(x) = 0 ⇔ x = 0

C. No symmetry D. limx→∞

xe−2x = limx→∞

x

e2xH= lim

x→∞1

2e2x= 0, so y = 0 is a HA.

E. f 0(x) = x(−2e−2x) + e−2x(1) = e−2x(−2x+ 1) > 0 ⇔ −2x+ 1 > 0 ⇔ x < 12

and f 0(x) < 0 ⇔ x > 12

,

so f is increasing on −∞, 12

and decreasing on 12,∞ . F. Local maximum value f 1

2= 1

2e−1 = 1/(2e);

no local minimum value

G. f 00(x) = e−2x(−2) + (−2x+ 1)(−2e−2x)= 2e−2x[−1− (−2x+ 1)] = 4 (x− 1) e−2x.

f 00(x) > 0 ⇔ x > 1 and f 00(x) < 0 ⇔ x < 1, so f is

CU on (1,∞) and CD on (−∞, 1). IP at (1, f(1)) = (1, e−2)

H.

85. From the graph, we estimate the points of inflection to be about (±0.82, 0.22).f(x) = e−1/x

2 ⇒ f 0(x) = 2x−3e−1/x2 ⇒

f 00(x) = 2[x−3(2x−3)e−1/x2

+ e−1/x2

(−3x−4)] = 2x−6e−1/x2 2− 3x2 .

This is 0 when 2− 3x2 = 0 ⇔ x = ± 23

, so the inflection points

are ± 23, e−3/2 .

87. s(t) = Ae−ct cos(ωt+ δ) ⇒v(t) = s0(t) = A{e−ct [−ω sin(ωt+ δ)] + cos(ωt+ δ)(−ce−ct)} = −Ae−ct [ω sin(ωt+ δ) + c cos(ωt+ δ)] ⇒a(t) = v0(t) = −A{e−ct[ω2 cos(ωt+ δ)− cω sin(ωt+ δ)] + [ω sin(ωt+ δ) + c cos(ωt+ δ)](−ce−ct)}

= −Ae−ct[ω2 cos(ωt+ δ)− cω sin(ωt+ δ)− cω sin(ωt+ δ)− c2 cos(ωt+ δ)]

= −Ae−ct[(ω2 − c2) cos(ωt+ δ)− 2cω sin(ωt+ δ)] = Ae−ct[(c2 − ω2) cos(ωt+ δ) + 2cω sin(ωt+ δ)]

89. (a) y(t) = y(0)ekt = 200ekt ⇒ y(0.5) = 200e0.5k = 360 ⇒ e0.5k = 1.8 ⇒ 0.5k = ln 1.8 ⇒k = 2 ln 1.8 = ln(1.8)2 = ln 3.24 ⇒ y(t) = 200e(ln 3.24)t = 200(3.24)t

(b) y(4) = 200(3.24)4 ≈ 22,040 bacteria


(c) y0(t) = 200(3.24)t · ln 3.24, so y0(4) = 200(3.24)4 · ln 3.24 ≈ 25,910 bacteria per hour

(d) 200(3.24)t = 10,000 ⇒ (3.24)t = 50 ⇒ t ln 3.24 = ln 50 ⇒ t = ln 50/ ln 3.24 ≈ 3.33 hours

91. Let P (t) = 64

1 + 31e−0.7944t=

A

1 +Bect= A(1 +Bect)−1, where A = 64, B = 31, and c = −0.7944.

P 0(t) = −A(1 +Bect)−2(Bcect) = −ABcect(1 +Bect)−2

P 00(t) = −ABcect[− 2(1 +Bect)−3(Bcect)] + (1 +Bect)−2(−ABc2ect)

= −ABc2ect(1 +Bect)−3[−2Bect + (1 +Bect)] = −ABc2ect(1−Bect)

(1 +Bect)3

The population is increasing most rapidly when its graph changes from CU to CD; that is, when P 00(t) = 0 in this case.

P 00(t) = 0 ⇒ Bect = 1 ⇒ ect =1

B⇒ ct = ln

1

B⇒ t =

ln(1/B)

c=ln(1/31)

−0.7944 ≈ 4.32 days. Note that

P1

cln1

B=

A

1 +Bec(1/c) ln(1/B)=

A

1 +Beln(1/B)=

A

1 +B(1/B)=

A

1 + 1=

A

2, one-half the limit of P as t→∞.

93. Let u = −2y2. Then du = −4y dy and 1

0ye−2y

2dy =

−20

eu − 14du = − 1

4eu

−20= − 1

4(e−2 − 1) = 1

4(1− e−2).

95. Let u = ex, so du = ex dx. When x = 0, u = 1; when x = 1, u = e. Thus,1

0

ex

1 + e2xdx =

e

1

1

1 + u2du = arctanu

e

1= arctan e− arctan 1 = arctan e− π

4.


2√x

⇒ e√x

√xdx = 2 eu du = 2eu +C = 2e

√x + C.

99. Let u = x2 + 2x. Then du = (2x + 2) dx = 2(x + 1) dx and

x+ 1

x2 + 2xdx =

12du

u= 1

2ln |u|+C = 1

2ln x2 + 2x +C.

101. Let u = ln(cosx). Then du =− sinxcosx

dx = − tanxdx ⇒

tanx ln(cosx) dx = − udu = − 12u

2 +C = − 12 [ ln(cosx)]

2 + C.

103. Let u = tan θ. Then du = sec2 θ dθ and 2tan θ sec2 θ dθ = 2udu =2u

ln 2+ C =

2tan θ

ln 2+ C.

105. 1− x

x

2

dx =1

x− 1

2

dx =1

x2− 2

x+ 1 dx = − 1

x− 2 ln |x|+ x+C

107. cosx ≤ 1 ⇒ ex cosx ≤ ex ⇒ 1

0ex cosxdx ≤ 1

0ex dx = ex

1

0= e− 1

109. f(x) =√x

1

es

sds ⇒ f 0(x) =

d

dx

√x

1

es

sds =

e√x

√x

d

dx

√x =

e√x

√x

1

2√x=

e√x

2x

111. fave =1

4− 14

1

1

xdx =

1

3ln |x| 4

1=1

3[ln 4− ln 1] = 1

3ln 4


113. V =1

0

2πx

1 + x4dx by cylindrical shells. Let u = x2 ⇒ du = 2xdx. Then

V =1

0

π

1 + u2du = π tan−1 u

1

0= π tan−1 1− tan−1 0 = π

π

4=

π2

4.

115. f(x) = lnx+ tan−1 x ⇒ f(1) = ln 1 + tan−1 1 = π4⇒ g π

4= 1 [where g = f−1].

f 0(x) =1

x+

1

1 + x2, so g0 π

4=

1

f 0(1)=

1

3/2=2

3.

117. We find the equation of a tangent to the curve y = e−x, so that we can find the

x- and y-intercepts of this tangent, and then we can find the area of the triangle.

The slope of the tangent at the point a, e−a is given by d

dxe−x

x=a

= −e−a,

and so the equation of the tangent is y − e−a = −e−a(x− a) ⇔

y = e−a(a− x+ 1).

The y-intercept of this line is y = e−a(a− 0 + 1) = e−a(a+ 1). To find the x-intercept we set y = 0 ⇒

e−a(a− x+ 1) = 0 ⇒ x = a+ 1. So the area of the triangle is A(a) = 12e−a(a+ 1) (a+ 1) = 1

2e−a(a+ 1)2. We

differentiate this with respect to a: A0(a) = 12e−a(2)(a+ 1) + (a+ 1)2e−a(−1) = 1

2e−a 1− a2 . This is 0

at a = ±1, and the root a = 1 gives a maximum, by the First Derivative Test. So the maximum area of the triangle is

A(1) = 12e−1(1 + 1)2 = 2e−1 = 2/e.

119. limx→−1

F (x) = limx→−1

bx+1 − ax+1

x+ 1H= lim

x→−1bx+1 ln b− ax+1 ln a

1= ln b− ln a = F (−1), so F is continuous at −1.

121. Using FTC1, we differentiate both sides of the given equation, x

0f(t) dt = xe2x +

x

0e−tf(t) dt, and get

f(x) = e2x + 2xe2x + e−xf(x) ⇒ f(x) 1− e−x = e2x + 2xe2x ⇒ f(x) =e2x(1 + 2x)

1− e−x.

PROBLEMS PLUS1. Let y = f(x) = e−x

2

. The area of the rectangle under the curve from−x to x is A(x) = 2xe−x2

where x ≥ 0. We maximize

A(x): A0(x) = 2e−x2 − 4x2e−x2 = 2e−x2 1− 2x2 = 0 ⇒ x = 1√

2. This gives a maximum since A0(x) > 0

for 0 ≤ x < 1√2

and A0(x) < 0 for x > 1√2

. We next determine the points of inflection of f(x). Notice that

f 0(x) = −2xe−x2 = −A(x). So f 00(x) = −A0(x) and hence, f 00(x) < 0 for − 1√2< x < 1√

2and f 00(x) > 0 for x < − 1√

2

and x > 1√2

. So f(x) changes concavity at x = ± 1√2

, and the two vertices of the rectangle of largest area are at the inflection

points.

3. Consider the statement that dn

dxn(eax sin bx) = rneax sin(bx + nθ). For n = 1,

d

dx(eax sin bx) = aeax sin bx+ beax cos bx, and

reax sin(bx+ θ) = reax[sin bx cos θ + cos bx sin θ] = reaxa

rsin bx+

b

rcos bx = aeax sin bx+ beax cos bx

since tan θ = b

a⇒ sin θ =

b

rand cos θ = a

r. So the statement is true for n = 1.

Assume it is true for n = k. Then

dk+1

dxk+1(eax sin bx) =

d

dxrkeax sin(bx+ kθ) = rkaeax sin(bx+ kθ) + rkeaxb cos(bx+ kθ)

= rkeax[a sin(bx+ kθ) + b cos(bx+ kθ)]

But

sin[bx+ (k + 1)θ] = sin[(bx+ kθ) + θ] = sin(bx+ kθ) cos θ + sin θ cos(bx+ kθ) = ar sin(bx+ kθ) + b

r cos(bx+ kθ).

Hence, a sin(bx+ kθ) + b cos(bx+ kθ) = r sin[bx+ (k + 1)θ]. So

dk+1

dxk+1(eax sin bx) = rkeax[a sin(bx+kθ)+b cos(bx+kθ)] = rkeax[r sin(bx+(k+1)θ)] = rk+1eax[sin(bx+(k+1)θ)].

Therefore, the statement is true for all n by mathematical induction.

5. We first show that x

1 + x2< tan−1 x for x > 0. Let f(x) = tan−1 x− x

1 + x2. Then

f 0(x) =1

1 + x2− 1(1 + x2)− x(2x)

(1 + x2)2=(1 + x2)− (1− x2)

(1 + x2)2=

2x2

(1 + x2)2> 0 for x > 0. So f(x) is increasing

on (0,∞). Hence, 0 < x ⇒ 0 = f(0) < f(x) = tan−1 x− x

1 + x2. So x

1 + x2< tan−1 x for 0 < x. We next show

that tan−1 x < x for x > 0. Let h(x) = x− tan−1 x. Then h0(x) = 1− 1

1 + x2=

x2

1 + x2> 0. Hence, h(x) is increasing

on (0,∞). So for 0 < x, 0 = h(0) < h(x) = x− tan−1 x. Hence, tan−1 x < x for x > 0, and we conclude thatx

1 + x2< tan−1 x < x for x > 0.

7. By the Fundamental Theorem of Calculus, f(x) = x

1

√1 + t3 dt ⇒ f 0(x) =

√1 + x3 > 0 for x > −1.

So f is increasing on (−1,∞) and hence is one-to-one. Note that f(1) = 0, so f−1(1) = 0 ⇒(f−1)0(0) = 1/f 0(1) = 1√

2.

305

306 ¤ CHAPTER 7 PROBLEMS PLUS

9. If L = limx→∞

x+ a

x− a

x

, then L has the indeterminate form 1∞, so

lnL = limx→∞

lnx+ a

x− a

x

= limx→∞

x lnx+ a

x− a= lim

x→∞ln(x+ a)− ln(x− a)

1/xH= lim

x→∞

1

x+ a− 1

x− a−1/x2

= limx→∞

(x− a)− (x+ a)

(x+ a)(x− a)· −x

2

1= lim

x→∞2ax2

x2 − a2= lim

x→∞2a

1− a2/x2= 2a

Hence, lnL = 2a, so L = e2a. From the original equation, we want L = e1 ⇒ 2a = 1 ⇒ a = 12 .

11. Both sides of the inequality are positive, so cosh(sinhx) < sinh(coshx)

⇔ cosh2(sinhx) < sinh2(coshx) ⇔ sinh2(sinhx) + 1 < sinh2(coshx)

⇔ 1 < [sinh(coshx)− sinh(sinhx)][sinh(coshx) + sinh(sinhx)]

⇔ 1 < sinhex + e−x

2− sinh ex − e−x

2sinh

ex + e−x

2+ sinh

ex − e−x

2

⇔ 1 < [2 cosh(ex/2) sinh(ex/2)][2 sinh(ex/2) cosh(ex/2)] [use the addition formulas and cancel]

⇔ 1 < [2 sinh(ex/2) cosh(ex/2)][2 sinh(ex/2) cosh(ex/2)] ⇔ 1 < sinh ex sinh e−x,

by the half-angle formula. Now both ex and e−x are positive, and sinh y > y for y > 0, since sinh 0 = 0 and

(sinh y − y)0 = cosh y − 1 > 0 for x > 0, so 1 = exe−x < sinh ex sinh e−x. So, following this chain of reasoning

backward, we arrive at the desired result.

13. Let f(x) = e2x and g(x) = k√x [k > 0]. From the graphs of f and g,

we see that f will intersect g exactly once when f and g share a tangent

line. Thus, we must have f = g and f 0 = g0 at x = a.

f(a) = g(a) ⇒ e2a = k√a ( )

and f 0(a) = g0(a) ⇒ 2e2a =k

2√a

⇒ e2a =k

4√a

.

So we must have k√a =

k

4√a

⇒√a

2

=k

4k⇒ a = 1

4. From ( ), e2(1/4) = k 1/4 ⇒

k = 2e1/2 = 2√e ≈ 3.297.

15. Suppose that the curve y = ax intersects the line y = x. Then ax0 = x0 for some x0 > 0, and hence a = x1/x00 . We find the

maximum value of g(x) = x1/x, x > 0, because if a is larger than the maximum value of this function, then the curve y = ax

does not intersect the line y = x. g0(x) = e(1/x) ln x − 1

x2lnx+

1

x· 1x

= x1/x1

x2(1− lnx). This is 0 only where

x = e, and for 0 < x < e, f 0(x) > 0, while for x > e, f 0(x) < 0, so g has an absolute maximum of g(e) = e1/e. So if

y = ax intersects y = x, we must have 0 < a ≤ e1/e. Conversely, suppose that 0 < a ≤ e1/e. Then ae ≤ e, so the graph of

y = ax lies below or touches the graph of y = x at x = e. Also a0 = 1 > 0, so the graph of y = ax lies above that of y = x

at x = 0. Therefore, by the Intermediate Value Theorem, the graphs of y = ax and y = x must intersect somewhere between

x = 0 and x = e.

7 inverse functions -...

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