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Instructor’s Solutions Manual, Section 4.3 Exercise 1

Solutions to Exercises, Section 4.3

1. For x = 7 and y = 13, evaluate each of the following:

(a) ln(x +y) (b) lnx + lny

[This exercise and the next one emphasize that ln(x +y) does not equallnx + lny .]

solution

(a) ln(7+ 13) = ln 20 ≈ 2.99573

(b) ln 7+ ln 13 ≈ 1.94591+ 2.56495

= 4.51086


2. For x = 0.4 and y = 3.5, evaluate each of the following:

(a) ln(x +y) (b) lnx + lny

solution

(a)ln(0.4+ 3.5) = ln 3.9 ≈ 1.36098

(b)

ln 0.4+ ln 3.5 ≈ −0.916291+ 1.25276

= 0.336472



(a) ln(xy) (b) (lnx)(lny)

[This exercise and the next one emphasize that ln(xy) does not equal(lnx)(lny).]

solution

(a) ln(3 · 8) = ln 24 ≈ 3.17805

(b) (ln 3)(ln 8) ≈ (1.09861)(2.07944)

≈ 2.2845


4. For x = 1.1 and y = 5, evaluate each of the following:

(a) ln(xy) (b) (lnx)(lny)

solution

(a)ln(1.1× 5) = ln 5.5 ≈ 1.70475

(b)

(ln 1.1)(ln 5) ≈ (0.0953102)(1.60944)

≈ 0.1533962



(a) ln xy (b) lnx

lny

[This exercise and the next one emphasize that ln xy does not equal lnx

lny .]

solution

(a) ln 122 = ln 6 ≈ 1.79176

(b)ln 12ln 2

≈ 2.484910.693147

≈ 3.58496


6. For x = 18 and y = 0.3, evaluate each of the following:

(a) ln xy (b) lnx

lny

solution

(a)ln 18

0.3 = ln 60 ≈ 4.09434

(b)ln 18ln 0.3

≈ 2.89037−1.20397

≈ −2.4007


7. Find a number y such that lny = 4.

solution Recall that lny is simply shorthand for loge y . Thus theequation lny = 4 can be rewritten as loge y = 4. The definition of alogarithm now implies that y = e4.


8. Find a number c such that ln c = 5.

solution Recall that ln c is simply shorthand for loge c. Thus theequation ln c = 5 can be rewritten as loge c = 5. The definition of alogarithm now implies that c = e5.


9. Find a number x such that lnx = −2.

solution Recall that lnx is simply shorthand for loge x. Thus theequation lnx = −2 can be rewritten as loge x = −2. The definition of alogarithm now implies that x = e−2.


10. Find a number x such that lnx = −3.

solution Recall that lnx is simply shorthand for loge x. Thus theequation lnx = −3 can be rewritten as loge x = −3. The definition of alogarithm now implies that x = e−3.


11. Find a number t such that ln(2t + 1) = −4.

solution The equation ln(2t + 1) = −4 implies that

e−4 = 2t + 1.

Solving this equation for t, we get

t = e−4 − 12

.


12. Find a number w such that ln(3w − 2) = 5.

solution The equation ln(3w − 2) = 5 implies that

e5 = 3w − 2.

Solving this equation for w, we get

w = e5 + 23

.


13. Find all numbers y such that ln(y2 + 1) = 3.

solution The equation ln(y2 + 1) = 3 implies that

e3 = y2 + 1.

Thus y2 = e3 − 1, which means that y = √e3 − 1 or y = −√e3 − 1.


14. Find all numbers r such that ln(2r 2 − 3) = −1.

solution The equation ln(2r 2 − 3) = −1 implies that

e−1 = 2r 2 − 3.

Thus r 2 = e−1+32 , which means that r =

√e−1+3

2 or r = −√e−1+3

2 .


15. Find a number x such that e3x−1 = 2.

solution The equation e3x−1 = 2 implies that

3x − 1 = ln 2.

Solving this equation for x, we get

x = 1+ ln 23

.


16. Find a number y such that e4y−3 = 5.

solution The equation e4y−3 = 5 implies that

4y − 3 = ln 5.

Solving this equation for y , we get

y = 3+ ln 54

.


For Exercises 17–24, find all numbers x that satisfy the given equation.

17. ln(x + 5)− ln(x − 1) = 2

solution Our equation can be rewritten as follows:

2 = ln(x + 5)− ln(x − 1)

= lnx + 5x − 1

.

Thusx + 5x − 1

= e2.

We can solve the equation above for x, getting x = e2 + 5e2 − 1

.


18. ln(x + 4)− ln(x − 2) = 3


3 = ln(x + 4)− ln(x − 2)

= lnx + 4x − 2

.

Thusx + 4x − 2

= e3.

We can solve the equation above for x, getting x = 2(e3 + 2)e3 − 1

.


19. ln(x + 5)+ ln(x − 1) = 2


2 = ln(x + 5)+ ln(x − 1)

= ln((x + 5)(x − 1)

)= ln(x2 + 4x − 5).

Thusx2 + 4x − 5 = e2,

which implies thatx2 + 4x − (e2 + 5) = 0.

We can solve the equation above using the quadratic formula, gettingx = −2+√9+ e2 or x = −2−√9+ e2. However, both x + 5 and x − 1are negative if x = −2−√9+ e2; because the logarithm of a negativenumber is undefined, we must discard this root of the equation above.We conclude that the only value of x satisfying the equationln(x + 5)+ ln(x − 1) = 2 is x = −2+√9+ e2.


20. ln(x + 4)+ ln(x + 2) = 2


2 = ln(x + 4)+ ln(x + 2)

= ln((x + 4)(x + 2)

)= ln(x2 + 6x + 8).

Thusx2 + 6x + 8 = e2,

which implies thatx2 + 6x + 8− e2 = 0.

We can solve the equation above using the quadratic formula, gettingx = −3+√1+ e2 or x = −3−√1+ e2. However, both x + 4 and x + 2are negative if x = −3−√1+ e2; because the logarithm of a negativenumber is undefined, we must discard this root of the equation above.We conclude that the only value of x satisfying the equationln(x + 4)+ ln(x + 2) = 2 is x = −3+√1+ e2.


21.ln(12x)ln(5x)

= 2


2 = ln(12x)ln(5x)

= ln 12+ lnxln 5+ lnx

.

Solving this equation for lnx (the first step in doing this is to multiplyboth sides by the denominator ln 5+ lnx), we get

lnx = ln 12− 2 ln 5

= ln 12− ln 25

= ln 1225 .

Thus x = 1225 .


22.ln(11x)ln(4x)

= 2


2 = ln(11x)ln(4x)

= ln 11+ lnxln 4+ lnx

.

Solving this equation for lnx (the first step in doing this is to multiplyboth sides by the denominator ln 4+ lnx), we get

lnx = ln 11− 2 ln 4

= ln 11− ln 16

= ln 1116 .

Thus x = 1116 .


23.(ln(3x)

)lnx = 4


4 = (ln(3x)) lnx

= (lnx + ln 3) lnx

= (lnx)2 + (ln 3)(lnx).

Letting y = lnx, we can rewrite the equation above as

y2 + (ln 3)y − 4 = 0.

Use the quadratic formula to solve the equation above for y , getting

y ≈ −2.62337 or y ≈ 1.52476.

Thuslnx ≈ −2.62337 or lnx ≈ 1.52476,

which means thatx ≈ e−2.62337 ≈ 0.072558

orx ≈ e1.52476 ≈ 4.59403.


24.(ln(6x)

)lnx = 5


5 = (ln(6x)) lnx

= (lnx + ln 6) lnx

= (lnx)2 + (ln 6)(lnx).

Letting y = lnx, we can rewrite the equation above as

y2 + (ln 6)y − 5 = 0.

Use the quadratic formula to solve the equation above for y , getting

y ≈ −3.30474 or y ≈ 1.51298.

Thuslnx ≈ −3.30474 or lnx ≈ 1.51298,

which means thatx ≈ e−3.30474 ≈ 0.0367088

orx ≈ e1.51298 ≈ 4.54024.


25. Find the number c such that area( 1x ,1, c) = 2.

solution Because 2 = area( 1x ,1, c) = ln c, we see that c = e2.


26. Find the number c such that area( 1x ,1, c) = 3.

solution Because 3 = area( 1x ,1, c) = ln c, we see that c = e3.


27. Find the number t that makes et2+6t as small as possible.[Here et2+6t means e(t2+6t).]

solution Because ex is an increasing function of x, the number et2+6t

will be as small as possible when t2 + 6t is as small as possible. To findwhen t2 + 6t is as small as possible, we complete the square:

t2 + 6t = (t + 3)2 − 9.

The equation above shows that t2 + 6t is as small as possible whent = −3.


28. Find the number t that makes et2+8t+3 as small as possible.

solution Because ex is an increasing function of x, the numberet2+8t+3 will be as small as possible when t2 + 8t + 3 is as small aspossible, which is equivalent to making t2 + 8t as small as possible. Tofind when t2 + 8t is as small as possible, we complete the square:

t2 + 8t = (t + 4)2 − 16.

The equation above shows that t2 + 8t is as small as possible whent = −4.


29. Find a number x such that

e2x + ex = 6.

solution Note that e2x = (ex)2. This suggests that we let t = ex .Then the equation above can be rewritten as

t2 + t − 6 = 0.

The solutions to this equation (which can be found either by using thequadratic formula or by factoring) are t = −3 and t = 2. Thus ex = −3or ex = 2. However, there is no real number x such that ex = −3(because ex is positive for every real number x), and thus we must haveex = 2. Thus x = ln 2 ≈ 0.693147.


30. Find a number x such that

e2x − 4ex = 12.

solution Note that e2x = (ex)2. This suggests that we let t = ex .Then the equation above can be rewritten as

t2 − 4t − 12 = 0.

The solutions to this equation (which can be found either by using thequadratic formula or by factoring) are t = −2 and t = 6. Thus ex = −2or ex = 6. However, there is no real number x such that ex = −2(because ex is positive for every real number x), and thus we must haveex = 6. Thus x = ln 6 ≈ 1.79176.


31. Find a number y such that

1+ lny2+ lny

= 0.9.

solution Multiplying both sides of the equation above by 2+ lny andthen solving for lny gives lny = 8. Thus y = e8 ≈ 2980.96.


32. Find a number w such that

4− lnw3− 5 lnw

= 3.6.

solution Multiplying both sides of the equation above by 3− 5 lnwand then solving for lnw gives lnw = 0.4. Thus w = e0.4 ≈ 1.49182.


For Exercises 33–36, find a formula for (f ◦ g)(x) assuming that f andg are the indicated functions.

33. f(x) = lnx and g(x) = e5x

solution

(f ◦ g)(x) = f (g(x)) = f(e5x) = ln e5x = 5x


34. f(x) = lnx and g(x) = e4−7x

solution

(f ◦ g)(x) = f (g(x)) = f(e4−7x)

= ln e4−7x = 4− 7x


35. f(x) = e2x and g(x) = lnx

solution

(f ◦ g)(x) = f (g(x)) = f(lnx)= e2 lnx = (elnx)

2 = x2


36. f(x) = e8−5x and g(x) = lnx

solution

(f ◦ g)(x) = f (g(x)) = f(lnx)= e8−5 lnx = e8e−5 lnx

= e8(elnx)−5 = e8x−5


For each of the functions f given in Exercises 37–46:

(a) Find the domain of f .

(b) Find the range of f .

(c) Find a formula for f−1.

(d) Find the domain of f−1.

(e) Find the range of f−1.

You can check your solutions to part (c) by verifying that f−1 ◦ f = Iand f ◦ f−1 = I. (Recall that I is the function defined by I(x) = x.)

37. f(x) = 2+ lnx

solution

(a) The expression 2+ lnx makes sense for all positive numbers x. Thusthe domain of f is the set of positive numbers.

(b) To find the range of f , we need to find the numbers y such that

y = 2+ lnx

for some x in the domain of f . In other words, we need to find thevalues of y such that the equation above can be solved for a positivenumber x. To solve this equation for x, subtract 2 from both sides,getting y − 2 = lnx, which implies that

x = ey−2.


The expression above on the right makes sense for every real number yand produces a positive number x (because e raised to any power ispositive). Thus the range of f is the set of real numbers.

(c) The expression above shows that f−1 is given by the expression

f−1(y) = ey−2.

(d) The domain of f−1 equals the range of f . Thus the domain of f−1 isthe set of real numbers.

(e) The range of f−1 equals the domain of f . Thus the range of f−1 is theset of positive numbers.


38. f(x) = 3− lnx

solution

(a) The expression 3− lnx makes sense for all positive numbers x. Thusthe domain of f is the set of positive numbers.


y = 3− lnx

for some x in the domain of f . In other words, we need to find thevalues of y such that the equation above can be solved for a positivenumber x. To solve this equation for x, subtract 3 from both sides andthen multiply by −1, getting 3−y = lnx, which implies that

x = e3−y.



f−1(y) = e3−y.




39. f(x) = 4− 5 lnx

solution

(a) The expression 4− 5 lnx makes sense for all positive numbers x. Thusthe domain of f is the set of positive numbers.


y = 4− 5 lnx

for some x in the domain of f . In other words, we need to find thevalues of y such that the equation above can be solved for a positivenumber x. To solve this equation for x, subtract 4 from both sides,then divide both sides by −5, getting 4−y

5 = lnx, which implies that

x = e(4−y)/5.



f−1(y) = e(4−y)/5.




40. f(x) = −6+ 7 lnx

solution

(a) The expression −6+ 7 lnx makes sense for all positive numbers x.Thus the domain of f is the set of positive numbers.


y = −6+ 7 lnx

for some x in the domain of f . In other words, we need to find thevalues of y such that the equation above can be solved for a positivenumber x. To solve this equation for x, add 6 to both sides, then divideboth sides by 7, getting y+6

7 = lnx, which implies that

x = e(y+6)/7.



f−1(y) = e(y+6)/7.




41. f(x) = 3e2x

solution

(a) The expression 3e2x makes sense for all real numbers x. Thus thedomain of f is the set of real numbers.


y = 3e2x

for some x in the domain of f . In other words, we need to find thevalues of y such that the equation above can be solved for a realnumber x. To solve this equation for x, divide both sides by 3, gettingy3 = e2x , which implies that 2x = ln y

3 . Thus

x = ln y3

2.

The expression above on the right makes sense for every positivenumber y and produces a real number x. Thus the range of f is the setof positive numbers.


f−1(y) = ln y3

2.

(d) The domain of f−1 equals the range of f . Thus the domain of f−1 isthe set of positive numbers.


(e) The range of f−1 equals the domain of f . Thus the range of f−1 is theset of real numbers.


42. f(x) = 5e9x

solution

(a) The expression 5e9x makes sense for all real numbers x. Thus thedomain of f is the set of real numbers.


y = 5e9x

for some x in the domain of f . In other words, we need to find thevalues of y such that the equation above can be solved for a realnumber x. To solve this equation for x, divide both sides by 5, gettingy5 = e9x , which implies that 9x = ln y

5 . Thus

x = ln y5

9.

The expression above on the right makes sense for every positivenumber y and produces a real number x. Thus the range of f is the setof positive numbers.


f−1(y) = ln y5

9.

(d) The domain of f−1 equals the range of f . Thus the domain of f−1 isthe set of positive numbers.


43. f(x) = 4+ ln(x − 2)

solution

(a) The expression 4+ ln(x − 2) makes sense when x > 2. Thus thedomain of f is the interval (2,∞).


y = 4+ ln(x − 2)

for some x in the domain of f . In other words, we need to find thevalues of y such that the equation above can be solved for a numberx > 2. To solve this equation for x, subtract 4 from both sides, gettingy − 4 = ln(x − 2), which implies that x − 2 = ey−4. Thus

x = 2+ ey−4.

The expression above on the right makes sense for every real number yand produces a number x > 2 (because e raised to any power ispositive). Thus the range of f is the set of real numbers.


f−1(y) = 2+ ey−4.


(e) The range of f−1 equals the domain of f . Thus the range of f−1 is theinterval (2,∞).


44. f(x) = 3+ ln(x + 5)

solution

(a) The expression 3+ ln(x + 5) makes sense when x > −5. Thus thedomain of f is the interval (−5,∞).


y = 3+ ln(x + 5)

for some x in the domain of f . In other words, we need to find thevalues of y such that the equation above can be solved for a numberx > −5. To solve this equation for x, subtract 3 from both sides,getting y − 3 = ln(x + 5), which implies that x + 5 = ey−3. Thus

x = ey−3 − 5.

The expression above on the right makes sense for every real number yand produces a number x > −5 (because e raised to any power ispositive). Thus the range of f is the set of real numbers.


f−1(y) = ey−3 − 5.


(e) The range of f−1 equals the domain of f . Thus the range of f−1 is theinterval (−5,∞).


45. f(x) = 5+ 6e7x

solution

(a) The expression 5+ 6e7x makes sense for all real numbers x. Thus thedomain of f is the set of real numbers.


y = 5+ 6e7x

for some x in the domain of f . In other words, we need to find thevalues of y such that the equation above can be solved for a realnumber x. To solve this equation for x, subtract 5 from both sides,then divide both sides by 6, getting y−5

6 = e7x , which implies that

7x = ln y−56 . Thus

x = ln y−56

7.

The expression above on the right makes sense for every y > 5 andproduces a real number x. Thus the range of f is the interval (5,∞).


f−1(y) = ln y−56

7.

(d) The domain of f−1 equals the range of f . Thus the domain of f−1 isthe interval (5,∞).


46. f(x) = 4− 2e8x

solution

(a) The expression 4− 2e8x makes sense for all real numbers x. Thus thedomain of f is the set of real numbers.


y = 4− 2e8x

for some x in the domain of f . In other words, we need to find thevalues of y such that the equation above can be solved for a realnumber x. To solve this equation for x, subtract 4 from both sides,then divide both sides by −2, getting 4−y

2 = e8x , which implies that

8x = ln 4−y2 . Thus

x = ln 4−y2

8.

The expression above on the right makes sense for every y < 4 andproduces a real number x. Thus the range of f is the interval (−∞,4).


f−1(y) = ln 4−y2

8.

(d) The domain of f−1 equals the range of f . Thus the domain of f−1 isthe interval (−∞,4).

Instructor’s Solutions Manual, Section 4.3 Problem 47

Solutions to Problems, Section 4.3

47. Verify that the last five rectangles in the figure in Example 2 have area18 , 1

9 , 110 , 1

11 , and 112 .

solution In Example 2 we computed the areas of the first threerectangles in the figure on that page. The areas of the other rectanglesare computed similarly:

The base of the fourth rectangle is the interval [74 ,2]. The height of the

fourth rectangle is 12 . Because the fourth rectangle has base 1

4 andheight 1

2 , the area of the fourth rectangle equals 14 · 1

2 , which equals 18 .

The base of the fifth rectangle is the interval [2, 94]. The height of the

fifth rectangle is 49 . Because the fifth rectangle has base 1

4 and height 49 ,

the area of the fifth rectangle equals 14 · 4


The base of the sixth rectangle is the interval [94 ,

52]. The height of the

sixth rectangle is 25 . Because the sixth rectangle has base 1

4 and height25 , the area of the sixth rectangle equals 1

4 · 25 , which equals 1

10 .

The base of the seventh rectangle is the interval [52 ,

114 ]. The height of

the seventh rectangle is 411 . Because the seventh rectangle has base 1

4and height 4

11 , the area of the seventh rectangle equals 14 · 4

11 , whichequals 1

11 .

The base of the eighth rectangle is the interval [114 ,3]. The height of the

eighth rectangle is 13 . Because the eighth rectangle has base 1

4 andheight 1

3 , the area of the eighth rectangle equals 14 · 1



48. Consider this figure:

y � 1�x

1 1.5 2 2.5x

1

y

The region under the curve y = 1x , above the x-axis, and between the

lines x = 1 and x = 2.5.

(a) Calculate the sum of the areas of all six rectangles shown in thefigure above.

(b) Explain why the calculation you did in part (a) shows that

area( 1x ,1,2.5) < 1.

(c) Explain why the inequality above shows that e > 2.5.

solution

(a) The base of the first rectangle is the interval [1, 54]. The height of the

first rectangle is 1. Because the first rectangle has base 14 and height 1,

the area of the first rectangle equals 14 · 1, which equals 1

4 .


The base of the second rectangle is the interval [54 ,


second rectangle is 45 . Because the second rectangle has base 1

4 andheight 4

5 , the area of the second rectangle equals 14 · 4


The base of the third rectangle is the interval [32 ,


third rectangle is 23 . Because the third rectangle has base 1

4 and height23 , the area of the third rectangle equals 1


6 .

The base of the fourth rectangle is the interval [74 ,2]. The height of the

fourth rectangle is 47 . Because the fourth rectangle has base 1

4 andheight 4

7 , the area of the fourth rectangle equals 14 · 4


The base of the fifth rectangle is the interval [2, 94]. The height of the

fifth rectangle is 12 . Because the fifth rectangle has base 1

4 and height 12 ,

the area of the fifth rectangle equals 14 · 1


The base of the sixth rectangle is the interval [94 ,

104 ]. The height of the

sixth rectangle is 49 . Because the sixth rectangle has base 1

4 and height49 , the area of the sixth rectangle equals 1


9 .

Thus the sum of the areas of all six rectangles shown in the figure is

14 + 1

5 + 16 + 1

7 + 18 + 1

9 ;

a calculator shows that this number is approximately 0.9956.

(b) The union of the rectangles in the figure above contains the regionunder the curve y = 1

x , above the x-axis, and between the lines x = 1

and x = 2.5. Thus area( 1x ,1,2.5) is less than the sum of the areas of all

six rectangles, which as we saw in part (a) is less than 0.996, which isless than 1. Thus area( 1

x ,1,2.5) < 1.


(c) We define e as the number such that the area of the region under thecurve y = 1

x , above the x-axis, and between the lines x = 1 and x = e is1. The inequality in part (b) shows that we need to go beyond 2.5 tohave the area of the region in question be as large as 1. Thus e > 2.5.


49. Explain whylnx ≈ 2.302585 logx

for every positive number x.

solution By the formula for change of base for logarithms (seeSection 3.2), we have

lnx = loge x =logxlog e

≈ 2.302585 logx,

where the term 2.302585 comes from using a calculator to approximate1

log e .


50. Explain why the solution to part (b) of Exercise 5 in this section is thesame as the solution to part (b) of Exercise 5 in Section 3.3.

solution In part (b) of Exercise 5 in this section, we used a calculatorto find that

ln 12ln 2

≈ 3.58496.

In part (b) of Exercise 5 in Section 3.3, we used a calculator to find that

log 12log 2

≈ 3.58496.

These two solutions are the same because the formula for the changeof base for logarithms shows that

loga yloga b

= logb y,

which implies that the ratio loga yloga b

does not depend on a (because theright side of the equation above does not contain a). In particular,

ln 12ln 2

= loge 12loge 2

= log10 12log10 2

= log 12log 2

.


51. Suppose c is a number such that area( 1x ,1, c) > 1000. Explain why

c > 21000.

solution We have

loge c = ln c = area( 1x ,1, c) > 1000,

which implies thatc > e1000 > 21000.

where the last inequality holds because e > 2.


The functions cosh and sinh are defined by

cosh x = ex + e−x2

and sinh x = ex − e−x2

for every real number x. For reasons that do not concern us here, thesefunctions are called the hyperbolic cosine and hyperbolic sine; they areuseful in engineering.

52. Show that cosh is an even function.

solution Note that

cosh(−x) = e−x + ex2

= ex + e−x2

= coshx.

Thus cosh is an even function.


53. Show that sinh is an odd function.

solution Note that

sinh(−x) = e−x − ex2

= −ex − e−x

2= sinhx.

Thus sinh is an odd function.


54. Show that(coshx)2 − (sinhx)2 = 1

for every real number x.

solution

(coshx)2 − (sinhx)2

=(ex + e−x

2

)2 −(ex − e−x

2

)2

= (ex)2 + 2exe−x + (e−x)24

− (ex)2 − 2exe−x + (e−x)2

4

= e2x + 2+ e−2x

4− e

2x − 2+ e−2x

4

= 1


55. Show that coshx ≥ 1 for every real number x.

solution Suppose x is a real number. From the previous problem, wehave

(coshx)2 = 1+ (sinhx)2 ≥ 1.

Because ex > 0 and e−x > 0, we see from the definition of coshx thatcoshx > 0. Thus taking square roots of both sides of the inequalityabove shows that coshx ≥ 1.


56. Show that

cosh(x +y) = coshx coshy + sinhx sinhy

for all real numbers x and y .

solution

coshx coshy + sinhx sinhy

= ex + e−x2

· ey + e−y

2+ e

x − e−x2

· ey − e−y

2

= ex+y + ex−y + e−x+y + e−x−y4

+ ex+y − ex−y − e−x+y + e−x−y

4

= ex+y + e−x−y2

= cosh(x +y)


57. Show that

sinh(x +y) = sinhx coshy + coshx sinhy

for all real numbers x and y .

solution

sinhx coshy + coshx sinhy

= ex − e−x2

· ey + e−y

2+ e

x + e−x2

· ey − e−y

2

= ex+y + ex−y − e−x+y − e−x−y4

+ ex+y − ex−y + e−x+y − e−x−y

4

= ex+y − e−x−y2

= sinh(x +y)


58. Show that(coshx + sinhx)t = cosh(tx)+ sinh(tx)

for all real numbers x and t.

solution Suppose x and t are real numbers. Then

(coshx + sinhx)t =(ex + e−x

2+ e

x − e−x2

)t= (ex)t

= etx

= etx + e−tx2

+ etx − e−tx

2

= cosh(tx)+ sinh(tx).


59. Show that if x is very large, then

coshx ≈ sinhh ≈ ex

2.

solution If x is very large, then e−x ≈ 0. Thus if x is very large wehave

coshx = ex + e−x2

≈ ex

2

and

sinhx = ex − e−x2

≈ ex

2.


60. Show that the range of sinh is the set of real numbers.

solution Suppose y is a real number. To show that y is in the rangeof sinh, we must find a number x such that sinhx = y . In other words,we must show that there is a real number x such that

ex − e−x2

= y,

which is equivalent to the equation

ex − 2y − e−x = 0.

Multiplying both sides of the equation above by ex gives the equivalentequation

(ex)2 − 2yex − 1 = 0.

Now use the quadratic formula to solve the equation above for ex ,getting the information that the equation above is satisfied if

ex =2y ±

√4y2 + 4

2= y ±

√y2 + 1.

Because√y2 + 1 > y , choosing the minus sign in the equation above

would lead to a negative value for ex , which is impossible. However, ifwe take

x = ln(y +

√y2 + 1

),

then


ex = y +√y2 + 1,

which implies that sinhx = y .

We have shown that an arbitrary real number y is in the range of sinh.Thus the range of sinh is the set of real numbers.


61. Show that sinh is a one-to-one function and that its inverse is given bythe formula

(sinh)−1(y) = ln(y +

√y2 + 1

)for every real number y .

solution In our solution to the previous problem, we showed that foreach real number y there is exactly one real number x such thatsinhx = y , which implies that sinh is a one-to-one function. In oursolution to the previous problem, we furthermore showed that theunique number x such that sinhx = y is given by the formula

x = ln(y +

√y2 + 1

).

Thus(sinh)−1(y) = ln

(y +

√y2 + 1

).


62. Show that the range of cosh is the interval [1,∞).solution Suppose y ≥ 1. To show that y is in the range of cosh, wemust find a number x such that coshx = y . In other words, we mustshow that there is a real number x such that

ex + e−x2

= y,

which is equivalent to the equation

ex − 2y + e−x = 0.

Multiplying both sides of the equation above by ex gives the equivalentequation

(ex)2 − 2yex + 1 = 0.

Now use the quadratic formula to solve the equation above for ex ,getting the information that the equation above is satisfied if

ex =2y ±

√4y2 − 4

2= y ±

√y2 − 1.

Because y ≥ 1, the square root above makes sense. We could chooseeither the plus or minus sign, but to be definite let’s choose the plussign, taking

x = ln(y +

√y2 − 1

).

Then


ex = y +√y2 − 1,

which implies that coshx = y .

We have shown that an arbitrary number y ≥ 1 is in the range of cosh.From Problem 55 we also know that no number less than 1 can be inthe range of cosh. Thus the range of cosh is the interval [1,∞).


63. Suppose f is the function defined by

f(x) = coshx

for every x ≥ 0. In other words, f is defined by the same formula ascosh, but the domain of f is the interval [0,∞) and the domain of coshis the set of real numbers. Show that f is a one-to-one function andthat its inverse is given by the formula

f−1(y) = ln(y +

√y2 − 1

)for every y ≥ 1.

solution Our solution to the previous problem shows that if y ≥ 1and coshx = y , then

x = ln(y +

√y2 − 1

)or x = ln

(y −

√y2 − 1

).

These two solutions are the same if y = 1, so let’s assume that y > 1.

Then we also have y +√y2 − 1 > 1 and thus

ln(y +

√y2 − 1

)> 0.

However

y −√y2 − 1 = y −

√y2 − 1 ·

y +√y2 − 1

y +√y2 − 1

= 1

y +√y2 − 1

< 1,


which implies that

ln(y −

√y2 − 1

)< 0,

which eliminates the second solution above if we want to consider onlyvalues of x in the interval [0,∞).In other words, if y ≥ 1, then there is exactly one x ≥ 0 such thatcoshx = y , which implies that f is a one-to-one function. Furthermore,the unique number x such that f(x) = y is given by the formula

x = ln(y −

√y2 + 1

).

Thusf−1(y) = ln

(y −

√y2 + 1

).


64. Write a description of how the shape of the St. Louis Gateway Arch,whose picture appears on the opening page of this chapter, is related tothe graph of coshx.[You should be able to find the necessary information using anappropriate web search.]

solution The shape of the St. Louis Gateway Arch is based on thegraph of coshx. Specifically, the shape of the St. Louis Gateway Arch isobtained by starting with the graph of coshx on a suitable interval,then stretching vertically by a certain amount, then stretchinghorizontally by a certain amount, then reflecting through the horizontalaxis, and finally shifting the graph up a suitable amount.

If we use feet as the units of measurement, then several web sitesreport that the shape of the St. Louis Gateway Arch is the graph of thefunction f defined by

f(x) = 693.8597− 68.7672 cosh(0.0100333x),

where the domain of f is the interval [−299.2239,299.2239] (thus theSt. Louis Gateway Arch is just slightly less than 600 feet wide at thebase).

Here is the graph of the function f defined above:


�300 300

600

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