Download Solutions to Exercises, Section 3.1

Instructor’s Solutions Manual, Section 3.1
Exercise 1
Solutions to Exercises, Section 3.1
For Exercises 1–8, evaluate the indicated quantities. Do not use a
calculator because otherwise you will not gain the understanding that
these exercises should help you attain.
1. 253/2
3
solution 253/2 = (251/2 ) = 53 = 125
Instructor’s Solutions Manual, Section 3.1
2. 85/3
5
solution 85/3 = (81/3 ) = 25 = 32
Exercise 2
Instructor’s Solutions Manual, Section 3.1
3. 323/5
3
solution 323/5 = (321/5 ) = 23 = 8
Exercise 3
Instructor’s Solutions Manual, Section 3.1
4. 813/4
3
solution 813/4 = (811/4 ) = 33 = 27
Exercise 4
Instructor’s Solutions Manual, Section 3.1
Exercise 5
5. 32−4/5
−4
solution 32−4/5 = (321/5 )
= 2−4 =
1
1
=
4
2
16
Instructor’s Solutions Manual, Section 3.1
Exercise 6
6. 8−5/3
−5
solution 8−5/3 = (81/3 )
= 2−5 =
1
1
=
5
2
32
Instructor’s Solutions Manual, Section 3.1
7. (−8)7/3
solution
7
(−8)7/3 = (−8)1/3
= (−2)7
= −128
Exercise 7
Instructor’s Solutions Manual, Section 3.1
8. (−27)4/3
solution
4
(−27)4/3 = (−27)1/3
= (−3)4
= 81
Exercise 8
Instructor’s Solutions Manual, Section 3.1
For Exercises 9–20, expand the indicated expression.
√
9. (2 + 3)2
solution
(2 +
√
√
√ 2
3)2 = 22 + 2 · 2 · 3 + 3
√
=4+4 3+3
√
=7+4 3
Exercise 9
Instructor’s Solutions Manual, Section 3.1
10. (3 +
√
2)2
solution
(3 +
√
√
√ 2
2)2 = 32 + 2 · 3 · 2 + 2
√
=9+6 2+2
√
= 11 + 6 2
Exercise 10
Instructor’s Solutions Manual, Section 3.1
√
11. (2 − 3 5)2
solution
√
√ 2
√
(2 − 3 5)2 = 22 − 2 · 2 · 3 · 5 + 32 · 5
√
= 4 − 12 5 + 9 · 5
√
= 49 − 12 5
Exercise 11
Instructor’s Solutions Manual, Section 3.1
√
12. (3 − 5 2)2
solution
√
√ 2
√
(3 − 5 2)2 = 32 − 2 · 3 · 5 · 2 + 52 · 2
√
= 9 − 30 2 + 25 · 2
√
= 59 − 30 2
Exercise 12
Instructor’s Solutions Manual, Section 3.1
13. (2 +
√
Exercise 13
3)4
solution Note that
(2 +
√
√
2
3)4 = (2 + 3)2 .
√
Thus first we need to compute (2 + 3)2 . We already did that in
Exercise 9, getting
√
√
(2 + 3)2 = 7 + 4 3.
Thus
(2 +
√
√
2
3)4 = (2 + 3)2
√
= (7 + 4 3)2
√
√ 2
= 72 + 2 · 7 · 4 · 3 + 4 2 · 3
√
= 49 + 56 3 + 16 · 3
√
= 97 + 56 3.
Instructor’s Solutions Manual, Section 3.1
14. (3 +
√
Exercise 14
2)4
solution Note that
(3 +
√
√
2
2)4 = (3 + 2)2 .
√
Thus first we need to compute (3 + 2)2 . We already did that in
Exercise 10, getting
√
√
(3 + 2)2 = 11 + 6 2.
Thus
(3 +
√
√
2
2)4 = (3 + 2)2
√
= (11 + 6 2)2
√
√ 2
= 112 + 2 · 11 · 6 · 2 + 62 · 2
√
= 121 + 132 2 + 36 · 2
√
= 193 + 132 2.
Instructor’s Solutions Manual, Section 3.1
15. (3 +
√
x)2
solution
(3 +
√
√
√ 2
x)2 = 32 + 2 · 3 · x + x
√
=9+6 x+x
Exercise 15
Instructor’s Solutions Manual, Section 3.1
16. (5 +
√
x)2
solution
(5 +
√
√
√ 2
x)2 = 52 + 2 · 5 · x + x
√
= 25 + 10 x + x
Exercise 16
Instructor’s Solutions Manual, Section 3.1
17. (3 −
√
2x)2
solution
(3 −
√
√
√ 2
2x)2 = 32 − 2 · 3 · 2x + 2x
√
= 9 − 6 2x + 2x
Exercise 17
Instructor’s Solutions Manual, Section 3.1
18. (5 −
√
3x)2
solution
(5 −
√
√
√ 2
3x)2 = 52 − 2 · 5 · 3x + 3x
√
= 25 − 10 3x + 3x
Exercise 18
Instructor’s Solutions Manual, Section 3.1
√
19. (1 + 2 3x)2
solution
√
√ 2
√
(1 + 2 3x)2 = 12 + 2 · 2 · 3x + 22 · 3x
√
= 1 + 4 3x + 4 · 3x
√
= 1 + 4 3x + 12x
Exercise 19
Instructor’s Solutions Manual, Section 3.1
√
20. (3 + 2 5x)2
solution
√
√ 2
√
(3 + 2 5x)2 = 32 + 2 · 3 · 2 · 5x + 22 · 5x
√
= 9 + 12 5x + 4 · 5x
√
= 9 + 12 5x + 20x
Exercise 20
Instructor’s Solutions Manual, Section 3.1
Exercise 21
For Exercises 21–32, find a formula for the inverse function f −1 of the
indicated function f .
21. f (x) = x 9
solution By the definition of roots, the inverse of f is the function
f −1 defined by
f −1 (y) = y 1/9 .
Instructor’s Solutions Manual, Section 3.1
Exercise 22
22. f (x) = x 12
solution By the definition of roots, the inverse of f is the function
f −1 defined by
f −1 (y) = y 1/12 .
Instructor’s Solutions Manual, Section 3.1
Exercise 23
23. f (x) = x 1/7
solution By the definition of roots, f = g −1 , where g is the function
−1
defined by g(y) = y 7 . Thus f −1 = (g −1 ) = g. In other words,
f −1 (y) = y 7 .
Instructor’s Solutions Manual, Section 3.1
Exercise 24
24. f (x) = x 1/11
solution By the definition of roots, f = g −1 , where g is the function
−1
defined by g(y) = y 11 . Thus f −1 = (g −1 ) = g. In other words,
f −1 (y) = y 11 .
Instructor’s Solutions Manual, Section 3.1
Exercise 25
25. f (x) = x −2/5
solution To find a formula for f −1 , we solve the equation x −2/5 = y
5
for x. Raising both sides of this equation to the power − 2 , we get
−5/2
. Hence
x=y
f −1 (y) = y −5/2 .
Instructor’s Solutions Manual, Section 3.1
Exercise 26
26. f (x) = x −17/7
solution To find a formula for f −1 , we solve the equation x −17/7 = y
7
for x. Raising both sides of this equation to the power − 17 , we get
−7/17
. Hence
x=y
f −1 (y) = y −7/17 .
Instructor’s Solutions Manual, Section 3.1
27. f (x) =
Exercise 27
x4
81
4
solution To find a formula for f −1 , we solve the equation x81 = y for
x. Multiplying both sides by 81 and then raising both sides of this
1
equation to the power 4 , we get x = (81y)1/4 = 811/4 y 1/4 = 3y 1/4 .
Hence
f −1 (y) = 3y 1/4 .
Instructor’s Solutions Manual, Section 3.1
Exercise 28
28. f (x) = 32x 5
solution To find a formula for f −1 , we solve the equation 32x 5 = y
for x. Dividing both sides by 32 and then raising both sides of this
1
equation to the power 5 , we get x = (y/32)1/5 = y 1/5 /321/5 = y 1/5 /2.
Hence
y 1/5
.
f −1 (y) =
2
Instructor’s Solutions Manual, Section 3.1
Exercise 29
29. f (x) = 6 + x 3
solution To find a formula for f −1 , we solve the equation 6 + x 3 = y
for x. Subtracting 6 from both sides and then raising both sides of this
1
equation to the power 3 , we get x = (y − 6)1/3 . Hence
f −1 (y) = (y − 6)1/3 .
Instructor’s Solutions Manual, Section 3.1
Exercise 30
30. f (x) = x 6 − 5
solution To find a formula for f −1 , we solve the equation x 6 − 5 = y
for x. Adding 5 to both sides and then raising both sides of this
1
equation to the power 6 , we get x = (y + 5)1/6 . Hence
f −1 (y) = (y + 5)1/6 .
Instructor’s Solutions Manual, Section 3.1
Exercise 31
31. f (x) = 4x 3/7 − 1
solution To find a formula for f −1 , we solve the equation
4x 3/7 − 1 = y for x. Adding 1 to both sides, then dividing both sides by
7
4, and then raising both sides of this equation to the power 3 , we get
y+1 7/3
. Hence
x= 4
y + 1 7/3
.
f −1 (y) =
4
Instructor’s Solutions Manual, Section 3.1
Exercise 32
32. f (x) = 7 + 8x 5/9
solution To find a formula for f −1 , we solve the equation
7 + 8x 5/9 = y for x. Subtracting 7 from both sides, then dividing both
9
sides by 8, and then raising both sides of this equation to the power 5 ,
y−7 9/5
. Hence
we get x = 8
f −1 (y) =
y − 7 9/5
.
8
Instructor’s Solutions Manual, Section 3.1
Exercise 33
For Exercises 33–38, find a formula for (f ◦ g)(x) assuming that f and
g are the indicated functions.
33. f (x) = x 1/2
and g(x) = x 3/7
solution
1/2
(f ◦ g)(x) = f g(x) = f (x 3/7 ) = (x 3/7 )
= x 3/14
Instructor’s Solutions Manual, Section 3.1
34. f (x) = x 5/3
Exercise 34
and g(x) = x 4/9
solution
5/3
(f ◦ g)(x) = f g(x) = f (x 4/9 ) = (x 4/9 )
= x 20/27
Instructor’s Solutions Manual, Section 3.1
35. f (x) = 3 + x 5/4
and
Exercise 35
g(x) = x 2/7
solution
5/4
(f ◦ g)(x) = f g(x) = f (x 2/7 ) = 3 + (x 2/7 )
= 3 + x 5/14
Instructor’s Solutions Manual, Section 3.1
36. f (x) = x 2/3 − 7
and
Exercise 36
g(x) = x 9/16
solution
2/3
−7
(f ◦ g)(x) = f g(x) = f (x 9/16 ) = (x 9/16 )
= x 3/8 − 7
Instructor’s Solutions Manual, Section 3.1
37. f (x) = 5x
√
2
and
g(x) = x
Exercise 37
√
8
solution
√
(f ◦ g)(x) = f g(x) = f (x 8 )
= 5(x
√
8
√
2
)
= 5x
√
16
= 5x 4
Instructor’s Solutions Manual, Section 3.1
38. f (x) = 7x
√
12
and g(x) = x
Exercise 38
√
3
solution
√
(f ◦ g)(x) = f g(x) = f (x 3 )
= 7(x
√
3
√
)
12
= 7x
√
36
= 7x 6
Instructor’s Solutions Manual, Section 3.1
Exercise 39
For Exercises 39–46, find all real numbers x that satisfy the indicated
equation.
√
39. x − 5 x + 6 = 0
√
solution This equation involves x; thus we make the substitution
√
√
x = y. Squaring both sides of the equation x = y gives x = y 2 .
With these substitutions, the equation above becomes
y 2 − 5y + 6 = 0.
This new equation can now be solved either by factoring the left side or
by using the quadratic formula (see Section 2.2). Let’s factor the left
side, getting
(y − 2)(y − 3) = 0.
Thus y = 2 or y = 3 (the same result could have been obtained by
using the quadratic formula).
√
√
√
Substituting x for y now shows that x = 2 or x = 3. Thus x = 4
or x = 9.
Instructor’s Solutions Manual, Section 3.1
Exercise 40
√
40. x − 7 x + 12 = 0
√
solution This equation involves x; thus we make the substitution
√
√
x = y. Squaring both sides of the equation x = y gives x = y 2 .
With these substitutions, the equation above becomes
y 2 − 7y + 12 = 0.
This new equation can now be solved either by factoring the left side or
by using the quadratic formula (see Section 2.2). Let’s factor the left
side, getting
(y − 3)(y − 4) = 0.
Thus y = 3 or y = 4 (the same result could have been obtained by
using the quadratic formula).
√
√
√
Substituting x for y now shows that x = 3 or x = 4. Thus x = 9
or x = 16.
Instructor’s Solutions Manual, Section 3.1
41. x −
√
Exercise 41
x=6
√
solution This equation involves x; thus we make the substitution
√
√
x = y. Squaring both sides of the equation x = y gives x = y 2 .
Making these substitutions and subtracting 6 from both sides, we have
y 2 − y − 6 = 0.
This new equation can now be solved either by factoring the left side or
by using the quadratic formula. Let’s use the quadratic formula, getting
y=
1±
√
1 + 24
1±5
=
.
2
2
Thus y = 3 or y = −2 (the same result could have been obtained by
factoring).
√
√
√
Substituting x for y now shows that x = 3 or x = −2. The first
possibility corresponds to the solution x = 9. There are no real
√
numbers x such that x = −2. Thus x = 9 is the only solution to this
equation.
Instructor’s Solutions Manual, Section 3.1
42. x −
√
Exercise 42
x = 12
√
solution This equation involves x; thus we make the substitution
√
√
x = y. Squaring both sides of the equation x = y gives x = y 2 .
Making these substitutions and subtracting 12 from both sides, we have
y 2 − y − 12 = 0.
This new equation can now be solved either by factoring the left side or
by using the quadratic formula. Let’s use the quadratic formula, getting
y=
1±
√
1 + 48
1±7
=
.
2
2
Thus y = 4 or y = −3 (the same result could have been obtained by
factoring).
√
√
√
Substituting x for y now shows that x = 4 or x = −3. The first
possibility corresponds to the solution x = 16. There are no real
√
numbers x such that x = −3. Thus x = 16 is the only solution to this
equation.
Instructor’s Solutions Manual, Section 3.1
Exercise 43
43. x 2/3 − 6x 1/3 = −8
solution This equation involves x 1/3 and x 2/3 ; thus we make the
substitution x 1/3 = y. Squaring both sides of the equation x 1/3 = y
gives x 2/3 = y 2 . Making these substitutions and adding 8 to both sides,
we have
y 2 − 6y + 8 = 0.
This new equation can now be solved either by factoring the left side or
by using the quadratic formula. Let’s factor the left side, getting
(y − 2)(y − 4) = 0.
Thus y = 2 or y = 4 (the same result could have been obtained by
using the quadratic formula).
Substituting x 1/3 for y now shows that x 1/3 = 2 or x 1/3 = 4. Thus
x = 23 or x = 43 . In other words, x = 8 or x = 64.
Instructor’s Solutions Manual, Section 3.1
Exercise 44
44. x 2/3 + 3x 1/3 = 10
solution This equation involves x 1/3 and x 2/3 ; thus we make the
substitution x 1/3 = y. Squaring both sides of the equation x 1/3 = y
gives x 2/3 = y 2 . Making these substitutions and subtracting 10 from
both sides, we have
y 2 + 3y − 10 = 0.
This new equation can now be solved either by factoring the left side or
by using the quadratic formula. Let’s factor the left side, getting
(y − 2)(y + 5) = 0.
Thus y = 2 or y = −5 (the same result could have been obtained by
using the quadratic formula).
Substituting x 1/3 for y now shows that x 1/3 = 2 or x 1/3 = −5. Thus
x = 23 or x = (−5)3 . In other words, x = 8 or x = −125.
Instructor’s Solutions Manual, Section 3.1
Exercise 45
45. x 4 − 3x 2 = 10
solution This equation involves x 2 and x 4 ; thus we make the
substitution x 2 = y. Squaring both sides of the equation x 2 = y gives
x 4 = y 2 . Making these substitutions and subtracting 10 from both
sides, we have
y 2 − 3y − 10 = 0.
This new equation can now be solved either by factoring the left side or
by using the quadratic formula. Let’s factor the left side, getting
(y − 5)(y + 2) = 0.
Thus y = 5 or y = −2 (the same result could have been obtained by
using the quadratic formula).
Substituting x 2 for y now shows that x 2 = 5 or x 2 = −2. The first of
√
√
these equations implies that x = 5 or x = − 5; the second equation
is not satisfied by any real value of x. In other words, the original
√
√
equation implies that x = 5 or x = − 5.
Instructor’s Solutions Manual, Section 3.1
Exercise 46
46. x 4 − 8x 2 = −15
solution This equation involves x 2 and x 4 ; thus we make the
substitution x 2 = y. Squaring both sides of the equation x 2 = y gives
x 4 = y 2 . Making these substitutions and adding 15 to both sides, we
have
y 2 − 8y + 15 = 0.
This new equation can now be solved either by factoring the left side or
by using the quadratic formula. Let’s factor the left side, getting
(y − 3)(y − 5) = 0.
Thus y = 3 or y = 5 (the same result could have been obtained by
using the quadratic formula).
Substituting x 2 for y now shows that x 2 = 3 or x 2 = 5. Thus the
√
√
√
original equation has four solutions: x = 3, x = − 3, x = 5, or
√
x = − 5.
Instructor’s Solutions Manual, Section 3.1
Exercise 47
47. Suppose x is a number such that 3x = 4. Evaluate 3−2x .
solution
−2
3−2x = (3x )
= 4−2
=
1
42
=
1
16
Instructor’s Solutions Manual, Section 3.1
Exercise 48
48. Suppose x is a number such that 2x = 13 . Evaluate 2−4x .
solution
−4
2−4x = (2x )
=
1 −4
3
= 34
= 81
Instructor’s Solutions Manual, Section 3.1
Exercise 49
49. Suppose x is a number such that 2x = 5. Evaluate 8x .
solution
x
8x = (23 )
= 23x
3
= (2x )
= 53
= 125
Instructor’s Solutions Manual, Section 3.1
Exercise 50
50. Suppose x is a number such that 3x = 5. Evaluate
1 x
x
= (3−2 )
solution
9
= 3−2x
−2
= (3x )
= 5−2
=
1
25
1 x
9
.
Instructor’s Solutions Manual, Section 3.1
Exercise 51
For Exercises 51–56, evaluate the indicated quantities assuming that f
and g are the functions defined by
f (x) = 2x
and
g(x) =
x+1
.
x+2
51. (f ◦ g)(−1)
solution
(f ◦ g)(−1) = f g(−1) = f (0) = 20 = 1
Instructor’s Solutions Manual, Section 3.1
Exercise 52
52. (g ◦ f )(0)
solution
(g ◦ f )(0) = g f (0) = g(20 ) = g(1) =
2
3
Instructor’s Solutions Manual, Section 3.1
53.
(f ◦ g)(0)
solution
1
(f ◦ g)(0) = f g(0) = f ( 2 ) = 21/2 ≈ 1.414
Exercise 53
Instructor’s Solutions Manual, Section 3.1
54.
Exercise 54
(g ◦ f )( 32 )
solution
3 23/2 + 1
3
(g ◦ f )( 2 ) = g f ( 2 ) = g(23/2 ) = 3/2
2
+2
≈ 0.793
Instructor’s Solutions Manual, Section 3.1
55.
Exercise 55
(f ◦ f )( 12 )
solution
1 1
(f ◦ f )( 2 ) = f f ( 2 ) = f (21/2 )
≈ f (1.41421)
= 21.41421
≈ 2.66514
Instructor’s Solutions Manual, Section 3.1
56.
Exercise 56
(f ◦ f )( 35 )
solution
3 3
(f ◦ f )( 5 ) = f f ( 5 ) = f (23/5 )
≈ f (1.51572)
= 21.51572
≈ 2.85941
Instructor’s Solutions Manual, Section 3.1
Exercise 57
57. Find an integer m such that
√
2
(3 + 2 5)2 − m
is an integer.
√
solution First we evaluate (3 + 2 5)2 :
√
√ 2
√
(3 + 2 5)2 = 32 + 2 · 3 · 2 · 5 + 22 · 5
√
= 9 + 12 5 + 4 · 5
√
= 29 + 12 5.
Thus
√
√
2
(3 + 2 5)2 − m = (29 + 12 5 − m)2 .
If we choose m = 29, then we have
√
√
2
(3 + 2 5)2 − m = (12 5)2
= 122 ·
√
2
5
= 122 · 5,
which is an integer. Any choice other than m = 29 will leave a term
√
√
involving 5 when (29 + 12 5 − m)2 is expanded. Thus m = 29 is the
only solution to this exercise.
Instructor’s Solutions Manual, Section 3.1
Exercise 58
58. Find an integer m such that
√ 2
2
(5 − 2 3) − m
is an integer.
√
solution First we evaluate (5 − 2 3)2 :
√
√ 2
√
(5 − 2 3)2 = 52 − 2 · 5 · 2 · 3 + 22 · 3
√
= 25 − 20 3 + 4 · 3
√
= 37 − 20 3.
Thus
√
√
2
(5 − 2 3)2 − m = (37 − 20 3 − m)2 .
If we choose m = 37, then we have
√
√
2
(5 − 2 3)2 − m = (−20 3)2
= (−20)2 ·
√
2
3
= (−20)2 · 3,
which is an integer. Any choice other than m = 37 will leave a term
√
√
involving 3 when (37 − 20 3 − m)2 is expanded. Thus m = 37 is the
only solution to this exercise.
Instructor’s Solutions Manual, Section 3.1
Problem 59
Solutions to Problems, Section 3.1
√
√
59. Sketch the graph of the functions x + 1 and x + 1 on the interval
[0, 4].
√
√
solution The graphs of x + 1 (blue) and x + 1 (red) on the interval
[0, 4] are shown below.
y
3
2
1
1
2
3
4
x
√
The graph of x + 1 is obtained by shifting up one unit the graph of
√
x on the interval [0, 4].
√
The graph of x + 1 is obtained by shifting left one unit the graph of
√
x on the interval [1, 5].
Instructor’s Solutions Manual, Section 3.1
Problem 60
60. Sketch the graph of the functions 2x 1/3 and (2x)1/3 on the interval
[0, 8].
solution The graphs of 2x 1/3 (blue) and (2x)1/3 (red) on the interval
[0, 8] are shown below.
y
4
3
2
1
2
4
6
8
x
The graph of 2x 1/3 is obtained by vertically stretching the graph of x 1/3
on the interval [0, 8] by a factor of 2.
Note that (2x)1/3 = 21/3 x 1/3 . Thus the graph of (2x)1/3 is obtained by
vertically stretching the graph of x 1/3 on the interval [0, 8] by a factor
of 21/3 (approximately 1.26).
Instructor’s Solutions Manual, Section 3.1
Problem 61
61. Sketch the graphs of the functions x 1/4 and x 1/5 on the interval [0, 81].
solution The graphs of x 1/4 (blue) and x 1/5 (red) on the interval
[0, 81] are shown below. For x in the interval (1, 81), we have the
inequality x 1/4 > x 1/5 . However, for x in the interval (0, 1), the reverse
inequality x 1/4 < x 1/5 holds (although the interval (0, 1) is such a small
part of the interval [0, 81] that this behavior is difficult to see in the
graph below):
y
3
8115
1
81
x
Instructor’s Solutions Manual, Section 3.1
62. Show that
Problem 62
√
2 + 3 = 32 + 12 .
solution We need to verify that the square of
Here is that computation:
3
2
+
2
1
2
=
3
2
+2
=2+
√
3
3
2
·
1
2
+
3
2
1
2
+
1
2
equals 2 +
√
3.
Instructor’s Solutions Manual, Section 3.1
63. Show that
Problem 63
√
2 − 3 = 32 − 12 .
solution We need to verify that the square of
Here is that computation:
3
2
−
2
1
2
=
3
2
−2
=2−
√
3
3
2
·
1
2
+
3
2
1
2
−
1
2
equals 2 −
√
3.
Instructor’s Solutions Manual, Section 3.1
64. Show that
Problem 64
√
√
9 − 4 5 = 5 − 2.
solution We need to verify that the square of
Here is that computation:
√
√
√
( 5 − 2)2 = 5 − 2 5 · 2 + 4
√
=9−4 5
√
5 − 2 equals 9 − 4 5.
Instructor’s Solutions Manual, Section 3.1
Problem 65
√
√
65. Show that (23 − 8 7)1/2 = 4 − 7.
√
√
solution We need to verify that the square of 4 − 7 equals 23 − 8 7.
Here is that computation:
(4 −
√
√
7)2 = 16 − 2 · 4 7 + 7
√
= 23 − 8 7
Instructor’s Solutions Manual, Section 3.1
Problem 66
66. Make up a problem similar in form to the problem above, without
duplicating anything in this book.
solution
In general there is no way to simplify an expression of the
√
form m + n k. Thus the way to make up a problem similar to the
previous one, start with an expression of the desired form and square
√
it. For example, if we start by squaring 5 − 3, then we have
(5 −
√
√
3)2 = 25 − 2 · 5 3 + 3
√
= 28 − 10 3.
Thus we can now make up the following problem:
√
√ 1
Show that 28 − 10 3 2 = 5 − 3.
Instructor’s Solutions Manual, Section 3.1
Problem 67
√
√
67. Show that (99 + 70 2)1/3 = 3 + 2 2.
√
solution We need to verify that the cube of 3 + 2 2 equals
√
99 + 70 2. Here is that computation:
√
√
√
(3 + 2 2)3 = (3 + 2 2)2 (3 + 2 2)
√
√
= (9 + 2 · 3 · 2 2 + 8)(3 + 2 2)
√
√
= (17 + 12 2)(3 + 2 2)
√
√
= 51 + 34 2 + 36 2 + 24 · 2
√
= 99 + 70 2
Instructor’s Solutions Manual, Section 3.1
Problem 68
√
√
68. Show that (−37 + 30 3)1/3 = −1 + 2 3.
√
solution We need to verify that the cube of −1 + 2 3 equals
√
−37 + 30 3. Here is that computation:
√
√
√
(−1 + 2 3)3 = (−1 + 2 3)2 (−1 + 2 3)
√
√
= (1 − 2 · 2 3 + 12)(−1 + 2 3)
√
√
= (13 − 4 3)(−1 + 2 3)
√
√
= −13 + 26 3 + 4 3 − 8 · 3
√
= −37 + 30 3
Instructor’s Solutions Manual, Section 3.1
69. Show that if x and y are positive numbers with x = y, then
√
x−y
√
√ = x + y.
x− y
solution
√
√
x+ y
x−y
x−y
√
√ = √
√ ·√
√
x− y
x− y
x+ y
=
=
√
√
(x − y)( x + y)
x−y
√
x+ y
Problem 69
Instructor’s Solutions Manual, Section 3.1
70. Explain why
Problem 70
10100 ( 10200 + 1 − 10100 )
is approximately equal to
1
2.
solution
10100 ( 10200 + 1 − 10100 )
√
( 10200 + 1 + 10100 )
= 10100 ( 10200 + 1 − 10100 ) · √
( 10200 + 1 + 10100 )
=
10100 (10200 + 1 − 10200 )
√
10200 + 1 + 10100
10100
= √
200
10
+ 1 + 10100
≈
=
10100
+ 10100
10100
1
2
Instructor’s Solutions Manual, Section 3.1
Problem 71
√
71. Explain why the equation x 2 = x is not valid for all real numbers x
√
and should be replaced by the equation x 2 = |x|.
√
solution Note that (−3)2 = 9 = 3. Thus for x = −3, it is not true
√
that x 2 = x.
More generally, if t is a a positive number, then there are two real
numbers whose square equals t; the positive number whose square
√
equals t is defined to be t.
Now for any real number x = 0, the number x 2 is positive, and we can
apply the remark above to t = x 2 . Specifically, there exist two real
numbers whose square equals x 2 ; these two numbers are x and −x. By
√
the remark above, x 2 equals x if x > 0 and equals −x if x < 0. In
√
other words, we have x 2 = |x| (derived here under the assumption
that x = 0, but the desired equation also obviously holds when x = 0).
Instructor’s Solutions Manual, Section 3.1
Problem 72
√
72. Explain why the equation x 8 = x 4 is valid for all real numbers x, with
no necessity for using absolute value.
solution Suppose x is a real number with x = 0. Then x 8 is a
positive number. The two numbers whose square equals x 8 are x 4 and
−x 4 . Regardless of whether x is positive or negative, x 4 is positive.
√
Thus x 8 = x 4 (derived here under the assumption that x = 0, but the
desired equation also obviously holds when x = 0).
Instructor’s Solutions Manual, Section 3.1
Problem 73
73. Show that if x and y are positive numbers, then
x+y <
√
x + y.
[In particular, if x and y are positive numbers, then
√
√
√
x + y = x + y.]
solution Suppose x and y are positive numbers. Then
√ x+y <x+2 x y +y
√
= ( x + y)2 .
Because everything in sight is positive, we can take square roots of both
sides of the inequality above, preserving the direction of the inequality,
getting
√
x + y < x + y.
Instructor’s Solutions Manual, Section 3.1
Problem 74
74. Show that if 0 < x < y, then
√
y − x < y − x.
solution Because x < y, we know that y − x > 0. Applying the result
from the previous problem, we have
Subtracting
√
√
y = x + (y − x) < x + y − x.
x from both sides of the inequality above shows that
√
y − x < y − x.
Instructor’s Solutions Manual, Section 3.1
Problem 75
75. Explain why the spoken phrase “the square root of x plus one” could be
interpreted in two different ways that would not give the same result.
solution The phrase “the square root of x plus one” could mean
“(the square root of x) plus one”, which in symbols is
√
x + 1.
Or the phrase “the square root of x plus one” could mean “the square
root of (x plus one)”, which in symbols is
√
x + 1.
These two interpretations of the spoken phrase do not give the same
result. For example, if x = 4, the first interpretation gives 3 and the
√
second interpretation gives 5.
Instructor’s Solutions Manual, Section 3.1
Problem 76
76. One of the graphs in this section suggests that
√
√
x< 3x
if
0<x<1
√
√
x> 3x
if
x > 1.
and
Explain why each of these inequalities holds.
solution First suppose 0 < x < 1. Then
x 1/6 < 1
because if x 1/6 were greater than or equal to 1, then multiplying x 1/6
by itself six times (which produces x) would give a number greater than
or equal to 1. Multiplying both sides of the inequality above by x 1/3
and using the identity x 1/6 · x 1/3 = x 1/2 gives
x 1/2 < x 1/3 ,
as desired.
Now suppose x > 1. Then
x 1/6 > 1
because if x 1/6 were less than or equal to 1, then multiplying x 1/6 by
itself six times (which produces x) would give a number less than or
Instructor’s Solutions Manual, Section 3.1
Problem 76
equal to 1. Multiplying both sides of the inequality above by x 1/3 and
using the identity x 1/6 · x 1/3 = x 1/2 gives
x 1/2 > x 1/3 ,
as desired.
Instructor’s Solutions Manual, Section 3.1
Problem 77
77. What is the domain of the function (3 + x)1/4 ?
solution For (3 + x)1/4 to be defined as a real number, we must have
3 + x ≥ 0, which is equivalent to the inequality x ≥ −3. Thus the
domain of the function (3 + x)1/4 is the interval [−3, ∞).
Instructor’s Solutions Manual, Section 3.1
Problem 78
78. What is the domain of the function (1 + x 2 )1/8 ?
solution Because x 2 ≥ 0 for every real number x, we have 1 + x 2 ≥ 1
for every real number x. In particular, 1 + x 2 is positive for every real
number x, which means that (1 + x 2 )1/8 is defined as a real number for
every real number x. Thus the domain of the function (1 + x 2 )1/8 is the
set of real numbers.
Instructor’s Solutions Manual, Section 3.1
Problem 79
79. Suppose x is a positive number. Using only the definitions of roots and
integer powers, explain why
3
6
(x 1/2 ) = (x 1/4 ) .
solution We have
(x 1/4 )2
2
= (x 1/4 )2 · (x 1/4 )2 = x 1/4 · x 1/4 · x 1/4 · x 1/4 = x,
where the last equality above holds because of the definition of x 1/4 .
2
The equation (x 1/4 )2 = x states that (x 1/4 )2 is a number whose
square equals x. Thus (x 1/4 )2 must equal x 1/2 (because everything in
sight is positive). Now if we cube both sides of the equality
x 1/2 = (x 1/4 )2 , we get
3
(x 1/2 ) = (x 1/4 )2 · (x 1/4 )2 · (x 1/4 )2
= x 1/4 · x 1/4 · x 1/4 · x 1/4 · x 1/4 · x 1/4
6
= (x 1/4 ) ,
as desired.
Instructor’s Solutions Manual, Section 3.1
Problem 80
80. Suppose x is a positive number and n is a positive integer. Using only
the definitions of roots and integer powers, explain why
n
2n
(x 1/2 ) = (x 1/4 )
.
solution In the solution to the previous problem we showed that
x 1/2 = (x 1/4 )2 .
Raise both sides of this equality to the nth power, getting
n
n
(x 1/2 ) = (x 1/4 )2
= (x 1/4 )2 · (x 1/4 )2 · · · · · (x 1/4 )2
n times
= x
1/4
·x
1/4
1/4
· · · · · x 2n times
= (x
as desired.
1/4 2n
)
,
Instructor’s Solutions Manual, Section 3.1
Problem 81
81. Suppose x is a positive number and n and p are positive integers.
Using only the definitions of roots and integer powers, explain why
np
n
(x 1/2 ) = (x 1/(2p) )
.
solution We have
(x 1/(2p) )p
2
= (x 1/(2p) )p · (x 1/(2p) )p
1/(2p)
= x 1/(2p) · x 1/(2p)
· · · · · x
2p times
= x,
where the last equality above holds because of the definition of x 1/(2p) .
2
The equation (x 1/(2p) )p = x states that (x 1/(2p) )p is a number
whose square equals x. Thus (x 1/(2p) )p must equal x 1/2 (because
everything in sight is positive). Raise both sides of the equality
x 1/2 = (x 1/(2p) )p to the nth power, getting
n
n
(x 1/2 ) = (x 1/(2p) )p
= (x 1/(2p) )p · (x 1/(2p) )p · · · · · (x 1/(2p) )p
n times
= x
1/(2p)
·x
1/(2p)
· · · · · x
np times
np
= (x 1/(2p) )
.
1/(2p)
Instructor’s Solutions Manual, Section 3.1
Problem 82
82. Suppose x is a positive number and m, n, and p are positive integers.
Using only the definitions of roots and integer powers, explain why
np
n
(x 1/m ) = (x 1/(mp) )
.
solution We have
(x 1/(mp) )p
m
= (x 1/(mp) )p · (x 1/(mp) )p · · · · · (x 1/(mp) )p
m times
1/(mp)
= x 1/(mp) · x 1/(mp)
· · · · · x
mp times
= x,
where the last equality above holds because of the definition of x 1/(mp) .
m
= x states that (x 1/(mp) )p is a number
The equation (x 1/(mp) )p
th
whose m power equals x. Thus (x 1/(mp) )p must equal x 1/m (because
everything in sight is positive). Raise both sides of the equality
x 1/m = (x 1/(mp) )p
to the nth power, getting
Instructor’s Solutions Manual, Section 3.1
Problem 82
n
n
(x 1/m ) = (x 1/(mp) )p
= (x 1/(mp) )p · (x 1/(mp) )p · · · · · (x 1/(mp) )p
n times
1/(mp)
= x 1/(mp) · x 1/(mp)
· · · · · x
np times
np
= (x 1/(mp) )
as desired.
,
Instructor’s Solutions Manual, Section 3.1
Problem 83
83. Using the result from the problem above, explain why the definition of
exponentiation of a positive number by a positive rational number
gives the same result even if the positive rational number is not
expressed in reduced form.
solution Suppose x is a positive number and m and n are positive
m
integers such that n is in reduced form (meaning that m and n have no
n
common factors other than 1). Then x n/m is defined to equal (x 1/m ) .
However, suppose p is a positive integer. Then
np
mp
np
mp
equals
n
m,
although
is not in reduced form. If we apply the definition of exponentiation
by a rational number to x (np)/(mp) , we see that x (np)/(mp) is defined to
np
equal (x 1/(mp) ) , which the previous problem tells us is equal to
n
(x 1/m ) .
Thus we get the same result even if we use a fraction not in reduced
form.
Instructor’s Solutions Manual, Section 3.1
84. Using the result that
25/2 is irrational.
Problem 84
√
2 is irrational (proved in Section 0.1), show that
solution Note that
√
2=
25/2
25/2
25/2
.
=
=
24/2
22
4
√
Thus if 25/2 were a rational number, then we would have written 2 as
√
the quotient of two rational numbers, which would mean that 2 would
be a rational number (which is not true). Thus 25/2 is not a rational
number.
Instructor’s Solutions Manual, Section 3.1
85. Using the result that
solution Note that
Problem 85
√
2 is irrational, explain why 21/6 is irrational.
√
3
2 = 21/6 .
√
Thus if 21/6 were a rational number, then we would have written 2 as
√
the cube of a rational number, which would mean that 2 would be a
rational number (which is not true). Thus 21/6 is not a rational number.
Instructor’s Solutions Manual, Section 3.1
Problem 86
86. Suppose you have a calculator that can only compute square roots.
Explain how you could use this calculator to compute 71/8 .
solution Note that
1/2 1/2
71/8 = (71/2 )
,
which can be written as
7
1/8
#
√
=
7.
Thus to calculate 71/8 on a calculator that can only compute square
roots, enter 7 and then press the square-root button three times.
Instructor’s Solutions Manual, Section 3.1
Problem 87
87. Suppose you have a calculator that can only compute square roots and
can multiply. Explain how you could use this calculator to compute
73/4 .
solution Note that
1/2
73/4 = 71/2 · 71/4 = 71/2 · (71/2 )
,
which can be written as
73/4 =
√ √
7·
7.
Thus to calculate 73/4 on a calculator that can only compute square
roots and can multiply, enter 7 and then press the square-root button,
then multiply that result by the number obtained by entering 7 and
pressing the square-root button twice.
Instructor’s Solutions Manual, Section 3.1
Problem 88
88. Give an example of three irrational numbers x, y, and z such that
xyz
is a rational number.
solution Note that
√
2
2 = 21/4 .
√
Thus if 21/4 were a rational number, then we would have written 2 as
√
the square of a rational number, which would mean that 2 would be a
rational number (which is not true). Thus 21/4 is not a rational number.
Now take x = 21/4 , y = 21/4 , and z = 21/2 . Then x, y, and z are
irrational numbers and
xyz = 21/4 · 21/4 · 21/2 = 2.
Instructor’s Solutions Manual, Section 3.1
Problem 89
89. Give an example of three irrational numbers x, y, and z such that
z
(x y )
is a rational number.
solution Let x = y = z =
z
√
2. Then
(x y ) = x yz =
√ √2·√2 √ 2
2
= 2 = 2.
Instructor’s Solutions Manual, Section 3.1
Problem 90
90. Is the function f defined by f (x) = 2x for every real number x an even
function, an odd function, or neither?
solution Note that f (−1) = 12 and f (1) = 2. Thus f (−1) = f (1) and
f (−1) = −f (1). Hence f is neither an even function nor an odd
function.
Instructor’s Solutions Manual, Section 3.1
Problem 91
91. What is wrong with the following string of equalities, which seems to
show that −1 = 1?
−1 = i · i = −1 −1 = (−1)(−1) = 1 = 1
solution The error in the string of equalities above is with the
“equality” −1 −1 = (−1)(−1), which would follow from the identity
a b = ab
by taking a = −1 and b = −1. However, as stated in the text, this
identity assumes that a and b are positive. Thus this identity cannot be
used with a = −1 and b = −1.