Download CHAPTER EIGHT

8.1 SOLUTIONS
253
CHAPTER EIGHT
Solutions for Section 8.1
EXERCISES
1. (a) Since 9 − x is defined for any value of x, the domain of m(x) = 9 − x is all real numbers.
(b) For any number y, we can solve the equation m(x) = y for x:
9−x = y
9 = y+x
9 − y = x.
So if we input 9 − y, we get m(9 − y) = y. Since y can be any number, the range is also all real numbers.
2. (a) The expression x2 is defined for all values of x. Thus,
Domain: All real numbers.
(b) To find the range, we want to know all possible output values. Since x2 ≥ 0 for all x, we see that the range is y ≥ 0.
Alternately, we solve the equation y = f (x) for x in terms of y. Since
y = x2 ,
taking the square root of both sides gives
Since
√
y is only defined for y ≥ 0, we have
√
y = x.
Range: y ≥ 0.
3. (a) The expression 7 is trivially defined for all values of x since no values of x make it undefined. Thus,
Domain: All real numbers.
(b) To find the range, we want to know all possible output values. Since the only possible y-value is 7, the range is the
single number 7. We can also see this by looking at a graph of this function, which is a horizontal line at a height of
7 on the y axis. We have
Range: y = 7.
4. (a) The expression x2 − 3 is defined for all values of x so we have
Domain: All real numbers.
(b) To find the range, we want to know all possible output values. Since x2 ≥ 0 for all x, we have x2 − 3 ≥ −3 for all
x, and we see that the range is y ≥ −3. Alternately, we solve the equation y = f (x) for x in terms of y. We have
y = x2 − 3
y + 3 = x2
Since
√
x=±
y + 3 is only defined for y ≥ −3, we have
p
y + 3.
Range: y ≥ −3.
254
Chapter Eight /SOLUTIONS
5. (a) Since x − 3 is defined for any value of x, the domain is all real numbers.
(b) For any y we can solve the equation x − 3 = y for x. The solution is x = y + 3. This tells us that f (y + 3) = y.
Since y can be any number, the range is all real numbers.
6. This is a linear function with nonzero slope, so the domain is all real numbers and the range is all real numbers. To see
this more explicitly, we discuss each in turn.
(a) The expression 5x − 1 is defined for all values of x so we have
Domain: All real numbers.
(b) To find the range, we want to know all possible output values. We solve the equation y = f (x) for x in terms of y.
We have
y = 5x − 1
y + 1 = 5x
y+1
x=
.
5
Since (y + 1)/5 is defined for all y, we have
Range: All real numbers.
√
7. (a) The expression x − 4 is undefined if x < 4, since then x − 4 is negative, and the square root of a negative number
is not
√ defined. Also, we can’t have x = 4, since then x − 4 = 0, so we would be dividing by zero in the expression
1/ x − 4. Thus, to evaluate f (x), we must have x − 4 > 0. Thus
Domain: x > 4.
(b) To find the range, we want to know all possible output values. We solve the equation y = f (x) for x in terms of y.
Since
1
y= √
,
x−4
squaring gives
1
y2 =
,
x−4
and multiplying by x − 4 gives
y 2 (x − 4) = 1
y 2 x − 4y 2 = 1
y 2 x = 1 + 4y 2
1 + 4y 2
x=
.
y2
This formula tells us how to find the x-value which corresponds to a given y-value. The √
formula works for any y
except y = 0 (which puts a 0 in the denominator). We know that y must be positive, since x − 4 is positive, so we
have
Range: y > 0.
√
8. (a) The expression x is only defined for x ≥ 0 so we have
Domain: x ≥ 0.
(b) To find the
√ range, we want to know all possible output values. Since
see that x + 1 ≥ 1 for all x. Thus, we have y ≥ 1 so
Range: y ≥ 1.
√
x ≥ 0 for all x, we can add 1 to both sides to
8.1 SOLUTIONS
9. (a) The expression
√
255
x + 1 is only defined for x + 1 ≥ 0. Subtracting 1 from both sides, we have x ≥ −1. Thus,
Domain: x ≥ −1.
(b) To find the range, we want to know all
√ possible output values. Since the square root of any expression is always
greater than or equal to zero, we have x + 1 ≥ 0 for all x, so y ≥ 0 and we have
Range: y ≥ 0.
10. (a) The expression 1/(x − 2) is undefined for x = 2 since we cannot divide by zero. We can divide by any other number
and so the expression is defined for all other x. Thus,
Domain: x 6= 2.
(b) To find the range, we want to know all possible output values. We solve the equation y = f (x) for x in terms of y.
1
x−2
(x − 2)y = 1
y=
xy − 2y = 1
xy = 1 + 2y
1 + 2y
x=
.
y
The expression (1 + 2y)/y is defined for all values of y except y = 0 since we can’t divide by zero. Thus,
Range: y 6= 0.
11. (a) Since we can divide by any number except for zero, we can have any x with x 6= −1 in 1/(x + 1), so the domain is
all real numbers 6= −1. We have
Domain: x 6= −1.
(b) To find the range, we want to know all possible output values. We solve the equation y = f (x) for x in terms of y.
1
+3
x+1
1
y−3 =
x+1
(x + 1)(y − 3) = 1
y=
xy + y − 3x − 3 = 1
xy − 3x = 4 − y
x(y − 3) = 4 − y
4−y
x=
.
y−3
The expression (4 − y)/(y − 3) is defined for all values of y except y = 3 since we can’t divide by zero. Thus,
12. (a) The expression
√
Range: y 6= 3.
2x − 4 is only defined when 2x − 4 ≥ 0 so we see
2x − 4 ≥ 0
2x ≥ 4
The expression
√
x ≥ 2.
2x − 4 is defined for all x ≥ 2, so we have
Domain: x ≥ 2.
256
Chapter Eight /SOLUTIONS
(b) To find the range, we want to know
√ all possible output values. Since the square root of any number is always greater
than or equal to zero, we have 2x − 4 ≥ 0 so y ≥ 0. Thus,
Range: y ≥ 0.
13. The square-root operation is undefined for negative values, so z − 25 must not be negative:
z − 25 ≥ 0
z ≥ 25.
Thus, the domain of h is z ≥ 25.
√
√
The square√root operation returns non-negative values. At z = 25, we have z − 25 = 0 = 0. For larger values
of z, the value z − 25 is positive. This means that
h(z) = 5 + A number greater than or equal to 0,
so the range of h is h(z) ≥ 5.
14. (a) Since the restaurant opens at 2 pm, t = 0, and closes at 2 am, t = 12, a reasonable domain is 0 ≤ t ≤ 12.
(b) Since there cannot be fewer than 0 clients in the restaurant and 200 can fit inside, the range is all numbers y such that
0 ≤ y ≤ 200.
15. The average speed one drives a car is unlikely to be less than 5 miles per hour (mph), and it is unlikely to be greater than
75 mph. So, a reasonable domain is 5 ≤ v ≤ 75. Answers may vary.
16. (a) The domain is 2 ≤ x ≤ 6.
(b) The range is all y such that 1 ≤ y ≤ 3.
17. Judging from the graph, the leftmost point is at x = −8, and the rightmost at x = 1. The lowest value is y = −5,
occurring at x = −8; the highest value is y = 20, occurring at or around x = −3. We conclude:
(a) The domain is −8 ≤ x ≤ 1.
(b) The range is −5 ≤ y ≤ 20.
18. Judging from the graph, the leftmost point is at x = 2, and the rightmost at x = 11. The lowest value is y = 3, occurring
at numerous values of x; the highest value is y = 11 occurring around x = 5.5. We conclude:
(a) The domain is 2 ≤ x ≤ 11.
(b) The range is 3 ≤ y ≤ 11.
19. The graph consists of two segments: one from x = −3 to x = 1 and one from x = 3 to x = 7. The lowest value of y is
y = 0, occurring at x = 1 and x = 3; the highest value is y = 7, occurring at x = 7. We conclude:
(a) The domain is −3 ≤ x ≤ 1 and 3 ≤ x ≤ 7.
(b) The range is 0 ≤ y ≤ 7.
20. (a) The age of the car cannot be negative, so a ≥ 0. Also, since the value cannot be negative, we have
V ≥0
18,000 − 3000a ≥ 0
18,000 ≥ 3000a
6 ≥ a.
We have a ≤ 6, and thus
Domain: 0 ≤ a ≤ 6.
This tells us that the age of the car ranges from 0 years to 6 years, or that the lifetime of the car for which it has value
is 6 years.
(b) The value of a car cannot be negative, so V ≥ 0. Also, since the age a cannot be negative, the largest value of the
expression 18000 − 3000a is 18000 and we have V ≤ 18,000. We have
Range: 0 ≤ V ≤ 18,000.
This tells that the car ranges in value from $18,000 to no value over the life of the car.
8.1 SOLUTIONS
257
21. (a) The number of units of a product cannot be negative, so we have x ≥ 0. Since cost cannot exceed $10,000, we have
C ≤ 10,000
2000 + 4x ≤ 10,000
4x ≤ 8000
x ≤ 2000.
The number of units produced cannot exceed 2000, so we have
Domain: 0 ≤ x ≤ 2000.
(b) The cost ranges from $0 to $10,000, so
Range: 0 ≤ C ≤ 10,000.
22. For any value of k we have
5 − 3x = k
−3x = k − 5
5−k
k−5
=
.
x=
−3
3
So the equation has a solution for all k. This means that every value of k is in the range of the function f (x) = 5 − 3k,
so its range is all real numbers.
23. The equation has no solutions unless k = −1. This means that the range of f is the set containing the single number −1.
PROBLEMS
24. We have
y = 10 − (x − 2)2
= 10 − (A nonnegative number)
= A number less than or equal to 10,
so the range is y ≤ 10.
25. We have
y = (x − 2)2 − 9
= −9 + (x − 2)2
= A number greater than or equal to −9,
so the range is y ≥ −9.
26. We have
y = 18 − 2(x − 2)2
= 2 9 − (x − 2)2
= 2 · (A number less than or equal to 9)
= A number less than or equal to 18,
so y ≤ 18.
27. Keeping in mind that
√
x is nowhere negative, we have
y=
=
p
9 − (x − 2)2
p
A number less than or equal to 9
= A nonnegative number less than or equal to 3,
so the range is 0 ≤ y ≤ 3.
258
Chapter Eight /SOLUTIONS
28. (a) The initial value of N , at t = 0, is 550, and N decreases at a rate of 100 for every increase of 1 in t, so N =
550 − 100t.
(b) People keep leaving until the theater is empty, that is, until N = 0. This happens when 100t = 550, so when t = 5.5,
so the express makes sense for t between 0 and 5.5 minutes.
29. We know that it can travel 24 miles for every gallon, so g(q) = 24q. Since the tank holds a maximum of 14 gallons,
the domain is 0 ≤ q ≤ 14. On an empty tank, the car cannot go anywhere, and on a full tank, the car can travel
g(14) = 24 · 14 = 336 miles, so the range is 0 ≤ d ≤ 336.
30. We know the domain is 10 ≤ H ≤ 40, because the temperature must be no less than 10◦ and no more than 40◦ . We know
that N ≥ 10 because at the highest tolerable temperature it develops in 10 days. At the lowest tolerable temperature of
10◦ , it takes 15 days longer (or 1 day for every 2◦ C), so the range is 10 ≤ N ≤ 25.
31. The minimum number of hours the employee can work in a week is 4 days at 7 hours per day or 28 hours total. The
maximum number of hours is 6 days at 9 hours per day or 54 hours total. Thus, the domain is 28 ≤ t ≤ 54. At $7.25 per
hour, this means the employee can earn between $203 and $391.50 per week, so the range is 203 ≤ P ≤ 391.5.
32. The domain is 40 ≤ r ≤ 100. At $40 per minute the company can afford 2500/40 = 62.5 minutes of advertisements,
whereas at $100 per minute it can afford only 25 minutes, so the range is 25 ≤ T ≤ 62.5.
33. (a) In order of evaluation:
(i) Operation: Subtract 3 from x. This does not restrict the domain, since we can subtract 3 from any number.
(ii) Operation: Divide 2 by the above result. This means that x − 3 can’t equal 0, so x 6= 3.
(b) The domain is x 6= 3.
34. (a) In order of evaluation:
(i) Operation: Subtract 5 from x. This does not restrict the domain, since we can subtract 5 from any number.
(ii) Operation: Take the square root of the above result. This means that x − 5 can’t be negative, so x ≥ 5.
(iii) Operation: Add 1 to the above result. This does not restrict the domain, since we can add 1 to any number.
(b) The domain is x ≥ 5.
35. (a) In order of evaluation:
(i) Operation: Subtract 3 from x. This does not restrict the domain, since we can subtract 3 from any number.
(ii) Operation: Square the above result. This does not restrict the domain, since we can square any number.
(iii) Operation: Subtract the above result from 4. This does not restrict the domain, since we can subtract any number
from 4.
(b) The domain is all real numbers.
36. (a) In order of evaluation:
(i) Operation: Subtract 3 from x. This does not restrict the domain, since we can subtract 3 from any number.
(ii) Operation: Square the above result. This does not restrict the domain, since we can square any number.
(iii) Operation: Subtract the above result from 4. This does not restrict the domain, since we can subtract any number
from 4.
(iv) Operation: Divide 7 by the above result. This means that the result of step (iii) can’t equal 0, so
(x − 3)2 6= 4.
Thus,
x − 3 6= 2
x 6= 5
and x − 3 6= −2
x 6= 1.
(b) The domain is all x except x = 1, 5.
8.1 SOLUTIONS
259
37. (a) In order of evaluation:
(i) Operation: Subtract 3 from x. This operation does not restrict the domain, since we can subtract 3 from any
number.
(ii) Operation: Raise the above result to the power of 1/2. This means that x − 3 can’t be negative, so x ≥ 3.
(iii) Operation: Subtract the above result from 4. This operation does not restrict the domain, since we can subtract
any number from 4.
(b) The domain is x ≥ 3.
38. (a) In order of evaluation:
(i) Operation: Subtract 3 from x. This operation does not restrict the domain, since we can subtract 3 from any
number.
(ii) Operation: Raise the above result to the power of 1/2. This means x − 3 can’t be negative, so x ≥ 3.
(iii) Operation: Subtract the above result from 4. This operation does not restrict the domain, since we can subtract
any number from 4.
(iv) Divide 7 by the above result. This means that the result of step (iii) can’t equal 0, so
(x − 3)1/2 − 4 6= 0
(x − 3)1/2 6= 4.
Thus,
x − 3 6= 16
x 6= 19.
(b) The domain is x ≥ 3, x 6= 19—that is, all numbers greater than or equal to 3 except 19.
39. To solve f (x) = a, we begin by undoing the algebraic operations.
(a) First we undo the operation of dividing by 2:
2a = 5x + 7.
We can undo this operation for any a-value.
(b) Next we undo the operation of adding 7:
We can undo this operation for any a-value.
(c) Last, we undo the operation of multiplying by 5:
2a − 7 = 5x.
2a − 7
= x.
5
We can undo this operation for any a-value.
So the equation can be solved for all values of a. Therefore, the range is all real numbers.
40. To solve f (x) = a, we begin by undoing the algebraic operations.
(a) We take reciprocals in order to undo the division:
1
5x + 7
=
.
a
2
The operation of taking reciprocals means that a must not equal 0.
(b) Next we must undo the operation of division by 2:
2
= 5x + 7.
a
This places no new restrictions on the value of a.
260
Chapter Eight /SOLUTIONS
(c) Next we must undo the operation of adding 7:
2
− 7 = 5x.
a
This places no new restrictions on the value of a.
(d) Finally we must undo the operation of multiplying by 5:
2
a
−7
= x.
5
This places no new restrictions on the value of a.
So the equation can be solved for all values of a except zero. So the range is y 6= 0, that is, all values of y except y = 0.
41. To solve f (x) = a, we begin by undoing the algebraic operations.
(a) First we must undo the operation of multiplying by 2:
a
= (x + 3)2 .
2
We can undo this operation for all values of a.
(b) Next we must undo the operation of squaring:
±
To undo this operation, a/2 can’t be negative, so
q
a
= x + 3.
2
a
≥0
2
a ≥ 0.
(c) Finally, we must undo the operation of adding 3:
±
q
a
− 3 = x.
2
We can undo this operation for any value of the left-hand expression.
So the equation can be solved for non-negative values of a, that is, a ≥ 0. Therefore, the range is y ≥ 0.
42. To solve f (x) = a, we begin by undoing the algebraic operations.
(a) We must first undo the operation of adding 5:
a − 5 = 2(4x + 3)2 .
We can undo this expression for any value of a.
(b) Next we must undo the operation of multiplying by 2:
a−5
= (4x + 3)2 .
2
We can undo this operation for any value of the left-hand expression.
(c) Next we must undo the operation of squaring:
±
r
a−5
= 4x + 3.
2
To undo this operation, the value of (a − 5)/2 can’t be negative:
a−5
≥0
2
a−5 ≥ 0
a ≥ 5,
so a must be greater than or equal to 5.
8.1 SOLUTIONS
261
(d) We must now undo the operation of adding 3:
±
r
a−5
− 3 = 4x.
2
We can undo this operation for any value of the left-hand expression.
(e) Finally we must undo the operation of multiplying by 4:
p a−5
−3
= x.
4
We can undo this operation for any value of the left-hand expression.
±
2
So the equation can be solved for all values of a ≥ 5. Therefore, the range is y ≥ 5.
43. We are told that the domain of z(t) is t ≤ 7. We infer that for values of t larger than 7, the input to the square root
operation is negative. This means b − 2t < 0 for t > 7, so b must equal 14.
Another way to see this is to find the domain of z in terms of b. For z(t) to be defined, we require b − 2t ≥ 0:
b − 2t ≥ 0
b ≥ 2t
t ≤ 0.5b.
Since we know the domain is t ≤ 7, we have 0.5b = 7, so b = 14.
44. Thinking in terms of output values, we see that the output of the square root operation must not be negative, so
z(t) = 2a + (A positive number or zero.)
Thus, the range of y = z(t) is y ≥ 2a. Since we know the range is y ≥ −8, this means 2a = −8, so a = −4.
45. See Figure 8.1. Using a computer or graphing calculator, we see that the graph appears to extend indefinitely to the left
and right. It appears to reach a peak of y = 100 and to draw close to the x-axis without crossing it as x grows large. We
conclude that the domain is all x and the range is 0 < y ≤ 100.
y
100
x
4
−4
Figure 8.1
46. See Figure 8.2. Using a computer or graphing calculator, we see that the graph appears to extend indefinitely to the left
and right. It appears to reach a peak of about y = 10.615 but to have no minimum value of y. We conclude that the
domain is all x and the range is y < 10.615.
y
12
x
−1
5
−12
Figure 8.2
10
262
Chapter Eight /SOLUTIONS
47. See Figure 8.3. Using a computer or graphing calculator, we see that the graph appears to extend from x = 1 to x = 9,
not to cross below the x-axis, and to reach a peak of y = 4 at x = 5. We conclude that the domain is 1 ≤ x ≤ 9 and the
range is 0 ≤ y ≤ 4.
y
y
6
4
1
x
x
1
9
8
−2
Figure 8.3
Figure 8.4
48. See Figure 8.3. Using a computer or graphing calculator, we see that the graph appears to extend indefinitely to the left
and right, to lie above the line y = 1, and to reach a peak of y = 6 at x = 3. We conclude that the domain is all x and the
range is 1 < y ≤ 6.
Solutions for Section 8.2
EXERCISES
1. We substitute u = x + 1 into y = u4 to get y = (x + 1)4 .
2. We substitute u = 3 − 4x into y = 5u3 to get y = 5(3 − 4x)3 .
3. We substitute r = t3 into w = r 2 + 5 to get w = (t3 )2 + 5 = t6 + 5.
4. We substitute p = 2q 4 into D = 5p − 1 to get D = 5(2q 4 ) − 1 = 10q 4 − 1.
5. We substitute w = 5s3 into q = 3 + 2w to get q = 3 + 2(5s3 ) = 3 + 10s3 .
6. We substitute q = 2r 3 into P = 3q 2 + 1 to get P = 3(2r 3 )2 + 1 = 3(4r 6 ) + 1 = 12r 6 + 1.
7. We substitute u = x2 into y = u2 + u + 1 to get y = (x2 )2 + (x2 ) + 1 = x4 + x2 + 1.
8. We substitute u = 3x3 into y = 2u2 +5u+7 to get y = 2(3x3 )2 +5(3x3 )+7 = 2(9x6 )+5(3x3 )+7 = 18x6 +15x3 +7.
√
9. We define √
u = x2 + 1 as the inside function and y = u as the outside function. Then composing these two functions
2
gives y = x + 1 as required. Other answers are possible.
10. We define u = x − 2 as the inside function and y = 5u3 as the outside function. Then composing these two functions
gives y = 5(x − 2)3 as required. Other answers are possible.
11. We define u = x3 as the inside function and y = 3u − 2 as the outside function. Then composing these two functions
gives y = 3x3 − 2 as required. Other answers are possible.
12. We can write
(2x + 1)5
3
1
= (2x + 1)5 ,
3
y=
so h(x) = 2x + 1, k = 1/3, p = 5.
8.2 SOLUTIONS
13. We can write
17
(1 − x3 )4
y=
= 17(1 − x3 )−4 ,
so h(x) = 1 − x3 , k = 17, p = −4.
14. We can write
y=
p
5 − x3
= (5 − x3 )1/2 ,
so h(x) = 5 − x3 , k = 1, p = 1/2.
15. We can write
y= r
2
1+
= 2 1+
so h(x) = 1 + 1/x, k = 2, p = −1/2.
1
x
1
x
−1/2
,
16. We can write
y = 100 1 +
so h(x) = 1 +
√
x, k = 100, p = 4.
√ 4
x ,
17. We can write
y = 1 + x + x2
so h(x) = 1 + x + x2 , k = 1, p = 3.
3
,
18. We can write
√
12 − 3 x
√ 1/2
= 5 12 − 3 x
,
p
y=5
so h(x) = 12 −
√
3
x, k = 5, p = 1/2.
19. Since(x + 3)2 = (3 + x)2 , we have
y = 0.5(x + 3)2 + 7(3 + x)2
= 7.5(x + 3)2 ,
so h(x) = x + 3, k = 7.5, p = 2.
PROBLEMS
20. These two functions share an inside function: to multiply by 2 and add 1.
21. These two functions share an outside function: to take the square root.
22. We substitute f (s) in for w in g(w) to obtain q as a function of s. The input variable is s and the output variable is q.
263
264
Chapter Eight /SOLUTIONS
23. We have
g(v) = f 1 − v 2 /c2
= 1 − v 2 /c2
This means
g(0.6c) =
=
1−
−1/2
(0.6c)2
c2
0.36c2
1−
c2
.
−1/2
−1/2
= (1 − 0.36)−1/2
= (0.64)−1/2
1
= √
0.64
1
=
0.8
= 1.25.
24. We have
p(t) = V (3t)
4
= π(3t)3
3
4
= π · 33 · t3
3
= 36πt3 .
This means
p(2) = 36π23
= 36π · 8
= 288π.
25. We see that
f x2 = 2x4 + 1
f (2x) = 8x2 + 1
= 2 x2
2
+1
= 2 · 4x2 + 1
= 2 (2x)2 + 1
f (x + 1) = 2x2 + 4x + 3 = 2(x2 + 2x + 1) + 1 = 2(x + 1)2 + 1,
so we conclude that f (x) = 2x2 + 1.
26. (a) There are many possible answers, such as
f (x) = (2x + 1)3 ,
f (x) = (3 + 5x)−3 ,
f (x) = (5 + x)1/2 .
Any function in the form f (x) = (linear function)p = (b + mx)p satisfies the requirement.
(b) We substitute a power function in for the variable in a linear function. There are many possible answers, such as
f (x) = 3 + 5(x3 ),
f (x) = 1 − 2(x5 ),
f (x) = 2(x1/3 ) + 7.
Any function in the form f (x) = b + m(power function) = b + m(xp ) satisfies the requirement.
27. (a) We have f (g(x)) = f (5 + 2x) = (5 + 2x)3 .
(b) We have g(f (x)) = g(x3 ) = 5 + 2x3 .
√
√ 3
3/2
28. (a) We have f (g(x)) = f ( x) = ( x)
+ 1.
√ +1=x
3
(b) We have g(f (x)) = g(x + 1) = x3 + 1.
265
8.3 SOLUTIONS
29. (a)
(b)
(c)
(d)
We have f (g(t)) = f (2t + 1) = 3(2t + 1)2 .
We have g(f (t)) = g(3t2 ) = 2(3t2 ) + 1 = 6t2 + 1.
We have f (f (t)) = f (3t2 ) = 3(3t2 )2 = 3(9t4 ) = 27t4 .
We have g(g(t)) = g(2t + 1) = 2(2t + 1) + 1 = (4t + 2) + 1 = 4t + 3.
30. In f (g(x)), we see that g(x) is the inside function and f (x) is the outside function. In this case, the outside function is
f (x) = 5x3 .
31. There are many possible answers such as y = (3x + 1)3 , or P = (x2 − 4)3 , or Q = (3t2 + 9)3 . Any three functions in
the form y = (f (x))3 are correct.
32. Using x as our input variable, the
√ inside expression is 5x + 2. The outside function takes the square root, so the resulting
function can by written as y = 5x + 2.
√
33. Using x as our input variable, the inside expression is x. The outside function multiplies by 5 and adds 2, so the resulting
√
function can by written as y = 5 x + 2.
34. We use composition to eliminate the quantity P as an intermediate quantity. We have A = 100P 0.3 = 100(10000 +
2000t)0.3 .
√
35. We use composition to eliminate the quantity Q as an intermediate quantity. We have R = 25 − 0.08Q = 25 − 0.08 t.
Solutions for Section 8.3
EXERCISES
1. See Figure 8.5. A formula for g(x) is:
g(x) = f (x) + 3
g(x) = x3 + 3.
y
y
6
2
4
x
2
2
−2
−2
x
2
−2
−4
−2
Figure 8.5
Figure 8.6
2. See Figure 8.6. A formula for g(x) is:
g(x) = f (x) − 2
g(x) = x3 − 2.
266
Chapter Eight /SOLUTIONS
3. See Figure 8.7. Since x = −1 on the graph of g should correspond to x = 0 on the graph of f , we see that a formula for
g(x) is:
g(x) = f (x + 1)
g(x) = (x + 1)3 .
y
2
x
2
−2
−2
Figure 8.7
4. See Figure 8.8. Since x = 2 on the graph of g should correspond to x = 0 on the graph of f , we see that a formula for
g(x) is:
g(x) = f (x − 2)
g(x) = (x − 2)3 .
y
2
x
2
−2
−2
Figure 8.8
5. See Figure 8.9. We have
g(x) = f (x + 1) − 3
g(x) = (x + 1)3 − 3.
y
2
x
2
−2
−2
−4
−6
Figure 8.9
267
8.3 SOLUTIONS
6. See Figure 8.10. We have
g(x) = f (x − 4) + 2
g(x) = (x − 4)3 + 2.
y
4
2
x
2
−2
4
6
−2
Figure 8.10
7. See Figure 8.11. We have
g(x) = f (x) − 3
= (5 − x) − 3
= 2 − x.
y
y
5
7
6
4
5
3
4
2
3
1
2
1
x
−2 −1
−1
1
2
3
4
5
x
−2 −1
−1
−2
1
2
3
4
−2
Figure 8.11
Figure 8.12
8. See Figure 8.12. We have
g(x) = f (x) + 1
= (5 − x) + 1
= 6 − x.
5
6
7
268
Chapter Eight /SOLUTIONS
9. See Figure 8.13. We have
g(x) = f (x − 2)
= 5 − (x − 2)
= 5−x+2
= 7 − x.
y
8
7
6
5
4
3
2
1
x
1
−2 −1
−1
2
3
4
5
6
7
8
−2
Figure 8.13
10. See Figure 8.14. We have
g(x) = f (x + 4)
= 5 − (x + 4)
= 5−x−4
= 1 − x.
y
5
4
3
2
1
x
−2 −1
−1
1
2
3
4
5
−2
Figure 8.14
11. We see that the graph of y = x2 has been shifted up 2 units, so a formula is y = x2 + 2.
12. We see that the graph of y = x2 has been shifted down 3 units, so a formula is y = x2 − 3.
8.3 SOLUTIONS
269
13. We see that the graph of y = x2 has been shifted to the right 1 unit, so a formula is y = (x − 1)2 .
14. We see that the graph of y = x2 has been shifted to the left 2 units, so a formula is y = (x + 2)2 .
15. We see that the graph of y = x2 has been shifted up 1 unit and to the right 3 units, so a formula is y = (x − 3)2 + 1.
16. We see that the graph of y = x2 has been shifted down 3 units and to the left 2 units, so a formula is y = (x + 2)2 − 3.
17. The graph is the graph of y = f (x) shifted vertically down 3 units. See Figure 8.15.
y
x
1
−1
2
3
4
5
6
7
−1
−2
−3
Figure 8.15
18. The graph is the graph of y = f (x) shifted vertically up 2 units. See Figure 8.16.
y
4
3
2
1
x
1
2
3
4
5
6
Figure 8.16
19. The graph is the graph of y = f (x) shifted horizontally to the right 1 unit. See Figure 8.17.
y
3
2
1
x
1
2
3
4
Figure 8.17
5
6
7
270
Chapter Eight /SOLUTIONS
20. The graph is the graph of y = f (x) shifted horizontally to the left 3 units. See Figure 8.18.
y
2
1
x
−3
−2
1
−1
2
3
Figure 8.18
21. The graph is the graph of y = f (x) shifted horizontally to the left 1 unit and vertically up 2 units. See Figure 8.19.
y
4
3
2
1
x
1
−1
2
3
4
5
Figure 8.19
22. The graph is the graph of y = f (x) shifted horizontally to the right 3 units and vertically down 1 unit. See Figure 8.20.
y
1
x
1
2
3
4
5
7
8
9
Figure 8.20
23. In every pound there are sixteen oz, so
Number
|
{z of oz} = 16 × Number of pounds
n
n = 16m.
|
{z
m
}
8.3 SOLUTIONS
271
24. In every foot there are twelve inches, so
Number
{zof inches} = 12 × Number
{zof feet}
|
|
m
n
m = 12n
1
· m.
n=
12
25. In every kilometer there are 1000 meters, so
Number
of meters} = 1000 × Number
of{zkilometers}
|
{z
|
m
n
m = 1000n
n = 0.001m.
26. In every week there are seven days, so
Number of days = 7 × Number
|
{zof weeks}
|
{z
n
}
m
n = 7m.
27. In every dollar there are 100 cents, so
of dollars}
Number
|
{zof cents} = 100 × Number
{z
|
m
n
m = 100n
n = 0.01m.
28. In every hour there are sixty minutes, so
Number
of minutes
|
|
{z
} = 60 × Number
{zof hours}
m
n
m = 60n
1
· m.
n=
60
PROBLEMS
29. (a) Since the object is taken out 5 minutes later, it has the same temperature as the original object had 5 minutes earlier,
so H = f (t − 5). A reasonable domain is t ≥ 5.
(b) Since the temperatures are all 10◦ colder, the formula is H = f (t) − 10.
30. (a) The erosion of the second dune starts 30 years earlier than the first. The height of the second sand dune in year 0 is
the same as the height of the first sand dune in 30 years.
(b) The height of the second sand dune is 50 cm higher than the first.
31. (a) The equation is y = x + 5.
(b) The equation is y = (x + 5) = x + 5.
(c) For this equation, shifting vertically up by any number of units is the same as shifting horizontally to the left by that
number of units. See Figure 8.21.
272
Chapter Eight /SOLUTIONS
y
y =x+5
y=x
8
6
5 units up
4
x
−8
4
−4
8
−2
5 units left
Figure 8.21
32. At H ◦ C, the temperature in Fahrenheit is
9
9
9
· H + 32, so the growth rate is P ( · H + 32). Thus Q(H) = P ( · H + 32).
5
5
5
33. For every foot there are 12 inches, so
Number
|
|
{zof inches} = 12 × Number
{zof feet}
g(t)
f (t)
g(t) = 12f (t).
34. For every gallon there are 3.785 liters, so
Number
|
{zof liters} = 3.785 × Number of gallons
|
f (t)
{z
}
g(t)
f (t) = 3.785g(t)
1
· f (t).
g(t) =
3.785
35. For every second there are 1,000,000 µs, so
of seconds}
Number of µs = 1,000,000 × Number
|
{z
|
{z
}
g(n)
f (n)
g(n) = 1,000,000f (n).
36. Since 1 kph equals 0.621 mph,
Speed in mph = 0.621 × Speed in kph
|
{z
f (t)
}
|
f (t) = 0.621g(t)
1
g(t) =
· f (t).
0.621
{z
g(t)
}
37. For every parsec there are 3.262 light years, so
Number of light years = 3.262 × Number of parsecs
|
{z
f (t)
}
|
f (t) = 3.262g(t)
1
g(t) =
· f (t).
3.262
{z
g(t)
}
8.3 SOLUTIONS
38. For every mi2 there are 2.59 km2 , so
2
2
Number
|
{zof km} = 2.59 × Number
{zof mi}
|
g(t)
f (t)
g(t) = 2.59f (t).
39. We see that
Number
of minutes} = 60 × Number
|
|
{z
{zof hours}
r
t
r = 60t
1
·r
t=
60
Inches of snow
Inches of snow
=
after r minutes
after t = (1/60)r hours
so
|
{z
|
}
w(r)
{z
f ((1/60)r)
w(r) = f ((1/60)r) .
}
40. We see that
Number of days = 7 × Number
|
{zof weeks}
|
so
{z
}
t
t = 7r,
Liters burned
{z
}
w(r)
Liters burned
=
in r weeks
|
r
in t = 7r days
|
{z
}
f (7r)
w(r) = f (7r).
41. We see that
Number
|
|
{zof MB} = 1024 × Number
{z of GB}
n
r
n = 1024r
so
Average time to
=
process r GB
|
{z
w(r)
}
Average time to
process n = 1024r MB
|
{z
f (1024r)
w(r) = f (1024r).
}
42. We see that
Number
of newtons} = 4.448 × Number
|
|
{z
{z of lbs}
v
u
v = 4.448u
1
v
u=
4.448
Speed at thrust
Speed at thrust
=
of v newtons
of u = (1/4.448)v lbs
so
|
{z
w(v)
}
|
{z
f ((1/4.448)v)
w(v) = f ((1/4.448)v) .
}
273
274
Chapter Eight /SOLUTIONS
43. We see that
Number
of hectares} = 2.471 × Number
{z
{zof acres}
|
|
s
r
s = 2.471r
1
·s
r=
2.471
Yield from
Yield from
=
s hectares
r = (1/2.471)s acres
so
|
{z
w(s)
}
{z
|
f ((1/2.471)s)
w(s) = f ((1/2.471)s) .
}
44. We see that
in
Price
in cents} = 100 × Price
|
|
{z
{zdollars}
q
p
q = 100p
p = 0.01q
so
Number of songs
at price q cents
|
{z
w(q)
Number of songs
=
}
at price p = 0.01q dollars
|
{z
f (0.01q)
w(q) = f (0.01q).
}
45. Since 1 in equals 2.54 cm, we see that
Number of centimeters = 2.54 × Number of centimeters.
This means that
g(t) =
Number of centimeters
after t hours
= 2.54 ×
Number of inches
after t hours
= 2.54f (t).
We see that by scaling the output of f by a factor of 2.54, we obtain the number of centimeters.
46. Since 1 acre equals 0.405 hectares, we see that
Number of hectares = 0.405 × Number of acres.
This means that
g(t) =
Number of hectares
lost after t years
= 0.405 ×
Number of acres
lost after t years
= 0.405f (t).
We see that by scaling the output of f by a factor of 0.405, we obtain the number of hectares.
8.3 SOLUTIONS
275
47. Since 1 gal equals 3.785 l, we see that
Number of liters = 3.785 × Number of gallons
1
× Number of liters.
3.785
so Number of gallons =
This means
g(s) =
Number of gallons
when depth is s
1
Number of liters
×
3.785
when depth is s
1
=
· f (s).
3.785
=
We see that by multiplying the output of f by a factor of 1/3.785 we obtain the number of gallons.
48. Since 1 m equals 100 cm, we see that
Depth in cm = 100 × Depth in m
d = 100s
s = 0.01d.
Since s = 0.01d and since f (s) is the number of liters at s meters, we see that
g(d) =
Number of liters
at depth d cm
= f (0.01d).
We see that by scaling the input of f by a factor of 0.01, we obtain the reservoir’s volume as a function of d, the number
of centimeters.
49. Since 1 lb equals 0.454 kg, we see that
Weight at birth
in kg
|
{z
m
Weight at birth
= 0.454 ×
}
in lbs
|
{z
}
w
m = 0.454w
1
· m.
w=
0.454
This means that if the birth weight is m kg,
Expected weight at
one month in lbs
=f
1
0.454
·m .
Converting the expected weight at one month from lbs to kg gives
g(m) = 0.454f
1
0.454
·m .
We see that by scaling the output by a factor of 0.454, and the input by a factor of 1/0.454, we obtain the one-month
weight in kg as a function of the birth weight in kg.
276
Chapter Eight /SOLUTIONS
50. There are 24 hours in a day, which means that that
Number of hours = 24 × Number of days
n = 24t
1
· n.
so t =
24
Since f (t) is the number of acres burned after t days, letting t = (1/24)n gives
Number of acres
burned after n hours
=f
1
·n .
24
Since 1 acre equals 0.405 hectares,
Number of hectares = 0.405 × Number of acres.
We conclude that
g(n) =
Number of hectares
burned after n hours
= 0.405 ×
Number of acres
burnedafter n hours
1
·n .
= 0.405f
24
We conclude that by scaling the input of f by a factor of 1/24, and scaling the output by a factor of 0.405, we obtain the
number of hectares burned as a function of n, the elapsed time in hours.
Solutions for Section 8.4
EXERCISES
1. Add 8.
2. Multiply by 10.
3. Raise to the 1/7th power (or take the 7th root.)
4. Raise to the 9th power.
5. We write this sequence of operations as the function y = 5x − 2, and undo the operations by solving for x:
y = 5x − 2
y + 2 = 5x
y+2
= x.
5
The resulting equation shows us that to undo the sequence of operations given, we add 2 and then divide by 5.
6. We write this sequence of operations as the function y = 3(x + 10), and undo the operations by solving for x:
y = 3(x + 10)
y
= x + 10
3
y
− 10 = x.
3
The resulting equation shows us that to undo the sequence of operations given, we divide by 3 and then subtract 10.
8.4 SOLUTIONS
277
7. We write this sequence of operations as the function y = 2x5 , and undo the operations by solving for x:
y = 2x5
y
= x5
2
1/5
y
2
= x.
The resulting equation shows us that to undo the sequence of operations given, we divide by 2 and raise to the 1/5th
power.
√
8. We write this sequence of operations as the function y = 3 6x + 10, and undo the operations by solving for x:
√
y = 3 6x + 10
y 3 = 6x + 10
3
y − 10 = 6x
y 3 − 10
= x.
6
The resulting equation shows us that to undo the sequence of operations given, we raise to the 3rd power, subtract 10, then
divide by 6.
9. We compose both ways. Starting with y and substituting in for x, we have:
y = 7x − 5 = 7
y + 5
7
− 5 = (y + 5) − 5 = y.
The operations undo each other as required. Starting with x and substituting in for y, we have:
(7x − 5) + 5
7x
y+5
=
=
= x.
7
7
7
In this order as well, the operations undo each other.
x=
10. We compose both ways. Starting with y and substituting in for x, we have:
y = 8x3 = 8
q 3
y
8
3
=8
y
8
= y.
The operations undo each other as required. Starting with x and substituting in for y, we have:
x=
q
3
y
=
8
r
3
√
8x3
3
= x3 = x.
8
In this order as well, the operations undo each other.
11. We compose both ways. Starting with y and substituting in for x, we have:
p
y = x5 + 1 = ( 5 y − 1)5 + 1 = (y − 1) + 1 = y.
The operations undo each other as required. Starting with x and substituting in for y, we have:
p
p
√
5
x = 5 y − 1 = 5 (x5 + 1) − 1 = x5 = x.
In this order as well, the operations undo each other.
12. We compose both ways. Starting with y and substituting in for x, we have:
10 + (3y − 10)
3y
10 + x
=
=
= y.
3
3
3
The operations undo each other as required. Starting with x and substituting in for y, we have:
y=
x = 3y − 10 = 3
10 + x
3
In this order as well, the operations undo each other.
− 10 = (10 + x) − 10 = x.
278
Chapter Eight /SOLUTIONS
13. (a) We have y = 8x5 + 4.
(b) We find the inverse by solving for x:
y = 8x5 + 4
r
5
y − 4 = 8x5
y−4
= x5
8
y−4
= x.
8
The sequence of operations in the inverse is to subtract 4, divide by 8, and then take the fifth root.
r
x−5
.
2
(b) We find the inverse by solving for x:
14. (a) We have y =
3
r
x−5
2
x−5
3
y =
2
2y 3 = x − 5
y=
3
2y 3 + 5 = x.
The sequence of operations in the inverse is to raise to the 3rd power, multiply by 2, and add 5.
15. The input of f is the number of MB, and the output is the time required. Switching the roles of input and output, we see
that n = g(T ) gives the amount in MB of data that a computer can process in T seconds, on average.
16. The input of f is the thrust in pounds-force, and the output is the jet’s maximum speed. Switching the roles of input and
output, we see that r = g(v) gives the thrust required for a jet to achieve a maximum speed of v.
17. The input of f is the number of acres, and the output is the yield in bushels. Switching the roles of input and output, we
see that r = g(Y ) gives the number of acres of corn required to yield Y bushels, on average.
18. The input of f is the number of people, and the output is the cost. Switching the roles of input and output, we see that
N = g(P ) gives the number of people that can be catered for a cost of $P .
19. The input of f is the number of kilometers, and the output is the pressure. Switching the roles of input and output, we see
that s = g(P ) gives the altitude at which the pressure is P .
20. The input of f is the concentration, and the output is the boiling point. Switching the roles of input and output, we see
that c = g(T ) gives the concentration at which the boiling point is T .
PROBLEMS
21. One way to check that these functions are inverses is to make sure they satisfy the identities f (g(t)) = t and g(f (x)) = x.
7
t
+
2
2
= t+7−7
f (g(t)) = 2
−7
= t.
Also,
7
2x − 7
+
2
2
2x
7
7
=
− +
2
2
2
= x.
g(f (x)) =
8.4 SOLUTIONS
279
22. One way to check that these functions are inverses is to make sure they satisfy the identities f (g(t)) = t and g(f (x)) = x.
f (g(t)) = 6
7
t − 4 1/7
6
t−4
+4
=6
6
= t−4+4
+4
= t.
Also,
g(f (x)) =
=
6x7 + 4 − 4
6
6x7
6
= x7
= x.
1/7
1/7
1/7
23. One way to check that these functions are inverses is to make sure they satisfy the identities f (g(t)) = t and g(f (x)) = x.
f (g(t)) = 32
(t + 2)1/5
2
5
((t + 2)1/5 )
25
t+2
−2
= 32
32
= t+2−2
= 32
−2
5
−2
= t.
Also,
g(f (x)) =
=
=
=
=
(32x5 − 2 + 2)1/5
2
(32x5 )1/5
2
1/5
32 (x5 )1/5
2
2x
2
x.
24. Since f (x) = (x/4) − (3/2) and g(t) = 4(t + 3/2), we have
4(t + 32 )
3
3
3
− =t+ − =t
4
2
2
2
x
x
3
3
g(f (x)) = 4
=4 =x
− +
4
2
2
4
f (g(t)) =
280
Chapter Eight /SOLUTIONS
25. One way to check that these functions are inverses is to make sure they satisfy the identities f (g(t)) = t and g(f (x)) = x.
r
3
f (g(t)) = 1 + 7
t−1
7
!3
t−1
7
= 1 + (t − 1) = t.
= 1+7
Also,
g(f (x)) =
=
r
3
√
3
1 + 7x3 − 1
7
x3 = x.
26. We solve:
2x3 + 7 = −9
2x3 = −9 − 7
2x3 = −16
16
x3 = −
2
x3 = −8.
Taking the cube root undoes cubing, so we take the cube root of both sides:
√
3
x3 = −8
√
x3 = 3 −8
x = −2.
27. We solve:
√
3 x+5
=5
4
√
3 x+5
√
3 x+5
√
3 x
√
3 x
√
x
√
x
Squaring undoes the square root, so we square both sides:
√
√
= 5·4
= 20
= 20 − 5
= 15
15
=
3
= 5.
x=5
( x)2 = 52
x = 25.
8.4 SOLUTIONS
28. Cubing undoes the cube root, so we start by cubing both sides:
p
√
3
30 − x =
p
√
3
( 30 − x)3 =
√
30 − x =
√
− x=
√
− x=
√
x=
3
33
27
27 − 30
−3
3.
Squaring undoes the square root, so we square both sides:
√
x=3
√ 2
( x) = 32
x = 9.
29. Squaring undoes the square root, so we start by squaring both sides:
(
p
p
x3 − 2 = 5
x3 − 2)2 = 52
x3 − 2 = 25
x3 = 25 + 2
x3 = 27.
Taking the cube root undoes cubing, so we take the cube root of both sides:
√
3
x3 = 27
√
3
x3 = 27
x = 3.
30. Solve for x in y = h(x) = 2x + 4:
y = 2x + 4
2x = y − 4
y−4
.
x=
2
Thus,
g(y) =
y−4
.
2
31. Solve for x in y = h(x) = 9x5 + 7:
y = 9x5 + 7
9x5 = y − 7
y−7
x5 =
9
x=
r
g(y) =
r
Thus,
5
5
y−7
.
9
y−7
.
9
281
282
Chapter Eight /SOLUTIONS
32. Solve for x in y = h(x) =
√
3
x + 3:
y=
√
3
x+3 = y
x+3
3
x = y 3 − 3.
Thus,
g(y) = y 3 − 3.
33. Let us call the the inverse function g, so that if y = p(x) then x = g(y). Solving for x, we obtain
√
5− x
y=
√
3+2 x
√
√ multiply by denominator
y 3+2 x = 5− x
√
√
3y + 2y x = 5 − x
distribute
√
√
3y − 5 = −2y x − x collect x-terms
√
3y − 5 = − x(2y + 1) factor
√
3y − 5
divide
− x=
2y
+ 1 2
3y − 5
x=
square both sides
2y + 1
2
3x − 5
.
so g(x) =
2x + 1
34. Let us call the inverse function g, so that if y = q(x), then x = g(y), so we solve:
4 − 3x
7 − 9x
y(7 − 9x) = 4 − 3x
y=
multiply by denominator
7y − 9xy = 4 − 3x
distribute
x(3 − 9y) = 4 − 7y
4 − 7y
x=
3 − 9y
4 − 7x
so g(x) =
.
3 − 9x
factor
3x − 9xy = 4 − 7y
collect x-terms at left
divide
Solutions for Chapter 8 Review
EXERCISES
1. The number of credit hours a part-time student takes can range from 0 to 10, so the domain is 0 ≤ C ≤ 10.
When C = 0, we have T = 0, and when C = 10, we have T = 300 + 200(10) = 2300. Tuition cost for a part-time
student ranges from $0 to $2300, so the range is 0 ≤ T ≤ 2300.
SOLUTIONS to Review Problems for Chapter Eight
283
2. The expression x4 is defined for all real x, so we have
Domain: All real numbers.
Since y is equal to something raised to an even power, we know that y cannot be negative. We have
Range: y ≥ 0.
3. The expression x−3 = 1/x3 is not defined at x = 0 since we cannot divide by zero. It is defined for all other real x, so
we have
Domain: x 6= 0.
Since 1/x3 can never be equal to 0, we have y 6= 0. We have
Range: y 6= 0.
4. The 2 in the denominator of the exponent in the expression 2x3/2 tells us to take the square root. Since we can’t take the
square root of a negative number, we need x ≥ 0. The other operations (cubing and multiplying by 2) do not put any
additional restrictions on the domain, so we have
Domain: x ≥ 0.
Since the square root of any number is never negative, we have y = 2x3/2 is never negative, so we have
Range: y ≥ 0.
5. The expression 5t−1 = 5/t is undefined at t = 0 since we can’t divide by zero. The expression is defined for all other
real values of t so we have
Domain: t 6= 0.
Since 5/t can never equal zero, we have A 6= 0. We have
Range: A 6= 0.
6. The exponent 2/3 means to square and to take the cube root. Since we can do these operations to any real number, we
have
Domain: All real numbers.
Since the 2 in the exponent 2/3 means to raise to the second power, the expression 6x2/3 can never be negative. We have
Range: P ≥ 0.
7. The exponent 5/3 means to raise to the fifth power and to take the cube root. Since we can do these operations to any real
number, we have
Domain: All real numbers.
When we take the cube root of a number, we can get any number, and when we raise a number to the fifth power, we can
get any real number. We have
Range: All real numbers.
284
Chapter Eight /SOLUTIONS
√
8. The exponent −1/2 means to take the square root and take the reciprocal, so we have x−1/2 = 1/ x. Since we can’t
take the square root of a negative number, we have x ≥ 0. Since we can’t divide by zero, we have x 6= 0. Thus,
Domain: x > 0.
The square root of any number is never negative, and 1 divided by anything can never be zero, so y cannot be negative
and cannot be zero. We have
Range: y > 0.
9. Since we can raise any number to the third power, we have
Domain: All real numbers.
We can get any number M by cubing another number, so we have
Range: All real numbers.
10. We know that:
√
• x ≤ 9, for otherwise 9 − x would be negative and 9 − x would
pbe undefined.
√
√
√
• 9 − x ≤ 4, for otherwise 4 − 9 − x would be negative and 4 − 9 − x would be undefined. This means:
√
9−x≤ 4
9 − x ≤ 16
x ≥ −7.
• y≥√
0, because the output of the square-root operation is√positive.
• y < 4 = 2, because the largest possible value of 4 − 9 − x is 4.
Thus, the domain is −7 ≤ x ≤ 9 and the range is 0 ≤ y ≤ 2.
11. We know that:
• x 6= 3 or otherwise the denominator would be zero.
• The fraction can not equal zero since the numerator does not equal zero; therefore, y 6= 4.
Thus, the domain is x 6= 3 and the range is y 6= 4.
12. For any number y, we can solve the equation m(x) = y for x:
9−x = y
9 = y+x
9 − y = x.
So if we input 9 − y, we get m(9 − y) = y. Since y can be any number, the range is also all real numbers.
13. To find the range, we want to know all possible output values. We solve the equation y = f (x) for x in terms of y. Since
y= √
squaring gives
y2 =
and multiplying by x − 4 gives
1
,
x−4
1
,
x−4
y 2 (x − 4) = 1
y 2 x − 4y 2 = 1
y 2 x = 1 + 4y 2
1 + 4y 2
.
x=
y2
SOLUTIONS to Review Problems for Chapter Eight
285
This formula tells us how to find the x-value which corresponds to a given y-value.√The formula works for any y except
y = 0 (which puts a 0 in the denominator). We know that y must be positive, since x − 4 is positive, so we have
Range: y > 0.
14. For any y we can solve the equation x − 3 = y for x. The solution is x = y + 3. This tells us that f (y + 3) = y. Since y
can be any number, the range is all real numbers.
15. To find the range, we want to know all possible output values. We solve the equation y = h(x) for x in terms of y. Since
y=
r
3
1
,
x
cubing both sides gives
y3 =
1
,
x
so
1
.
y3
This formula tells us how to find the x-value which corresponds to a given y-value. The formula works for any y except
y = 0 (which puts a 0 in the denominator). Thus
x=
Range: y 6= 0.
16. We have
√
3+ x
2
1
= (3 + A nonnegative number)
2
1
= (A number greater than or equal to 3)
2
3
= A number greater than or equal to ,
2
y=
so the range is y ≥ 3/2
17. We have
y = 2+
√ 3
x
= (2 + A nonnegative number)3
= (A number greater than or equal to 2)3
= A number greater than or equal to 8,
so y ≥ 8.
18. Recall that in order for a fraction to equal zero, its numerator must equal zero. Thus, we have
nonzero
z }| {
5
+9
x−3
= A number not equal to 9,
y=
so y 6= 9.
286
Chapter Eight /SOLUTIONS
19. We have
y = (x −
√
2)3
= Any number −
= (Any number)3 ,
√ 3
2
so the range is all real numbers.
20. (a) The domain is 1 ≤ x ≤ 7.
(b) The range is all y such that 2 ≤ y ≤ 18.
21. Judging from the graph, the leftmost point is at x = −1 and the rightmost at x = 9. The lowest value is y = 0, occurring
at x = −1; the highest value is y = 4, occurring somewhere between x = 1 and x = 2. We conclude:
(a) The domain is −1 ≤ x ≤ 9.
(b) The range is 0 ≤ y ≤ 4.
5
as the outside function. Then composing these two functions
22. We define u = x2 + 1 as the inside function and y =
u
5
as required. Other answers are possible.
gives y = 2
x +1
23. We define u = x − 1 as the inside function and y = 1 + 2u + 5u2 as the outside function. Then composing these two
functions gives y = 1 + 2(x − 1) + 5(x − 1)2 as required. Other answers are possible.
24. We define u = 3x − 2 as the inside function and y = 25u5 + 100 as the outside function. Then composing these two
functions gives y = 25(3x − 2)5 + 100 as required. Other answers are possible.
25. (a) We have f (g(x)) = f (3x) = 5(3x)2 = 5(9x2 ) = 45x2 .
(b) We have g(f (x)) = g(5x2 ) = 3(5x2 ) = 15x2 .
1
26. (a) We have f (g(x)) = f (x2 + 1) = 2
.
x
2 + 1
1
1
1
+ 1 = 2 + 1.
(b) We have g(f (x)) = g( ) =
x
x
x
27. See Figure 8.22. We have
g(x) = f (x + 1)
= 2(x + 1)2 − 1.
y
3
2
1
x
−4 −3 −2 −1
−1
1
2
−2
−3
Figure 8.22
28. See Figure 8.23. We have
g(x) = f (x) − 3
= (2x2 − 1) − 3
= 2x2 − 4.
SOLUTIONS to Review Problems for Chapter Eight
y
3
x
3
−3
−5
Figure 8.23
29. See Figure 8.24. We have
g(x) = f (x − 3) + 2
= (2(x − 3)2 − 1) + 2
= 2(x − 3)2 + 1.
y
5
4
3
2
1
x
1
−2
2
3
4
5
−1
−2
Figure 8.24
30. See Figure 8.25. We have
g(x) = f (x + 2) − 1
= (2(x + 2)2 − 1) − 1
= 2(x + 2)2 − 2.
y
2
1
x
−4
2
−2
−2
−4
Figure 8.25
287
288
Chapter Eight /SOLUTIONS
31. Since 1 hour is 3600 seconds, there are s/3600 hours in s seconds, the distance traveled in s seconds is f (s/3600) miles.
Thus g(s) = f (s/3600).
32. There are 5280 feet in a mile. So in t hours the car travels 5280f (t) feet. Thus F (t) = 5280f (t).
33. There are 60 minutes in an hour. So in m minutes, there are m/60 hours, and the car has traveled f (m/60) miles. Since
there are 1.609 kilometers in a mile, the car has traveled 1.609f (m/60) kilometers. So h(m) = 1.609f (m/60).
34. There are 3600 seconds in an hour, so s seconds is s/3600 hours. So in s seconds the car travels f (s/3600) miles. Since
there are 1609 meters in a mile, the car travels 1609f (s/3600) meters. So r(s) = 1609f (s/3600).
35. Solve for x in y = h(x) = 7x + 5:
y = 7x + 5
7x = y − 5
y−5
.
x=
7
Thus,
y−5
.
7
g(y) =
36. Solve for x in y = h(x) = 11x3 − 2:
y = 11x3 − 2
11x3 = y + 2
y+2
x3 =
11
x=
r
g(y) =
r
Thus,
37. Solve for x in y = h(x) =
3
3
y+2
.
11
y+2
.
11
7x5 + 2
:
7x5 + 3
7x5 + 2
7x5 + 3
7x5 + 2 = y(7x5 + 3)
y=
7x5 + 2 = 7x5 y + 3y
5
7x − 7x5 y = 3y − 2
7(1 − y)x5 = 3y − 2
3y − 2
x5 =
7(1 − y)
x=
Thus,
g(y) =
r
5
r
5
3y − 2
.
7 − 7y
3y − 2
.
7 − 7y
SOLUTIONS to Review Problems for Chapter Eight
38. Solve for x in y = h(x) =
289
4x
:
5x + 4
4x
5x + 4
4x = y(5x + 4)
y=
4x − 5yx = 4y
x(4 − 5y) = 4y
x=
Thus,
g(y) =
4y
.
4 − 5y
4y
.
4 − 5y
39. Divide by 77.
40. Subtract −10.
41. Raise to the 7th power.
42. Raise to the 7th power.
PROBLEMS
43. We know that the theater can hold anywhere from 0 to 200 people. Therefore the domain of the function is the integers,
n, such that 0 ≤ n ≤ 200.
We know that each person who enters the theater must pay $4.00. Therefore, the theater makes (0) · ($4.00) = 0
dollars if there is no one in the theater, and (200) · ($4.00) = $800.00 if the theater is completely filled. Thus the range
of the function would be the integer multiples of 4 from 0 to 800. (That is, 0, 4, 8, . . ..)
The graph of this function is shown in Figure 8.26.
money ($)
800
700
600
500
400
300
200
100
n, number of people
50 100 150 200 in attendance
Figure 8.26
44. The domain is 6380 ≤ r ≤ 7380 and since we know that
r = RE
= 6380
r = RE + A = 6380 + 1000 = 7380
minimum Orbiter altitude
maximum Orbiter altitude.
Since F decreases with increasing r, we can use the minimum and maximum values of r to determine the range:
Force at minimum distance = GmME (6380)−2
290
Chapter Eight /SOLUTIONS
= 1.027 × 1012 Newtons
Force at maximum distance = GmME (7380)−2
= 0.768 × 1012 Newtons.
The range is 0.768 × 1012 ≤ F ≤ 1.027 × 1012 . (This corresponds to a force ranging from 86.3 million tons to 115
million tons.)
45. (a) The domain is all real numbers, as s can take on any value.
(b) Since P cannot be less than 0, we first find the value of s that makes P equal to 0:
0 = −100,000 + 50,000s
100,000 = 50,000s
2 = s.
So the mine is profitable (P is positive) if s > 2. The mine also stays open when profits are 0, at s = 2, but for any
lower values, the mine closes. Thus, the domain of the function is s ≥ 2.
46. We have
y = 5 − 2−x
2
= 5 − (A number between 0 and 1)
= A number between 4 and 5,
so the range is 4 ≤ y < 5. Note that since the range of the original function includes 1 but not 0, the range of this function
includes 4 but not 5.
47. We have
y = 3 · 2−x
2
= 3 (A number between 0 and 1)
= A number between 0 and 3,
so the range is 0 < y ≤ 3. Note that since the range of the original function includes 1 but not 0, the range of this function
includes 3 but not 0.
48. We have
2
y = 7 · 2−x + 3
= 7 (A number between 0 and 1) + 3
= (A number between 0 and 7) + 3
= A number between 3 and 10,
so the range is 3 < y ≤ 10. Note that since the range of the original function includes 1 but not 0, the range of this
function includes 10 but not 3.
49. Using exponent rules, we can rewrite this as
y = 22−x
2
= 22 · 2−x
2
2
= 4 · 2−x .
Thus, we have
y = 4 (A number between 0 and 1)
= A number between 0 and 4,
so the range is 0 < y ≤ 4. Note that since the range of the original function includes 1 but not 0, the range of this function
includes 4 but not 0.
SOLUTIONS to Review Problems for Chapter Eight
291
50. To find the range, we want to know all possible output values. We solve the equation y = g(x) for x in terms of y. Since
y=
cubing both sides gives
p
3
x2 − 8,
y 3 = x2 − 8.
At this point, we notice that x2 − 8 ≥ −8, because x2 ≥ 0. Thus y 3 can take on any value greater than −8, which means
y can take on any value greater than the cube root of −8. Thus,
√
Range: y ≥ 3 −8
Range: y ≥ −2.
51. To find the range, we want to know all possible output values. However, we cannot easily solve y = p(x) for x in terms
of y to give us the possible output values.
Inspecting
the function, we see that x2 − 1 < 0 when −1 < x < 1. Thus, for −1 < x < 1, p(x) is undefined,
√
2
because x − 1 is undefined when x2 − 1 < 0. When x = ±1, x2 − 1 = 0, so p(x) = 0. Otherwise, if x > 1 or
x < −1, x2 − 1 > 0 and can be as large as we want, giving p(x) > 0 as large as we want. Thus
Range: y ≥ 0.
52. We have
5x
=k
x−1
5x = kx − k
(5 − k)x = k
We can complete the last line of the solution by dividing by 5 − k as long as k 6= 5. On the other hand, if k = 5, then the
last line says 0 = 5, which has no solution. So the equation has a solution for all values of k such that k 6= 5. This means
that the range of the function is all real numbers except 5.
53. To find the range, we find reasonable values for G. Since G represents a number of gallons, we must have G ≥ 0. Since
the number of hours must also be nonnegative, the expression 0.75 − 0.3h can never be larger than 0.75, so we have
G ≤ 0.75. We have
Range: 0 ≤ G ≤ 0.75.
54. (a) This is a linear function with nonzero slope, so the domain is all real numbers and the range is all real numbers.
(b) Since x cannot be negative, we have x ≥ 0. Since it is also true that y cannot be negative, we have
y≥0
100 − 25x ≥ 0
100 ≥ 25x
4 ≥ x.
We have
Domain: 0 ≤ x ≤ 4.
What about the range? We know y cannot be negative so we have y ≥ 0. Also, we know x must be nonnegative so
the quantity 100 − 25x is less than or equal to 100. Thus, we have y ≤ 100 and thus
Range: 0 ≤ y ≤ 100.
We can see this domain and range on a graph by graphing the line y = 100 − 25x and looking at the line segment in
the first quadrant.
292
Chapter Eight /SOLUTIONS
55. These two functions share an outside function: to raise to the second power and then multiply by 5.
56. These two functions share an inside function: to raise to the second power and then add 1.
57. We have y = u5 and u = x + 3, so
y = u5 = (x + 3)5 .
Therefore g(h(x)) = (x + 3)5 . To find h(g(x)) we put y = u + 3 and u = x5 , so
y = u + 3 = x5 + 3.
Therefore h(g(x)) = x5 + 3.
√
58. We have y = u and u = 2x + 1, so
√
y=
√
u=
√
2x + 1.
√
2x + 1. This time we put y = 2u + 1 and u = x, so
√
y = 2u + 1 = 2 x + 1.
√
Therefore h(g(x)) = 2 x + 1.
Therefore g(h(x)) =
59. (a)
(b)
(c)
(d)
We have f (g(t)) = f (t2 + 1) = 2 − 5(t2 + 1) = 2 − 5t2 − 5 = −3 − 5t2 .
We have g(f (t)) = g(2 − 5t) = (2 − 5t)2 + 1.
We have f (f (t)) = f (2 − 5t) = 2 − 5(2 − 5t) = 2 − 10 + 25t = 25t − 8.
We have g(g(t)) = g(t2 + 1) = (t2 + 1)2 + 1.
60. In f (g(x)), we see that g(x) is the inside function and f (x) is the outside function. In this case, the inside function is
g(x) = 5x + 1.
√
√
√
2
2
61. There are many
p possible answers such as y = 3x + 1, or P = x − 4, or Q = 3t + 9. Any three functions in the
form y = f (x) are correct.
62. Using x as our input variable, the inside expression is x5 . The outside function multiplies by 2 and subtracts 1, so the
resulting function can by written as y = 2x5 − 1.
63. Using x as our input variable, the inside expression is 2x − 1. The outside function raises this to the fifth power, so the
resulting function can by written as y = (2x − 1)5 .
64. We substitute g(W ) in for V in f (V ) to obtain U as a function of W . The input variable is W and the output variable is
U.
65. (a) The equation is y = (b + mx) + k = (b + k) + mx. The y-intercept is b + k. This makes sense since the old
y-intercept b has been shifted up by k units.
(b) The equation is y = b + m(x − k) = b + mx − mk = (b − mk) + mx. The y-intercept of this line is b − mk.
66. (a) If the entry fee increases by $5, the cost, C, increases by $5 so we are shifting the graph on the cost axis, which is a
vertical shift. The graph is shifted vertically up by 5. The formula is C = f (r) + 5.
(b) If the entry fee includes 3 free rides, the cost for the new function when r = 3 is the same as the cost for the old
function when r = 0. This is a horizontal shift to the right, and the formula is C = f (r − 3).
67. One way to check that these functions are inverses is to make sure they satisfy the identities g(f (t)) = t and f (g(x)) = x.
g(f (t)) = 1 − 1+
1
1
−1
1−t
1
1
1−t
= 1 − (1 − t)
= 1− = t.
Also,
f (g(x)) = 1 +
1
1− 1−
1
x−1
SOLUTIONS to Review Problems for Chapter Eight
1
1
x−1
= 1 + x − 1 = x.
= 1+
68. Since h(x) =
√
2x and k(t) = t2 /2, we have
h(k(t)) =
s 2
t2
2
=
√
t2 = t
√
( 2x)2
2x
k(h(x)) =
=
=x
2
2
provided t ≥ 0.
provided x ≥ 0.
69. Solving y = f (x) for x gives
x−2
2x + 3
(2x + 3)y = x − 2
y=
2xy + 3y = x − 2
3y + 2 = x − 2xy
x(1 − 2y) = 3y + 2
3y + 2
.
x=
1 − 2y
Thus,
3y + 2
.
1 − 2y
g(y) =
70. Solving y = f (x) for x gives
r
4 − 7x
4−x
4 − 7x
y2 =
4−x
2
y (4 − x) = 4 − 7x
y=
4y 2 − y 2 x = 4 − 7x
4y 2 − 4 = y 2 x − 7x
4y 2 − 4 = x(y 2 − 7)
x=
Thus,
g(y) =
4y 2 − 4
.
y2 − 7
4y 2 − 4
.
y2 − 7
71. Solving y = f (x) for x gives
√
x+3
√
11 − x
√
√
y(11 − x) = x + 3
√
√
11y − y x = x + 3
y=
293
294
Chapter Eight /SOLUTIONS
11y − 3 =
√
√
√
x+y x
11y − 3 = x(1 + y)
√
11y − 3
x=
1+y
x=
Thus,
g(y) =
11y − 3
1+y
11y − 3
1+y
2
2
.
.
72. (a) We have y = (2x)3 + 5 = 8x3 + 5.
(b) We find the inverse by solving for x:
y = (2x)3 + 5
y − 5 = (2x)3
p
3
√
3
y − 5 = 2x
y−5
= x.
2
The sequence of operations in the inverse is to subtract 5, take the cube root, and divide by 2. Alternatively,
y = 8x3 + 5
r
3
√
3
y − 5 = 8xe
y−5
= x3
8
y−5
=x
8
y−5
= x.
2
The sequence of operations in the inverse is to subtract 5, divide by 8, and then take the cube root.
q
x
73. (a) We have y = 9
+ 4.
7
(b) We find the inverse by solving for x:
y=
q
9
x
+4
7
x
+4
7
x
y9 − 4 =
7
7(y 9 − 4) = x.
y9 =
The sequence of operations in the inverse is to raise to the ninth power, subtract 4, and multiply by 7.