1. science
2. mathematics
3. algebra

11.1 SOLUTIONS
629
CHAPTER ELEVEN
Solutions for Section 11.1
Skill Refresher
S1.
√
36t2 = (36t2 )1/2 = 361/2 · (t2 )1/2 = 6|t1 | = 6|t|
S2. Inside the parenthesis we write the radical as an exponent, which results in
√ 2
2
3x x3 = 3x · x3/2 .
Then within the parenthesis we write
3x1 · x3/2
2
= 3x5/2
2
= 32 (x5/2 )2 = 9x5 .
S3. Raising (0.1) and 4xy 2 to the second power yields (0.1)2 = (0.01) and 4xy 2
(0.01) 16x2 y 4 = 0.16x2 y 4 .
2
= 16x2 y 4 . Therefore (0.1)2 4xy 2
2
S4. In this example the same variable base w occurs in two separate factors: w1/2 and w1/3 . Since we are multiplying these
factors, we need to add the exponents, namely 1/2 and 1/3. This requires a common denominator of 6. Therefore
7 5w1/2
S5. We have
2w1/3 = 70 · w1/2 · w1/3 = 70w3/6 · w2/6 = 70w5/6 .
10x5−2 = 2
10x3 = 2
x3 = 0.2
x = (0.2)1/3 = 0.585.
S6. We have
5
= 500
x2
1
= 100
x2
2
x = 1/100
x = ±(1/100)1/2 = ±0.1.
S7. False
S8. False
S9. False
S10. False
Exercises
1. Yes. Writing the function as
g(x) =
we have k = −1/6 and p = 9.
(−x3 )3
(−1)3 (x3 )3
−x9
1
=
=
= − x9 ,
6
6
6
6
=
630
Chapter Eleven /SOLUTIONS
2. Yes. Writing the function as
R(t) = √
4
4
1
1
4
= √
√ = √ = √ = 1/2 = t−1/2 ,
t
4 t
t
16t
16 · t
we have k = 1 and p = −1/2.
3. Although y is a power function of (x + 7), it is not a power function of x and cannot be written in the form f (x) = kxp .
4. Yes. Writing the function as
T (s) = (6s−2 )(es−3 ) = 6es−2 s−3 = 6es−5 ,
we have k = 6e and p = −5.
5. No. This function cannot be written in the form kxp because the variable is in the exponent.
6. Yes. Writing the function as
w4
1
K(w) = √
=
4
4 w3
w4
√
w3
=
1
4
w4
w3/2
=
1
1 4−(3/2)
w
= w5/2 ,
4
4
we have k = 1/4 and p = 5/2.
7. This is a power function in the form y = axp :
y=
48
· x−2 ,
30625
48
,
30625
a=
p = −2.
We have
y=3
= 3·
2
√
5 7x
54
4
24
√ √ 4
7 x
=
48
625 · 49x2
=
48
· x−2 .
30625
8. This is a power function in the form y = axp :
√ 1.5
y=
8π x ,
a=
√
8π,
p = 1.5.
We see that
y=
p
√
π(2x)3
π · 8x3
√
= 8π x3
√ 1.5
8π x .
=
=
√
9. Since the graph is symmetric about the y-axis, the power function is even.
10. Since the graph is symmetric about the origin, the power function is odd.
11. Since the graph is steeper near the origin and less steep away from the origin, the power function is fractional.
12. Since the graph is symmetric about the origin, the power function is odd.
11.1 SOLUTIONS
631
13. We use the form y = kxp and solve for k and p. Using the point (1, 3), we have 3 = k1p . Since 1p is 1 for any p, we
know that k = 3. Using our other point, we see that
13
13
133
ln
3
ln(13/3)
ln 4
1.058
= 3 · 4p
= 4p
= p ln 4
=p
≈ p.
So y = 3x1.058 .
14. We use the form y = kxp and solve for k and p. Using the point (1, 0.7), we have 0.7 = k1p . Since 1p is 1 for any p, we
know that k = 0.7. Using our other point, we see that
8
8
0.7
8
ln
0.7
ln(8/0.7)
ln 7
1.252
= 0.7 · 7p
= 7p
= p ln 7
=p
≈ p.
So y = 0.7x1.252 .
15. Since f (1) = k · 1p = k, we know k = f (1) = 23
Since f (2) = k · 2p = 83 , and since k = 32 , we know
3
2
which implies
16. Taking the ratio of
1)
g(− 5
,
g(2)
3
2
· x−2 .
we have
(− 51 )p
g(− 51 )
k(− 15 )p
=
=
=
g(2)
k(2)p
(2)p
Since
we have
3
8
1
3 2
· = .
8 3
4
2p =
Thus p = −2, and f (x) =
· 2p =
− 51
2
p
= −
g(− 51 )
25
=
= 25(−40) = −1000,
1
g(2)
− 40
−
1
10
p
= −1000,
which implies that (−10)−p = (−10)3 . So p = −3.
1
Then, using either point, for example, g(2) = − 40
, we have
g(2) = k(2)−3 = −
so
k
1
8
=−
1
,
40
1
,
40
1
10
p
.
632
Chapter Eleven /SOLUTIONS
so
k=−
Thus,
1
1
·8=− .
40
5
1
g(x) = − x−3 .
5
17. Substituting into the general formula c = kd2 , we have 45 = k(3)2 or k = 45/9 = 5. So the formula for c is
c = 5d2 .
When d = 5, we get c = 5(5)2 = 125.
18. Substituting into the general formula c = k/d2 , we have 45 = k/32 or k = 405. So the formula for c is
c=
405
.
d2
When d = 5, we get c = 405/52 = 16.2.
19. Substituting into the general formula y = kx, we have 6 = k(4) or k = 23 . So the formula for y is
y=
When y = 8, we get 8 = 23 x, so x =
2
3
·8=
16
3
3
x.
2
= 5.33.
20. Substituting into the general formula y = k/x, we have 6 = k/4 or k = 24. So the formula for y is
y=
24
.
x
When y = 8, we get 8 = 24/x, so x = 24/8 = 3.
21. We need to solve f (x) = kxp for p and k. To solve for p, take the ratio of any two values of f (x), say f (3) over f (2):
f (3)
27
9
=
= .
f (2)
12
4
Since f (3) = k · 3p and f (2) = k · 2p , we have
f (3)
k · 3p
3p
=
= p =
p
f (2)
k·2
2
p
3
2
=
9
.
4
Since ( 23 )p = 94 , we know p = 2. Thus, f (x) = kx2 . To solve for k, use any point from the table. Note that f (2) =
k · 22 = 4k = 12, so k = 3. Thus, f (x) = 3 · x2 .
22. Solve for g(x) by taking the ratio of (say) g(4) to g(3):
−32/3
g(4)
−32 −2
64
=
=
·
=
.
g(3)
−9/2
3
9
27
We know g(4) = k · 4p and g(3) = k · 3p . Thus,
g(4)
k · 4p
4p
=
= p =
p
g(3)
k·3
3
p
4
3
=
64
.
27
9
= − 61 . This gives
Thus p = 3. To solve for k, note that g(3) = k · 33 = 27k. Thus, 27k = g(3) = − 29 . Thus, k = − 54
g(x) = − 16 x3 .
11.1 SOLUTIONS
633
23. We need to solve j(x) = kxp for p and k. We know that j(x) = 2 when x = 1. Since j(1) = k · 1p = k, we have k = 2.
To solve for p, use the fact that j(2) = 16 and also j(2) = 2 · 2p , so
2 · 2p = 16,
giving 2p = 8, so p = 3. Thus, j(x) = 2x3 .
24. Solve for h(x) by taking the ratio of, say, h(4) to h( 14 ):
−1/8
h(4)
−1 −1
1
=
=
·
=
.
−32
8
32
256
h( 41 )
We know h(4) = k · 4p and h( 41 ) = k · ( 41 )p . Thus,
h(4)
4p
1
k · 4p
.
= 1 p = 16p =
=
1
1 p
256
h( 4 )
k · (4)
(4)
1
Since 16p = 256
= 1612 = 16−2 , p = −2. To solve for k, note that h(4) = k · 4p = k · 4−2 =
1
k
we have 16 = − 8 . Thus, k = −2,which gives h(x) = −2x−2 .
k
.
16
Since h(4) = − 18 ,
25. (a) lim x−4 = lim (1/x4 ) = 0.
x→∞
(b)
x→∞
lim 2x−1 = lim (2/x) = 0.
x→−∞
x→−∞
26. (a) lim (t−3 + 2) = lim (1/t3 + 2) = 0 + 2 = 2.
t→∞
(b)
t→∞
lim (5 − 7y −2 ) = lim (5 − 7/y 2 ) = 5 − 0 = 5.
y→−∞
y→−∞
Problems
27. The graphs are shown in Figure 11.1.
y
x6
x4
x2
2
1
x
1
−1
Figure 11.1
Some observations we can make are:
• All three curves pass through the points (−1, 1), (0, 0), and (1, 1).
• All three curves have even symmetry and are more or less “U” shaped.
• x4 and x6 are much “flatter” near the origin than x2 , but climb much more steeply for x < −1 or x > 1.
28. (a) Note that y = x−10 = ( x1 )10 is undefined at x = 0. Since y = ( x1 )10 is raised to an even power, the graph
“explodes” in the same direction as x approaches zero from the left and the right. Thus, as x −→ 0, x−10 −→ +∞,
−x10 −→ 0.
(b) As x −→ ∞, x−10 −→ 0, −x10 −→ −∞.
(c) As x −→ −∞, x−10 −→ 0, −x10 −→ −∞.
29. (a) As x −→ 0 from the right, x−3 −→ +∞, and x1/3 −→ 0.
(b) As x −→ ∞, x−3 −→ 0, and x1/3 −→ ∞.
634
Chapter Eleven /SOLUTIONS
30. (a) The power function will be of the form g(x) = kxp , and from the graph we know p must be odd and k must be
negative. Using (−1, 3), we have
3 = k(−1)p ,
so
or
3 = −k
(since p is odd)
k = −3.
We do not have enough information to solve for p, since any odd p will work. Therefore, we have g(x) = −3xp , p
odd.
(b) Since the function is of the form g(x) = −3xp , with p odd, we know that the the graph of this function is symmetric
about the origin. This implies that if (a, b) is a point on the graph, then (−a, −b) is also a point on the graph. Thus
the information that the point (1, −3) is on the graph does not help us.
(c) Since we know that the function is symmetric about the origin it will follow that the points (−2, 96) and (0, 0) also
lie on the graph. To get other points lying on the graph, we can find the formula for this function. We know that
g(x) = −3xp
so plugging in the point (2, −96) we get
Solving for p
−96 = −3(2)p .
−96 = −3(2)p
32 = 2p
p = 5.
Thus the formula for the function is given by
g(x) = −3x5 .
√
√
Any values satisfying this formula will describe points on the graph: e.g. (3, −729) or (−0.1, 0.00003) or ( 7, −147 7)
etc.
31. c(t) = 1t is indeed one possible formula. It is not, however, the only one. Because the vertical and horizontal axes are
asymptotes for this function, we know that the power p is a negative number and
c(t) = ktp .
If p = −3 then c(t) = kt−3 . Since (2, 21 ) lies on the curve, 21 = k(2)−3 or k = 4. So, c(t) = 4t−3 could describe this
function. Similarly, so could c(t) = 16x−5 or c(t) = 64x−7 ...
32. (a) We have A = x3 , B = x2 , C = x3/2 , D = x, E = x1/2 , F = x1/3 . See Figure 11.2. For x > 1, the value of
y = x3 is greater than the value of y = x2 , which is greater than y = x, which is a line.
For large x (in fact, all x > 1), the graph of y = x1/2 is below the graph of y = x, and y = x1/3 is below
y = x1/2 . This is reasonable since, for x > 1, squaring x makes it bigger and cubing it makes it bigger still; thus
taking the square root of x makes it smaller and taking its cube root makes it smaller still. Between x = 0 and x = 1,
the situation is reversed and y = x1/3 is on top. (Why?) Not surprisingly, y = x3/2 is between y = x and y = x2
for all x.
y
x3x2 x3/2 x
2
x1/2
x1/3
1
0
x
1
Figure 11.2
2
11.1 SOLUTIONS
635
(b) We see that y = x2 is concave up, whereas y = x1/2 is concave down. For x positive, y = x3 is concave up, whereas
y = x1/3 is concave down. This occurs because y = x2 and y = x1/2 are inverse functions (as are y = x3 and
y = x1/3 ). The graphs of y = x2 and y = x1/2 are reflections in the line y = x, which changes the concavity.
33. All four graphs contain the point (1, 1), so they all four have k = 1. Judging from their graphs, f and g appear to have
positive values of p. The graph of g climbs faster as x → ∞ so its value of p is larger. The graphs of v and w have
asymptotes, suggesting their values of p are negative. The graph of v approaches its horizontal asymptote faster than the
graph of w, so its value of p is “more negative.”
Therefore, ranking these functions in order of their power, p, gives: v, w, f, g.
34. The graphs of g and w exhibit odd symmetry (symmetry across the origin), so these functions have an odd value of p.
35. We have
1
F (x) = √
3
7x
1
= √
√
3
73x
n
z }| {
1
= √
·x−1/3
3
7
|{z}
k
1
so k = √
3
7
1
n=−
3
which means
f (x) = nkxn−1
1
1 1
· √ · x− 3 −1
3 37
4
1
· x− 3 .
=− √
3
3 7
=−
36. We have
f (x) =
=
so
k=
n=
which means
F (x) =
=
=
=
√
5
x2
4
2
1
· x5
4
1
4
2
5
kxn+1
n+1
1 2
x 5 +1
4
2
+1
5
1 7
x5
4
7
5
7
5
· x5 .
28
37. (a) Since the cost of the fabric, C(x), is directly proportional to the amount purchased, x, we know that the formula will
be of the form
C(x) = kx.
636
Chapter Eleven /SOLUTIONS
(b) Since 3 yards cost $28.50, we know that C(3) = $28.50. Thus, we have
28.50 = 3k
k = 9.5
Our formula for the cost of x yards of fabric is
C(x) = 9.5x.
(c) Notice that the graph in Figure 11.3 goes through the origin.
cost ($)
C(x) = 9.5x
50
5
x (yards)
Figure 11.3
(d) To find the cost of 5.5 yards of fabric, we evaluate C(x) for x = 5.5:
C(5.5) = 9.5(5.5) = $52.25.
38. Calories are directly proportional to ounces because as the amount of beef, x, increases, the number of calories, c, also
increases. Substituting into the general formula c = kx, we have 245 = k(3) or k = 81.67. So the formula is
c = 81.67x.
When x = 4, c = 81.67(4) = 326.68. Therefore, 4 ounces of hamburger contain 326.68 calories.
39. (a) Table 11.1 shows the circulation times in seconds for various mammals.
Table 11.1
Animal
Body mass (kg)
Circulation time (sec)
Blue whale
91000
302
African elephant
5450
150
White rhinoceros
3000
129
Hippopotamus
2520
123
Black rhinoceros
1170
102
Horse
700
90
Lion
180
64
Human
70
50
(b) If a mammal of mass m has a circulation time of T , then
T = 17.4m1/4 .
If a mammal of mass M has twice the circulation time, then
2T = 17.4M 1/4 .
11.1 SOLUTIONS
637
We want to find the relationship between m and M , so we divide these two equations, giving
17.4M 1/4
2T
=
.
T
17.4m1/4
Simplifying, we have
2=
M 1/4
.
m1/4
Taking the fourth power of both sides, we get
24 =
M
,
m
and thus
M
.
m
The body mass of the animal with the larger circulation time is 16 times the body mass of the other animal.
16 =
40. (a) Since B is proportional to the three-quarters power of M, we have
3
B = kM 4 .
(b) If ME is the average mass of an elephant, and Mm that of a mouse then
3
3
3
kME4
(4.6 · 1,000,000 grams) 4
Metabolic rate of an elephant
= 230,000 4 = 10,502.6,
=
3 =
3
Metabolic rate of a mouse
4
4
(20
grams)
kMm
so an elephant at rest uses over 10,000 times as much energy as a mouse!
41. Since the pitch is inversely proportional to the square root of the density, we have
1
1
P = k √ = kρ− 2 ,
ρ
for some constant k.
42. (a) We assume that the cost, c, in dollars, of a commercial is proportional to its time t, in seconds, so c = k · t. The
constant k is the cost per second that we want.
In Super Bowl XLIV in 2010, the cost was $3 million = $3 × 106 for 30 seconds, so
3 × 106 = k · 30.
So
3,000,000
= 100,000 dollars/sec.
30
In Super Bowl I in 1967, for the same cost, $3 million, advertisers could buy 34.439 minutes= 34.439 × 60
seconds.
k=
3 × 106 = k · 34.439 × 60
3 × 106
k=
≈ 1451.84 dollars/sec.
34.439 × 60
(b) The cost per second in 2010 was $90,000 and in 1967, it was $1451.84. Thus the cost per second has increased by a
factor of
$90,000
≈ 61.99.
$1451.84
So cost has increased by a factor of about 62.
638
Chapter Eleven /SOLUTIONS
43. The number of hours is inversely proportional to the speed, because as the speed, v, increases, the number of hours, h,
decreases.
Substituting the given values into the general formula h = k/v, we get 3.5 = k/55, so k = 192.5 and the formula
is
192.5
.
h=
v
When h = 3, we get 3 = 192.5/v, or v = 64.167. So getting to Albany in 3 hours would require the speed of
64.167 mph.
√
√
44. (a) With u = 9 and l = 225 we find k = 9/ 225 = 3/5. With k = 0.6 and l = 4, we find u = (0.6) 4 = 1.2
meters/sec.
(b) Suppose the existing ship has speed u and length l, so
√
u = k l.
The new ship has speed increased by 10%, so the new speed is 1.1u. If the new length is L, since the constant remains
the same, we have
√
1.1u = k L.
Dividing these two equations we get
√
1.1u
k L
= √ .
u
k l
Simplifying and squaring we get
√
L
1.1 = √
l
L
(1.1)2 =
l
so
L = (1.1)2 l = 1.21l.
Thus, the new hull length should be 21% longer than the hull length of the existing ship.
45. (a) Since the speed of sound is 340 meters/sec = 0.34 km/sec, if t is in seconds,
d = f (t) = 0.34t.
Thus, when t = 5 sec, d = 0.34 · 5 = 1.7 km. Other values in the second row of Table 11.2 are calculated in a
similar manner.
Table 11.2
Time, t
5 sec
10 sec
1 min
5 min
Distance, d (km)
1.7
3.4
20.4
102
Area, A (km2 )
9.1
36.3
1307
32685
(b) Using d = 0.34t, we want to calculate t when d = 200, so
200 = 0.34t
200
t=
= 588.24 sec = 9.8 mins.
0.34
(c) The values for A are listed in Table 11.2. These were calculated using the fact that the area of a circle of radius r is
A = πr 2 . At time t, the radius of the circle of people who have heard the explosion is d = 0.34t. Thus
A = πd2 = π(0.34t)2 = π(0.34)2 t2 = 0.363t2 .
This formula was used to calculate the values of A in Table 11.2.
11.1 SOLUTIONS
639
(d) Since the population density is 31 people/km2 , the population, P , who have heard the explosion is given by
P = 31A.
2
Since A = 0.363t , we have
P = 31 · 0.363t2 = 11.25t2 .
2
So P = f (t) = 11.25t .
(e) The graph of P = f (t) = 11.25t2 is in Figure 11.4.
P
3000000
2000000
1000000
100
300
500
t (sec)
Figure 11.4
To find when 1 million people have heard the explosion, we find t when P = 1,000,000:
1000000 = 11.25t2
1000000
t2 =
11.25
t=
r
1000000
= 298 sec ≈ 5 min.
11.25
46. (a) The radius, r, is 0 when t = 0 and increases by 200 meters per hour. Thus, after t hours, the radius in meters is given
by
r = 200t.
(b) Since the spill is circular, its area, A, in square meters, is given by
A = πr 2 .
Substituting for r from part (a), we have
A = π(r)2 = π(200t)2 = 40,000πt2 .
(c) When t = 7, the area (in square meters) is
A = 40,000 · π · 72 ≈ 6,157,521.601.
47. (a) Since cooking time, t, is inversely proportional to power level, w, we know that for some constant k
t=
k
.
w
We know that t = 6.5 minutes when w = 750 watts, so we solve for k:
k
750
k = 6.5(750) = 4875.
6.5 =
Thus, the function is
t = f (w) =
4875
.
w
640
Chapter Eleven /SOLUTIONS
(b) When w = 250 watts
4875
= 19.5 minutes.
250
Continuing in the same way gives the result in Table 11.3
t=
Table 11.3
Power, w (watts)
250
300
500
650
Time, t (mins)
19.5
16.25
9.75
7.5
(c) The graph of t = f (w) = 4875/w is shown in Figure 11.5
t (mins)
20
10
500
1000
w (watts)
Figure 11.5
(d) For a new dish, there is a new value of k. Since t = 2 when w = 250
k
250
k = 500,
2=
Calculating t when w = 500,
t=
500
= 1 minute.
500
48. (a) Since hailstones are approximately spherical, we expect their volume to be proportional to the cube of their radius.
Since mass is proportional to volume, mass should be proportional to the cube of radius. To check, for each hailstone
we calculate the ratio Mass/(Radius)3 .
1.835
= 2.140,
(0.95)3
Mass
0.058
= 2.148,
=
(0.3)3
(Radius)3
750
= 2.140.
(7.05)3
The fact that all three ratios are approximately 2.1 shows that m could be proportional to the cube of the radius.
(b) The constant of proportionality is about 2.1, so m = 2.1r 3 .
(c) To find the radius, substitute m = 3.4 kg = 3.4(1000 gm) = 3400 gm and solve for r:
3400 = 2.1r 3
r3 =
3400
2.1
so
r=
r
3
3400
= 11.7 cm.
2.1
(d) Pick one of the hailstones and calculate its volume. Using the formula for the volume of a sphere, for the average
hailstone, we have
4
4 3
πr = π(0.3)3 .
3
3
Mass
0.058
Density =
= 0.513 gm/cm3 .
=
4
Volume
π(0.3)3
3
Volume =
Thus, each cubic centimeter of ice weighs about 0.513 grams.
11.1 SOLUTIONS
641
49. (a) The domain of y = x−p is all real numbers except x = 0. To find the range we must consider separately the cases
when p is even and when p is odd. In the case that p is even, y is always positive. This is because if p = 2k (where k
is a positive integer) then
1
y = x−p = 2 k
(x )
and x2 is never negative. Also there is no value of x for which y is equal to zero, because y =
2 k
zero. Since the function (x ) ranges over all positive numbers, so will the function y =
range is all positive numbers.
If p is odd then we again note that zero is not in the range. We can rewrite
y = x−p =
(x2 )k
1
(x2 )k
can never be
. Thus if p is even the
1
.
xp
The range of the function xp is all real numbers. Therefore, the range of y =
excluding 0, or all real numbers except 0.
(b) If p is even we again write p = 2k (where k is a positive integer) and
y = x−p =
1
1
xp
will be the range of the function xp
1
.
(x2 )k
Now we note that y is symmetric with respect to the y-axis, since
1
1
=
((−x)2 )k
((−1)2 (x2 ))k
1
= 2 k
(x )
If p is odd,
1
1
1
=
= − p,
(−x)p
(−1)p xp
x
so y(−x) = −y(x). Thus, when p is odd, y = x−p is symmetric with respect to the origin.
(c) When p is even,
x−p → ∞
as
x → 0+
x−p → ∞
as
x → 0− .
Thus, values of the function show the same pattern on each side of the y-axis – a property we would certainly expect
for a function with even symmetry.
When p is odd, we found in part (b) that y = x−p is symmetric about the origin. Therefore, if y → +∞ as
x → 0+ we would expect y → −∞ as x → 0− . This behavior is consistent with the behavior of y = 1/x and
y = 1/x3 .
(d) Again we will write
1
y = x−p = p .
x
When x is a large positive number, xp is a large positive number, so its reciprocal is a small positive number. If p is
even (and thus the function is symmetric with respect to the y-axis) then y will be a small positive number for large
negative values of x. If p is an odd number (and thus y is symmetric with respect to the origin) then y will be a small
negative number for large negative values of x.
50. (a) We have
f (x) = g (h(x)) = 16x4 .
2
Since g(x) = 4x , we know that
g (h(x)) = 4 (h(x))2 = 16x4
(h(x))2 = 4x4 .
Thus,
Since h(x) ≤ 0 for all x, we know that
h(x) = 2x2 or − 2x2 .
h(x) = −2x2 .
642
Chapter Eleven /SOLUTIONS
(b) We have
f (x) = j (2g(x)) = 16x4 ,
j(x) a power function.
2
Since g(x) = 4x , we know that
j (2g(x)) = j(8x2 ) = 16x4 .
p
Since j(x) is a power function, j(x) = kx . Thus,
j(8x2 ) = k(8x2 )p = 16x4
k · 8p x2p = 16x4 .
Since x2p = x4 if p = 2, letting p = 2, we have
k · 64x4 = 16 · x4
64k = 16
1
k=
4
Thus, j(x) = 41 x2 .
51. (a) If p is negative, the domain is all x except x = 0.
There are no domain restrictions if p is positive.
(b) If p is even and positive, the range is y ≥ 0.
If p is even and negative, the range is y > 0.
If p is odd and positive, the range is all real numbers.
If p is odd and negative, the range is all real numbers except y = 0.
(c) If p is even, the graph is symmetric about the y-axis.
If p is odd, the graph is symmetric about the origin.
Solutions for Section 11.2
Exercises
1. Since 5x is not a power function, this is not a polynomial.
2. This is a polynomial of degree one (since x = x1 ).
3. This is a polynomial of degree two.
4. Yes, this is a polynomial (with variable t) of degree 6.
5. This is not a polynomial because 2ex is not a power function.
6. Since y = 4x2 − 7x9/2 + 10 and 9/2 is not a nonnegative integer, this is not a polynomial.
7. y = 2x3 − 3x + 7 is a third-degree polynomial with three terms. Its long-run behavior is that of y = 2x3 : as x →
−∞, y → −∞, as x → +∞, y → +∞.
8. y = 1 − 2x4 + x3 is a fourth degree polynomial with three terms. Its long-run behavior is that of y = −2x4 : as
x → ±∞, y → −∞.
9. y = (x + 4)(2x − 3)(5 − x) = −2x3 + 5x2 + 37x − 60 is a third-degree polynomial with four terms. Its long-run
behavior is that of y = −2x3 : as x → −∞, y → +∞, as x → +∞, y → −∞.
10. (a) lim (3x2 − 5x + 7) = lim (3x2 ) = ∞.
x→∞
(b)
x→∞
lim (7x2 − 9x3 ) = lim (−9x3 ) = ∞.
x→−∞
x→−∞
11.2 SOLUTIONS
643
Problems
11. See Figure 11.6. There are two real zeros at x ≈ 0.718 and x ≈ 1.702.
0.718
x
1.702
Figure 11.6
12. The graph of y = g(x) is shown in Figure 11.7 on the window −5 ≤ x ≤ 5 by −20 ≤ y ≤ 10. The minimum value of
g occurs at point B as shown in the figure. Using either a table feature or trace on a graphing calculator, we approximate
the minimum value of g to be −16.543 (to three decimal places).
y
x
B
Figure 11.7
13. The window −10 ≤ x ≤ 10 by −20 ≤ y ≤ 20 gives a reasonable picture of both functions. See Figure 11.8. The
functions cross the x-axis in the same places, which indicates the zeros of f and g are the same. The y-intercepts are
different, since f (0) = −5 and g(0) = 10. In addition, the end behaviors of the functions differ. The function g has been
flipped about the x-axis by the negative coefficient of x3 .
20
f (x)
x
10
−10
g(x)
−20
Figure 11.8
14. (a) The graphs of u and v have the same end behavior. As x → −∞, both u(x) and v(x) → ∞, and as x → ∞, both
u(x) and v(x) → −∞.
The graphs have different y-intercepts, and u has three distinct zeros. The function v has a multiple zero at
x = 0. See Figure 11.9.
644
Chapter Eleven /SOLUTIONS
10
v(x)
u(x)
x
10
−10
−10
Figure 11.9
(b) On the window −20 ≤ x ≤ 20 by −1600 ≤ y ≤ 1600, the peaks and valleys of both functions are not distinguishable. Near the origin, the behavior of both functions looks the same. The functions are still distinguishable from one
another on the ends.
On the window −50 ≤ x ≤ 50 by −25,000 ≤ y ≤ 25,000, the functions are still slightly distinct from one
another on the ends—but barely.
On the last window the graphs of both functions appear identical. Both functions look like the function y =
− 15 x3 .
1,600
25,000
25,000,000
x
20
−20
x
50
−50
x
500
v(x)
−500
v(x)
v(x)
u(x)
−1,600
u(x)
u(x)
−25,000
−25,000,000
Figure 11.10
15. To find the x-intercept for y = 2x − 4 let y = 0. We have
0 = 2x − 4
2x = 4
x = 2.
4
5
2
When x = 0 on y = x − 3x − 1 + x , then y = −1. This gives the y-intercept for y = x4 − 3x5 − 1 + x2 . Thus, we
have the points (2, 0) and (0, −1). The line through these points will have the same y-intercept, so the linear function is
of the form
y = mx − 1.
The slope, m, is found by taking
Thus,
0 − (−1)
1
= .
2−0
2
y=
is the line through the required points.
1
x−1
2
11.2 SOLUTIONS
645
16. (a) If x = 10, for example, we have :
1
1
(10)3 + (10)
50,000
2
1
=
+ 5 = 5.02
50
f (x) = f (10) =
The most significant term is the linear term 12 x.
y
(b)
y = f (x)
5
x
10
−10
−5
Figure 11.11
On this scale, the graph of f (x) in Figure 11.11 is difficult to distinguish from the line y = 21 x; that is, the graph
of the linear term alone. However, f itself is definitely not linear; it just looks this way near the origin.
(c) We would like to find the value of x for which the linear term is equal to the cubic term. That is, we want:
1
1
x3 = x
50,000
2
which implies that:
x3 = 25,000x
x3 − 25,000x = 0
x(x2 − 25,000) = 0
so
x=0
or
x=±
p
25,000 ≈ ±158.114.
Thus, the cubic term becomes “important” for x > 158.114 or x < −158.114. (Neither term is “important” at
x = 0, the origin.)
17. Here are the viewing windows used to create these figures. Your viewing windows may differ to a certain extent and still
give similar-looking graphs.
(a)
(b)
(c)
(d)
−3 ≤ x ≤ −1, −5 ≤ y ≤ 5
−3 ≤ x ≤ 4, −35 ≤ y ≤ 15
1.25 ≤ x ≤ 2.35, −0 ≤ y ≤ 6
−8 ≤ x ≤ 8, −50 ≤ y ≤ 2000
18. We have −0.1 ≤ x ≤ 0.1, 0 ≤ y ≤ f (0.1) or 0 ≤ y ≤ 0.011.
19. We have −1.1 ≤ x ≤ −0.9, f (−1.1) ≤ y ≤ f (−0.9) or −0.121 ≤ y ≤ 0.081.
20. We have −1.1 ≤ x ≤ 0.3, f (−1.1) ≤ y ≤ f (0.3) or −0.121 ≤ y ≤ 0.117.
21. We have −20 ≤ x ≤ 20, f (−20) ≤ f (20) or −7600 ≤ y ≤ 8400.
22. (a) Using a computer or a graphing calculator, we can get a picture of f (x) like the one in Figure 11.12. On this window
f appears to be invertible because it passes the horizontal line test.
(b) Substituting gives
f (0.5) = (0.5)3 + 0.5 + 1 = 1.625.
646
Chapter Eleven /SOLUTIONS
To find f −1 (0.5), we solve f (x) = 0.5. With a computer or graphing calculator, we trace along the graph of f in
Figure 11.13 to find
f −1 (0.5) ≈ −0.424.
y
2
1.625 = f (0.5)
y
1
30
0.5
x
x
−1 f −1 (0.5)
= −0.424
3
−3
0.5
1
−1
−30
Figure 11.12: f (x) = x3 + x + 1
Figure 11.13
23. Use a graphing calculator or computer to approximate values where f (x) = g(x) or to find the zeros for f (x) − g(x). In
either case, we find the points of intersection for f and g to be x ≈ −1.764, x ≈ 0.875 and x ≈ 3.889. The values of x
for which f (x) < g(x) are on the interval −1.764 < x < 0.875 or x > 3.889.
24. We see that F is a degree n = 4 polynomial with: a4 = 3, a3 = −4, a2 = 0, a1 = 5, a0 = −4. This means:
f (x) = n · an xn−1
|
{z
4·3x3
}
because n = 4 and a4 = 3
+ (n − 1) · an−1 xn−2
|
{z
3(−4)x5
+ 2a2 x
|{z}
}
because n − 1 = 3 and a3 = −4
because n − 2 = 2 and a2 = 0
2·0·x
+ a1
|{z}
because a1 = 5
5
= 4 · 3x3 + 3(−4)x2 + 2 · 0 · x + 5
= 12x3 − 12x2 + 5.
25. (a) The graph of the function in the suggested window is shown in Figure 11.14.
(b) At t = 0 (when Liddleville was founded), the population was 1 hundred people.
(c) The t-intercept for t > 0 will show when the population was zero. This occurs at t ≈ 7.54. Thus, Liddleville’s
population reached zero in July of 1897.
(d) The graph of y has a peak at t ≈ 3.12. The population at that point is ≈ 10.1 hundred. So the maximum population
was ≈ 1010 in February of 1893.
(e) Population predicted is −1.157 hundred. Since the actual population cannot be negative, we see the model does not
predict well at t = 8.
647
11.2 SOLUTIONS
y
10
8
6
4
2
2
−2
4
6
8
t
Figure 11.14
26. (a) The graph of C(x) = 4(x − 1)2 + 4 is the graph of y = x2 shifted right one unit, a vertical stretch by a factor of 4,
and then up by 4 units. The graph is shown in Figure 11.15.
(b) The price is $10,000 per unit, since R(1) = 10 means selling 1000 units yields $10,000,000.
(c) We have
Profit = R(x) − C(x)
= 10x − [4(x − 1)2 + 4]
= 10x − 4(x − 1)2 − 4
= −8 + 18x − 4x2
The graph of R(x) − C(x) is shown in Figure 11.16. Profit is negative for x < 0.5 and for x > 4. Profit = 0 at
x = 0.5 and x = 4. Thus, the firm will break even with either 500 or 4000 units, make a profit for 500 < x < 4000
units, and lose money for any number of units between 0 and 500 or greater than 4000.
C
50
Profit
10
x
25
−10
−20
−30
−40
x
1
2
3
2
4
6
4
Figure 11.15
Figure 11.16
27. (a) A graph of V is shown in Figure 11.17 for 0 ≤ t ≤ 5, 0 ≤ V ≤ 1.
Volume
1
V (t)
0.8
0.6
0.4
0.2
1
2
3
4
5
t (sec)
Figure 11.17
(b) The maximum value of V for 0 ≤ t ≤ 5 occurs when t ≈ 3.195, V ≈ .886. Thus, at just over 3 seconds into the
cycle, the lungs contain ≈ 0.88 liters of air.
(c) The volume is zero at t = 0 and again at t ≈ 5. This indicates that at the beginning and end of the 5 second cycle the
lungs are empty.
648
Chapter Eleven /SOLUTIONS
28. (a) We are interested in V for 0 ≤ T ≤ 30, and the y-intercept of V occurs at (0, 999.87). If we look at the graph of V
on the window 0 ≤ x ≤ 30 by 0 ≤ y ≤ 1500, the graph looks like a horizontal line. Since V is a cubic polynomial,
we suspect more interesting behavior with a better choice of window. Note that V (30) ≈ 1003.77, so we know V
varies (at least) from V = 999.87 to V ≈ 1003.77. Change the range to 998 ≤ y ≤ 1004. On this window we see a
more appropriate view of the behavior of V for 0 ≤ T ≤ 30. See Figure 11.18. [Note: To view the function V as a
cubic, a much larger window is needed. Try −500 ≤ x ≤ 500 by −3000 ≤ y ≤ 5000.]
Volume
1004
1003
V (T )
1002
1001
1000
10
20
30
T (◦ C)
Figure 11.18
(b) The value of V decreases for 0 ≤ T ≤ 3.961 and then increases for 3.961 < T < 30. The graph is concave up on the
interval 0 ≤ T ≤ 30 (i.e., the graph bends upward). Thus, the volume of 1 kg of water decreases as T increases from
0◦ C to 3.961◦ C and increases thereafter. The volume increases at an increasing rate as the temperature increases.
(c) If density, d, is given by d = m/V and m is constant, then the maximum density occurs when V is minimum.
Thus, the maximum density occurs when T ≈ 3.961◦ C. [Note: We have m = 1, but a graph of y = 1/V is very
difficult to distinguish from a horizontal line. One possible choice of window to view y = 1/V is 0 ≤ x ≤ 30,
0.000996 ≤ y ≤ 0.001001.]
29. Yes. For the sake of illustration, suppose f (x) = x2 + x + 1, a second-degree polynomial. Then
f (g(x)) = (g(x))2 + g(x) + 1
= g(x) · g(x) + g(x) + 1.
Since f (g(x)) is formed from products and sums involving the polynomial g, the composition f (g(x)) is also a polynomial. In general, f (g(x)) will be a sum of powers of g(x), and thus f (g(x)) will be formed from sums and products
involving the polynomial g(x). A similar situation holds for g(f (x)), which will be formed from sums and products
involving the polynomial f (x). Thus, either expression will yield a polynomial.
30. (a) If f is even, then for all x
f (x) = f (−x).
Since f (x) = ax2 + bx + c, this implies
ax2 + bx + c = a(−x)2 + b(−x) + c
= ax2 − bx + c
We can cancel the ax2 term and the constant term c from both sides of this equation, giving
bx = −bx
bx + bx = 0
2bx = 0.
Since x is not necessarily zero, we conclude that b must equal zero, so that if f is even,
f (x) = ax2 + c.
11.2 SOLUTIONS
649
(b) If g is odd, then for all x
3
2
Since g(x) = ax + bx + cx + d, this implies
g(−x) = −g(x).
ax3 + bx2 + cx + d = −(a(−x)3 + b(−x)2 + c(−x) + d)
= −(−ax3 + bx2 − cx + d)
= ax3 − bx2 + cx − d.
The odd-powered terms cancel, leaving
bx2 + d = −bx2 − d
2bx2 + 2d = 0
bx2 + d = 0.
Since this must hold true for any value of x, we know that both b and d must equal zero. Therefore,
g(x) = ax3 + cx.
31. (a) False. For example, g(x) = x3 + x2 is not odd.
(b) False. For example, g(x) = (x − 1)(x − 2)(x − 3) is not invertible, since there are three values (x = 1, 2, 3) where
g(x) = 0.
(c) False. For example, g(x) = −x3 , where lim g(x) = −∞.
x→∞
(d) True. If lim g(x) = −∞, the leading coefficient is positive; that is, an > 0. Thus
x→−∞
lim g(x) = lim an xn = ∞.
x→∞
x→∞
32. (a) See Figure 11.19.
3
f (x)
2
2π
−π
π
−2π
x
sin(x)
−2
−3
Figure 11.19
(b) The graphs are very similar on the interval
−π
π
≤x≤ ,
2
2
and even slightly larger intervals show close similarity.
(c)
As you can see, f
π
8
π
= 0.382683432 · · ·
8
π π 1 π 3
π 5
1
f
= −
+
8
8
6 8
120 8
= 0.382683717 · · · .
sin
differs from sin
π
8
only in the 7th decimal place—that is, by less than 0.0001%.
650
Chapter Eleven /SOLUTIONS
(d) Since sin x is periodic with period 2π, we know that sin 18 = sin(18 − 2π) = sin(18 − 4π) = sin(18 − 6π) = · · ·.
Notice that 18 − 6π = −0.8495 · · · is within the interval − π2 ≤ x ≤ π2 on which f resembles sin x. Thus,
f (18 − 6π) ≈ sin(18 − 6π), and since sin(18 − 6π) = sin 18, then
f (18 − 6π) ≈ sin 18.
Using a calculator, we find f (18 − 6π) = −0.7510 · · · , and sin(18) = −0.75098 · · ·. Thus, f (18 − 6π) is an excellent approximation for sin 18. (In fact, your calculator evaluates trigonometric functions internally using a method
similar to the one presented in this problem.)
33. (a) Substituting x = 0.5 into p, we have
p(0.5) = 1 − 0.5 + 0.52 − 0.53 + 0.54 − 0.55 ≈ 0.65625.
Since f (0.5) = 2/3 = 0.6666...., the approximation is accurate to 2 decimal places.
(b) We have p(1) = 1 − 1 + 1 − 1 + 1 − 1 = 0, but f (1) = 0.5. Thus p(1) is a poor approximation to f (1).
(c) See Figure 11.20. The two graphs are difficult to tell apart for −0.5 ≤ x ≤ 0.5, but for x but outside this region the
fit is not good.
y
6
4
2
x
1
−1
Figure 11.20
34. See Table 11.4.
Table 11.4 Approximations for the speed of sound (in m/s) in water at
temperature T (◦ C)
T , ◦C
0
5
10
15
20
25
30
Linear
1402.4
1427.6
1452.8
1478.0
1503.2
1528.4
1553.5
Quadratic
1402.4
1426.1
1447.0
1464.9
1480.0
1492.1
1501.4
Cubic
1402.7
1426.5
1447.6
1466.3
1482.9
1497.5
1510.5
(a) From Table 11.4, we see that at 5◦ C, the linear approximation already differs from the actual temperature by more
than 1◦ C. Thus, this approximation is valid only for 0 ≤ T < 5.
(b) We see that the quadratic approximation is valid for 0 ≤ T < 15.
(c) We see that the cubic approximation is valid for 0 ≤ T < 30.
(d) At T = 50◦ C, the cubic approximation gives a velocity of 1551.0 m/sec, which is much higher than the actual
value of 1542.6 m/sec. Thus, in order for the quartic term to give a better approximation at this temperature, it
should be negative, so that it will lower the estimate. In fact, a better approximation can be given by adding the
term −1.398845 · 10−6 T 4 . An even better approximation can be given by also adding the quintic (fifth-degree) term
2.787860 · 10−9 T 5 . The resulting formula is known as the Marczak formula:
v = 1.402385 · 103 + 5.038813T − 5.799136 · 10−2 T 2
+ 3.287156 · 10−4 T 3 − 1.398845 · 10−6 T 4 + 2.787860 · 10−9 T 5 .
11.3 SOLUTIONS
651
Solutions for Section 11.3
Exercises
1. Zeros occur where y = 0, which we can find by factoring:
x3 + 7x2 + 12x = 0
x(x2 + 7x + 12) = 0
x(x + 4)(x + 3) = 0.
Zeros are at x = 0, x = −4, and x = −3.
2. Zeros occur where y = 0. First we factor:
(x2 + 2x − 7)(x3 + 4x2 − 21x) = 0
(x2 + 2x − 7)x(x2 + 4x − 21) = 0
x(x2 + 2x − 7)(x + 7)(x − 3) = 0.
To find the points where x2 + 2x − 7 = 0, we use the quadratic formula, so
√
−2 ± 22 − 4 · 1 · −7
x=
2·1
√
−2 ± 32
x=
2
√
−2 ± 2 8
x=
2
√
x = −1 ± 8.
√
√
Thus, zeros are at x = 0, x = −7, x = 3, x = −1 + 8, and x = −1 − 8.
3. Zeros occur where y = 0, at x = −3, x = 2, and x = −7.
4. Zeros occur where y = 0, at x = −2 and x = b.
5. The graph of h shows zeros at x = 0, x = 3, and a multiple zero at x = −2. Thus
h(x) = x(x + 2)2 (x − 3).
Check by multiplying and gathering like terms.
6. The graph shows that g(x) has zeros at x = −2, x = 0, x = 2, x = 4. Thus, g(x) has factors of (x + 2), x, (x − 2), and
(x − 4), so
g(x) = k(x + 2)x(x − 2)(x − 4).
Since g(x) = x4 − 4x3 − 4x2 + 16x, we see that k = 1, so
g(x) = x(x + 2)(x − 2)(x − 4).
7. The graph shows zeros at x = −2 and x = 2. The fact that f “lingers” at x = 2 before crossing the x-axis indicates a
multiple zero at x = 2. Since the function changes sign at x = 2, the factor (x − 2) is raised to an odd power. Thus, try
f (x) = (x + 2)(x − 2)3
(Check this answer by expanding and gathering like terms.)
652
Chapter Eleven /SOLUTIONS
8. The function has a common factor of 4x which gives
f (x) = 4x(2x2 − x − 15),
and the quadratic factor reduces further giving
f (x) = 4x(2x + 5)(x − 3).
Thus, the zeros of f are x = 0, x =
−5
,
2
and x = 3.
9. This polynomial must be of fourth (or higher even-powered) degree, so either (but not both) of the zeros at x = 2 or
x = 5 could be doubled. One possible formula is y = k(x + 2)(x − 2)2 (x − 5). Solving for k gives
k(0 + 2)(0 − 2)2 (0 − 5) = 5
−40k = 5
1
k=− ,
8
so
1
y = − (x + 2)(x − 2)2 (x − 5).
8
Another possible formula is y = k(x + 2)(x − 2)(x − 5)2 . Solving for k gives
k(0 + 2)(0 − 2)(0 − 5)2 = 5
−100k = 5
k=−
1
,
20
so
1
(x + 2)(x − 2)(x − 5)2 .
20
There are other possible polynomials, but all are of degree higher than 4, so these are the simplest.
y=−
10. Factoring f gives f (x) = −5(x + 2)(x − 2)(5 − x)(5 + x), so the
x intercepts are at x = −2, 2, 5, −5.
The y intercept is at: y = f (0) = −5(2)(−2)(5)(5) = 500.
The polynomial is of fourth degree with the highest powered term 5x4 . Thus, both ends point upward. A graph of
y = f (x) is shown in Figure 11.21.
500
500
−5
−2
2
5
4 5
−5
Figure 11.21
Figure 11.22
11. Factoring g gives g(x) = 5(x − 4)(x + 5)(x − 5), so the
x intercepts are at x = 4, −5, 5.
The y intercept is at: y = g(0) = 5(−4)5(−5) = 500.
The polynomial is third degree with 5x3 the highest powered term. Thus, the end behavior is g(x) → ∞ as x → ∞
and g(x) → −∞ as x → −∞. A graph of y = g(x) is shown in Figure 11.22.
11.3 SOLUTIONS
653
Problems
12. (a) Setting each of the four factors equal to zero, we find that f has four zeros: x, = 1/2x = 1/3, x = 7, and x = 9.
(b) It is not possible to find a single viewing window which shows all zeros and all turning points because of the fact that
the turning point between x = 1/2 and x = 7 occurs at a very extreme y value (over 1000). The window required to
see this turning point makes it difficult to see the zeros at x = 1/2 and x = 1/3 because they occur so close together.
(c) The viewing window 0 ≤ x ≤ 10, −500 ≤ y ≤ 1500 clearly shows the turning points of f and the zeros x = 7 and
x = 9, while the window 0 ≤ x ≤ 1, −2 ≤ y ≤ 2 shows the zeros x = 1/2 and x = 1/3.
13. (a) The viewing window −6 ≤ x ≤ 2, −10 ≤ y ≤ 10 indicates that the zeros of f occur at approximately x = −5, x =
−1, x = 1/2, and x = 1. We therefore guess that
f (x) = (x + 5)(x + 1)(2x − 1)(x − 1).
We check this by expanding the product to obtain f (x).
(b) The window −7 ≤ x ≤ 2, −150 ≤ y ≤ 10 clearly shows all the turning points of f.
14. First, the viewing window 0 ≤ x ≤ 5, −25 ≤ y ≤ 25 shows that p has zeros at approximately x = 1 and at x = 3.
Also, since this view of p suggests that the graph of p “bounces off” the x-axis at x = 1, we guess that the factorization
of p(x) contains a positive even power of (x − 1). Next, the viewing window −20 ≤ x ≤ 10, −10000 ≤ y ≤ 10000
indicates another zero at approximately x = −15. Putting all of this information together and noting that p is a fourth
degree polynomial, we guess that the factorization of p(x) must contain (x − 3), (x − 1)2 , and (x + 15). Thus, we have
p(x) = k(x + 15)(x − 1)2 (x − 3),
where k is some constant. Since the leading coefficient of p must equal 1, we take k = 1. We check this factorization by
expanding the product to obtain p(x).
15. The graph represents a polynomial of even degree, at least 4th . Zeros are shown at x = −2, x = −1, x = 2, and x = 3.
Since f (x) → −∞ as x → ∞ or x → −∞, the leading coefficient must be negative. Thus, of the choices in the table,
only C and E are possibilities. When x = 0, function C gives
1
1
y = − (2)(1)(−2)(−3) = − (12) = −6,
2
2
and function E gives
y = −(2)(1)(−2)(−3) = −12.
Since the y-intercept appears to be (0, −6) rather than (0, −12), function C best fits the polynomial shown.
16. Clearly f (x) = x works. However, the solution is not unique. If f is of the form f (x) = ax2 + bx + c, then f (0) = 0
gives c = 0, and f (1) = 1 gives
a(1)2 + b(1) + 0 = 1,
so
a + b = 1,
or
b = 1 − a.
Since these are the only conditions which must be satisfied, any polynomial of the form
f (x) = ax2 + (1 − a)x
will work. If a = 0, we get f (x) = x.
17. Since all the three points fall on a horizontal line, the constant function f (x) = 1 (degree zero) is the only polynomial of
degree ≤ 2 to satisfy the given conditions.
654
Chapter Eleven /SOLUTIONS
18. To pass through the given points, the polynomial must be of at least degree 2. Thus, let f be of the form
f (x) = ax2 + bx + c.
Then using f (0) = 0 gives
a(0)2 + b(0) + c = 0,
so c = 0. Then, with f (2) = 0, we have
a(2)2 + b(2) + 0 = 0
4a + 2b = 0
so
Using f (3) = 3 and b = −2a gives
b = −2a.
a(3)2 + (−2a)(3) + 0 = 3
so
9a − 6a = 3
3a = 3
a = 1.
Thus, b = −2a gives b = −2. The unique polynomial of degree ≤ 2 which satisfies the given conditions is f (x) =
x2 − 2x.
19. The function f has zeros at x = −3, 1, 4. Thus, let f (x) = k(x + 3)(x − 1)(x − 4). Use f (2) = 5 to solve for k;
f (2) = k(2 + 3)(2 − 1)(2 − 4) = −10k. Thus −10k = 5 and k = − 21 . This gives
1
f (x) = − (x + 3)(x − 1)(x − 4).
2
20. The function g has zeros at x = −1 and x = 5, and a double zero at x = 3. Thus, let g(x) = k(x − 3)2 (x − 5)(x + 1).
1
. So
Use g(0) = 3 to solve for k; g(0) = k(−3)2 (−5)(1) = −45k. Thus −45k = 3 and k = − 15
g(x) = −
1
(x − 3)2 (x − 5)(x + 1).
15
21. The points (−3, 0) and (1, 0) indicate two zeros for the polynomial. Thus, the polynomial must be of at least degree 2.
We could let p(x) = k(x + 3)(x − 1) as in the previous problems, and then use the point (0, −3) to solve for k. An
alternative method would be to let p(x) be of the form
p(x) = ax2 + bx + c
and solve for a, b, and c using the given points.
The point (0, −3) gives
a · 0 + b · 0 + c = −3,
so
c = −3.
Using (1, 0), we have
a(1)2 + b(1) − 3 = 0
which gives
a + b = 3.
The point (−3, 0) gives
a(−3)2 + b(−3) − 3 = 0
or
9a − 3b = 3
3a − b = 1.
11.3 SOLUTIONS
655
From a + b = 3, substitute
a = 3−b
into
3a − b = 1.
Then
3(3 − b) − b = 1
9 − 3b − b = 1
so
−4b = −8
b = 2.
Then a = 3 − 2 = 1. Therefore,
p(x) = x2 + 2x − 3
is the polynomial of least degree through the given points.
22. (a) We could think of f (x) = (x − 2)3 + 4 as y = x3 shifted right 2 units and up 4. Thus, since y = x3 is invertible, f
should also be. Algebraically, we let
y = f (x) = (x − 2)3 + 4.
Thus,
y − 4 = (x − 2)3
p
3
p
3
y−4 = x−2
y−4+2 = x
So f (x) is invertible with an inverse
√
f −1 (x) = 3 x − 4 + 2.
(b) Since g is not so obvious, we might begin by graphing y = g(x). Figure 11.23 shows that the function g(x) =
x3 − 4x2 + 2 does not satisfy the horizontal-line-test, so g is not invertible.
y
x
−2
−1
−2
1
2
−4
−6
−8
−10
−12
−14
Figure 11.23
23. Try f (x) = k(x + 1)(x − 1)2 because f has a zero at x = −1 and a double zero at x = 1. Since f (0) = −1, we have
f (0) = k(0 + 1)(0 − 1)2 = k; thus k = −1. So
f (x) = −(x + 1)(x − 1)2 .
24. Let h(x) = k(x + 2)(x + 1)(x − 1)2 (x − 3), since h has zeros at x = −2, −1, 3 and a double zero at x = 1. To solve
for k, use h(2) = −1. Since, h(2) = k(2 + 2)(2 + 1)(2 − 1)2 (2 − 3) = k(4)(3)(1)(−1) = −12k, then −12k = −1,
1
or k = 12
. Thus
1
(x + 2)(x + 1)(x − 1)2 (x − 3)
h(x) =
12
is a possible choice.
656
Chapter Eleven /SOLUTIONS
25. To obtain the flattened effect of the graph near x = 0, let x = 0 be a multiple zero (of odd multiplicity). Thus, a possible
choice would be f (x) = kx3 (x + 1)(x − 2) for k > 0.
26. We know that g(−2) = 0, g(−1) = −3, g(2) = 0, and g(3) = 0. We also know that x = −2 is a multiple zero. Thus,
let
g(x) = k(x + 2)2 (x − 2)(x − 3).
Then, using g(−1) = −3, gives
g(−1) = k(−1 + 2)2 (−1 − 2)(−1 − 3) = k(1)2 (−3)(−4) = 12k,
so 12k = −3, and k = − 14 . Thus,
1
g(x) = − (x + 2)2 (x − 2)(x − 3)
4
is a possible formula for g.
27. The function f has zeros at x = −1, x = 0, and at x = 1. The zero at x = 1 is at least double, so let f (x) =
27
. Then,
kx(x + 1)(x − 1)2 . To determine the value of k, use the fact that f (− 21 ) = − 16
k −
so k = 3. Thus,
is a possible formula for f .
1
2
−
2
1
27
−1 = − ,
2
16
2
1
3
27
k −
−
=− ,
4
2
16
27
9
=− ,
k −
16
16
1
+1
2
−
f (x) = 3x(x + 1)(x − 1)2
28. This polynomial has double zeros at x = −4 and at x = 3. The y-intercept is at y = 4. So
f (x) = k(x + 4)2 (x − 3)2
where we want f (0) = 4. This implies that
k(0 + 4)2 (0 − 3)2 = 4,
giving
k=
So
f (x) =
4
1
=
.
42 · 32
36
1
(x + 4)2 (x − 3)2 .
36
29. We see that h has zeros at x = −2, x = −1 (a double zero), and x = 1. Thus, h(x) = k(x + 2)(x + 1)2 (x − 1). Then
h(0) = (2)(1)2 (−1)k = −2k, and since h(0) = −2, −2k = −2 and k = 1. Thus,
h(x) = (x + 2)(x + 1)2 (x − 1)
is a possible formula for h.
30. Notice that we can think of g as a vertically shifted polynomial. That is, if we let g(x) = h(x) + 4, then h(x) is a
polynomial with zeros at x = −1, x = 2, and x = 4; furthermore, since g(−2) = 0, h(−2) = 0 − 4 = −4. Thus,
h(x) = k(x + 1)(x − 2)(x − 4).
To find k, note that h(−2) = k(−2 + 1)(−2 − 2)(−2 − 4) = k(−1)(−4)(−6) = −24k. Since h(−2) = −24k = −4,
we have k = 61 , which gives
1
h(x) = (x + 1)(x − 2)(x − 4).
6
11.3 SOLUTIONS
657
Thus since g(x) = h(x) + 4, we have
g(x) =
1
(x + 1)(x − 2)(x − 4) + 4.
6
31. Let g(x) = k(x + 2)(x2 )(x − 2), since g has zeros at x = ±2, and a double zero at x = 0. Since g(1) = 1, we have
k(1 + 2)(12 )(1 − 2) = 1; thus −3k = 1 and k = − 13 . So
1
g(x) = − (x2 )(x + 2)(x − 2)
3
is a possible formula.
32. We see that g has zeros at x = −2, x = 2 and at x = 0. The zero at x = 0 is at least double, so let g(x) =
k(x + 2)(x − 2)x2 . We have g(1) = k(1 + 2)(1 − 2)(1)2 = k · 3(−1)(1)2 = −3k. Since g(1) = 1, −3k = 1, so
k = − 13 and
1
g(x) = − (x + 2)(x − 2)x2
3
is a possible formula for g.
33. y = 4x2 − 1 = (2x − 1)(2x + 1), which implies that y = 0 for x = ± 21 .
34. Note that y = x4 + 6x2 + 9 = (x2 + 3)2 . This implies that y = 0 if x2 = −3, but x2 = −3 has no real solutions. Thus,
there are no zeros.
35. Zeros occur where y = 0, which we can find by factoring:
(x2 − 8x + 12)(x − 3) = y
(x − 6)(x − 2)(x − 3) = y.
Zeros are at x = 6, x = 2, and x = 3.
36. Factoring y = x2 + 5x + 6 gives y = (x + 2)(x + 3). Thus y = 0 for x = −2 or x = −3.
37. y = 4x2 + 1 = 0 implies that x2 = − 41 , which has no solutions. There are no real zeros.
38. Zeros occur where y = 0, at x = 0 and x = −3. Since x2 is never less than zero, x2 + 4 is never less than 4, so x2 + 4
has no zeros.
39. From its formula, we know that f has double zeros at x = 5 and x = 3 and a single zero at x = 1. Since it is an even
function, we know the graph of f has even symmetry—that is, its graph is symmetrical across the y-axis. So we can use
the unknowns r, s, and g(x) to balance the zeros we do know. Here is one possibility:
• To balance the double zero at x = 3 with a double zero at x = −3, we can let s = 2.
• To balance the single zero at x = 1 with a single zero at x = −1, we can let r = −1.
• To balance the double zero at x = 5, we can let the second-degree polynomial g(x) = k(x + 5)2 .
Putting this together gives
f (x) = (x − 5)2 (x − 3)2 (x − 1) (x + 1) (x + 3)2 · k(x + 5)2
| {z } | {z } | {z }
(x−r)
Another possibility:
(x+3)s
g(x)
• To balance the double zero at x = 3 with a double zero at x = −3, we again let s = 2.
• If we let g(x) = k(x + 1)(x + 5), then g is a second-degree polynomial that balances the single zero at x = 1 and
one of the repeated zeros at x = 5.
• We can balance the other of the repeated zeros at x = 5 by letting r = −5.
Putting this together gives
f (x) = (x − 5)2 (x − 3)2 (x − 1) (x + 5) (x + 3)2 · k(x + 5)(x + 1)
| {z } | {z } |
(x−r)
(x+3)s
{z
g(x)
}
658
Chapter Eleven /SOLUTIONS
Note that k is an arbitrary non-zero constant. In Figure 11.24, we assume k = 1.
y
100000
−5
x
1
−3 −1
3
5
−50000
Figure 11.24
40. Two possibilities are
y = k1 (x + 1)(x − 2)2
y = k2 (x + 1)2 (x − 2).
Given the y-intercept of y = 3, we can find the values of k1 and k2 :
3 = k1 (0 + 1)(0 − 2)2
4k1 = 3
3
k1 = ,
4
3 = k2 (0 + 1)2 (0 − 2)
−2k2 = 3
3
k2 = − .
2
This means our two possibilities are
3
(x + 1)(x − 2)2
4
3
y = − (x + 1)2 (x − 2).
2
y=
41. (a) V (x) = x(6 − 2x)(8 − 2x)
(b) Values of x for which V (x) makes sense are 0 < x < 3, since if x < 0 or x > 3 the volume is negative.
(c) See Figure 11.25.
y
25
V (x) = x(6 − 2x)(8 − 2x)
20
15
10
5
x
1
2
3
Figure 11.25
(d) Using a graphing calculator, we find the peak between x = 0 and x = 3 to occur at x ≈ 1.13. The maximum volume
is ≈ 24.26 in3 .
11.3 SOLUTIONS
659
42. Let V be the amount of packing material you will need, then
V = (Volume of crate) − (Volume of box)
= Vc − Vb
where Vc and Vb are the volumes of the crate and box, respectively. We have for the box’s volume,
x+2
z }| {
Vb = length · width · depth = x(x + 2)(x − 1).
| {z }
x
| {z }
x−1
The wooden crate must be 1 ft longer than the cardboard box, so its length is (x + 1). This gives the required 0.5-ft
clearance between the crate and the front and back of the box. Similarly, the crate’s width must be 1 ft greater than the
box’s width of (x + 2), and its depth must be 2 ft greater than the box’s depth of (x − 1). See Figure 11.26.
Figure 11.26: Packing a box inside a crate
We have for the crate’s volume
(x+2)+1
z }| {
Vc = length · width · depth = (x + 1)(x + 3)(x + 1).
| {z }
x+1
Thus, the total amount of packing material will be
| {z }
(x−1)+2
V = Vc − Vb
= (x + 1)(x + 3)(x + 1) − x(x + 2)(x − 1).
The formula for V is a difference of two third-degree polynomials. The format is not terribly convenient, so we simplify
the formula by multiplying the factors for Vb and Vc and gathering like terms. Then for V we have
V = (x + 1)(x + 3)(x + 1) − x(x + 2)(x − 1)
= (x3 + 5x2 + 7x + 3) − (x3 + x2 − 2x)
= 4x2 + 9x + 3.
The formula V (x) = 4x2 + 9x + 3 gives the necessary information for appropriate values of x. Note that the quadratic
function y = 4x2 + 9x + 3 is defined for all values of x. However, since x represents the length of a box and (x − 1) is
the depth of the box, the formula only makes sense as a model for x > 1. In this case, an understanding of the component
polynomials representing Vb and Vc is necessary in order to determine the logical domain for V (x).
660
Chapter Eleven /SOLUTIONS
43. We express the volume as a function of the length, x, of the square’s side that is cut out in Figure 11.27.
11′′
11 − 2x
6
x
x
6
-
-
x
x
8.5 − 2x
8.5′′
?
?
Figure 11.27
by
Since the sides of the base are (11 − 2x) and (8.5 − 2x) inches and the depth is x inches, the volume, V (x), is given
V (x) = x(11 − 2x)(8.5 − 2x).
The graph of V (x) in Figure 11.28 suggest that the maximum volume occurs when x ≈ 1.585 inches. (A good viewing
window is 0 ≤ x ≤ 5 and 0 ≤ y ≤ 70.) So one side is x = 1.585, and therefore the others are 11 − 2(1.585) = 7.83
and 8.5 − 2(1.585) = 5.33.
The dimensions of the box are 7.83 by 5.33 by 1.585 inches.
y
70
(1.59, 66.1)
V (x)
x
5
Figure 11.28
44. The ln function is defined only for positive inputs. Here, the input of the ln function is the polynomial y = (x−3)2 (x+2).
This polynomial has zeros at x = 3, where the graph bounces, and x = −2. Its long-run behavior is like y = x3 . From
Figure 11.29, we see that for this polynomial, y is non-positive for x ≤ −2 and x = 3. Thus, the domain of g is all
x > −2 except x = 3.
y
20
15
10
5
x
−3 −2 −1
−5
1
2
Figure 11.29
3
4
11.3 SOLUTIONS
661
45. The domain is x ≥ c and a ≤ x ≤ b. Taking the hint, we see that the function y = (x − a)(x − b)(x − c) has zeros
at a, b, c and long-run behavior like y = x3 . Thus, the graph of y = (x − a)(x − b)(x − c) looks something like the
graph in the figure (though of coursep
the zeros may be spaced differently). We see that y < 0 on the intervals x < a and
b < x < c. Thus, the function y = (x − a)(x − b)(x − c) is undefined on these intervals, but is defined everywhere
else.
y
a
x
c
b
Figure 11.30
46. (a) Some things we know about the graph of a, are:
• As x → ∞, a(x) → ∞. As x → −∞, a(x) → −∞.
• a is an odd function, so it must be symmetric about the origin.
• There is a zero at (0, 0) on the graph of a.
(b) The zeros occur at x = 0, 1.112 and − 1.112. Since the function is odd we already knew that for every positive zero
there would be a corresponding negative zero.
(c) Since the function is symmetric about the origin, one only needs to concentrate on positive values of x. For values
of x between zero and one, x5 and 2x3 are very small, so −4x dominates and f (x) < 0. But, for values of x larger
than one, x5 and 2x3 get large very quickly and f (x) > 0 soon after x = 1. Although −4x becomes more and more
negative, its magnitude is less and less important in relation to the other two terms. There is no chance that the graph
is suddenly going to turn around and cross the x-axis once more.
We can also analyze a algebraically if we note that a(x) can be rewritten as a(x) = x(x4 + 2x2 − 4). Thus the
zeros of a(x) occur at zero and at the zeros of (x4 + 2x2 − 4). Using the quadratic formula we get
p
22 − 4(−4)
2
√
20
= −1 ±
p 2 √
so x = ± −1 ± 5.
−2 ±
x2 =
Since we are only interested in the positive solutions we will look at
x=
Now
x=
is not defined, so the only positive solution is
x=
p
p
−1 ±
p
−1 −
−1 +
√
√
5.
√
5
5 ≈ 1.112.
(d) The zeros of b(x) occur at 0, 1.112 and −1.112. This should not be a surprise since b(x) = 2a(x). To get the graph
of b(x), stretch the graph of a(x) by a factor of two in a vertical direction. Note that the x-intercepts do not change.
662
Chapter Eleven /SOLUTIONS
47. (a) We could let f (x) = k(x + 2)(x − 3)(x − 5) to obtain the given zeros. For a y-intercept of 4, f (0) = 4 =
2
. One possible formula is
k(0 + 2)(0 − 3)(0 − 5) = 30k. Thus 30k = 4, so k = 15
f (x) =
2
(x + 2)(x − 3)(x − 5).
15
(b) One possibility is that f looks like the function in Figure 11.31 and has a double zero at x = 5.
y
y
4
4
x
x
3
−2
3
−2
5
5
f (x)
f (x)
Figure 11.31
Figure 11.32
Then a formula for f is
f (x) = k(x + 2)(x − 3)(x − 5)2
and
f (0) = k(2)(−3)(−5)2 .
Thus,
−150k = 4
which gives us
k=−
2
.
75
Thus
2
(x + 2)(x − 3)(x − 5)2 .
75
Another possibility is that f has a double-zero at x = 3 instead of at x = 5. In this case f looks like the function in
Figure 11.32. This gives the formula
f (x) = −
f (x) = −
2
(x + 2)(x − 3)2 (x − 5).
45
Note that if f had a double zero at x = −2, there must be another zero for −2 < x < 0 in order for f to satisfy
f (0) = 4 and y → −∞ as x → ±∞.
(c) One possibility is that f looks like the graph in Figure 11.33, with a double zero at x = −2 and single zeros at x = 3
and x = 5.
y
y
f (x)
f (x)
4
−2
4
3
5
x
x
−2
Figure 11.33
3
Figure 11.34
5
11.4 SOLUTIONS
663
A formula for f is f (x) = k(x + 2)2 (x − 3)(x − 5), which gives us
k=
1
.
15
Thus,
1
(x + 2)2 (x − 3)(x − 5).
15
It is also possible that f has 3 double-zeros at x = −2, x = 3 and x = 5. This leads to a 6th degree polynomial
which looks like Figure 11.34. This gives the formula
f (x) =
f (x) =
48. (a)
(b)
(c)
(d)
(e)
(f)
1
(x + 2)2 (x − 3)2 (x − 5)2 .
225
Never true, because f (x) → −∞ as x → ±∞, which means f (x) must be of even degree.
Sometimes true, since f could be an even-degree polynomial without being symmetric to the y-axis.
Sometimes true, since f could have a multiple zero.
Never true, because f must be of even degree.
True, because, since f is of even degree, f (−x) must have the same long-run behavior as f (x).
Never true, because, since f (x) → −∞ as x → ±∞, f will fail the horizontal-line test.
Solutions for Section 11.4
Skill Refresher
6
7
6y 2 + 7
+ 3 =
y
y
y3
13
14
13
14
13 · 2 + 14
40
20
S2.
+
=
+
=
=
=
x−1
2x − 2
x−1
2(x − 1)
2(x − 1)
2(x − 1)
x−1
2
1
− 2
x−2
x5
x3
x
x
=
·
=
S3.
2
2x − 4
x
2(x − 2)
2
x5
S4.
9
12
9
12
+
=
+
x2 + 5x + 6
x+3
(x + 3)(x + 2)
x+3
9 + 12(x + 2)
=
(x + 3)(x + 2)
12x + 33
=
(x + 3)(x + 2)
3(4x + 11)
=
(x + 3)(x + 2)
S1.
S5.
5 − 18(x − 2)(x + 1)
5
18
−
=
(x − 2)2 (x + 1)
(x − 2)
(x − 2)2 (x + 1)
=
=
5 − 18x2 + 18x + 36
(x − 2)2 (x + 1)0
−18x2 + 18x + 41
(x − 2)2 (x + 1)
664
Chapter Eleven /SOLUTIONS
S6. Dividing by (x + y) is the same as multiplying by its reciprocal,
1
x+y
x+y
=
1
:
x+y
1
1
1
·
=
.
x+y x+y
(x + y)2
S7. We write this complex fraction as a multiplication problem. Therefore,
w+2
2
w+2
=
w+2
1
1
·
= .
2
w+2
2
S8. In this example, the numerator and denominator have no common factor. Therefore the fraction cannot be simplified any
further.
1
1
x+1
+ 2
2
x+1
x2
x+1
1
x−1 + x−2
x
x
=
=
· 2
=
=
.
= 2x
S9.
−2
2
1
1−x
x
x −1
(x + 1)(x − 1)
x−1
x
−
1
1− 2
x
x2
Exercises
1. This is a rational function, and it is already in the form of one polynomial divided by another.
2. This is not a rational function, as we cannot put it in the form of one polynomial divided by another, since 4x and 3x are
exponential, not power functions. Thus, f (x) is not the ratio of polynomials.
3. This is a rational function, as we can put it in the form of one polynomial divided by another:
f (x) =
1
x3
2
x3 + 2
x2
+ =
+
=
.
2
x
2x
2x
2x
4. This is not a rational function. We cannot put it in the form of one polynomial divided by another, since 3x is an exponential, not a power function.
√
5. This is not a rational function, as we cannot put it in the form of one polynomial divided by another, since x + 1 is not
a polynomial.
6. This is a rational function, as we can put it in the form of one polynomial divided by another:
f (x) =
x3
x
1
3x
1
3x + 1
1
+ =
.
+ = + =
2x2
6
2
6
6
6
6
√
7. This is not a rational function, as we cannot put it in the form of one polynomial divided by another, since 4 x + 7 is not
a polynomial.
8. We have
lim (2x−3 + 4) = lim (2/x3 + 4) = 0 + 4 = 4.
x→∞
x→∞
9. We have
lim (3x−2 + 5x + 7) = lim (3/x2 + 5x + 7) = ∞.
x→∞
x→∞
10. We have
lim
x→∞
11. We have
lim
x→−∞
3
3x2
4x + 3x2
= lim
= .
2
4x + 3x
x→∞ 4x2
4
3x2 + x
3x2
= lim
= 0.
x→−∞ 5x3
2x2 + 5x3
12. As x → ±∞, 1/x → 0 and x/(x + 1) → 1, so h(x) approaches 3 − 0 + 1 = 4.
Therefore y = 4 is the horizontal asymptote.
11.4 SOLUTIONS
665
13. As x → ±∞, 1/x → 0, so f (x) → 1. Therefore y = 1 is the horizontal asymptote.
−3x2
−3
−3x2 + x + 2
=
=
as x → ±∞.
2
2x + 1
2x2
2
3
Thus, y = − 2 is the horizontal asymptote.
14. g(x) =
15. For the function f , f (x) → 1 as x → ±∞ since for large values of x, f (x) ≈
3
x2
x2
= 1.
x
x2
= x for large values of x. Thus, as x → ±∞, g(x) approaches the line y = x.
The function g(x) ≈
The function h will behave like y = xx2 = x1 for large values of x. Thus, h(x) → 0 as x → ±∞.
Problems
16. Let y = f (x). Then x = f −1 (y). Solving for x,
4 − 3x
5x − 4
y(5x − 4) = 4 − 3x
y=
5xy − 4y = 4 − 3x
5xy + 3x = 4y + 4
x(5y + 3) = 4y + 4
4y + 4
x=
5y + 3
Therefore,
f −1 (x) =
(factor out an x)
4x + 4
.
5x + 3
17. Note: There are many examples to fit these descriptions. Some choices are:
Even: f (x) =
x2
+1
x2
Odd: f (x) =
x3
+1
x2
Neither: f (x) =
x+1
.
x−1
If f (x) is even, then f (−x) = f (x). This will be true if and only if both p(x) and q(x) are even or both are odd. If one
is even and one is odd, then and only then will f (x) be odd. If one is neither then f (x) is neither.
18. (a) If lim r(x) = 0, then the degree of the denominator is greater than the degree of the numerator, so n > m.
x→∞
(b) If lim r(x) = k, with k 6= 0, the degree of the numerator and denominator must be equal, so n = m.
x→∞
19. (a) A graph of f for 0 ≤ t ≤ 10 and 0 ≤ y ≤ 1.5 is shown in Figure 11.35.
y
y = f (t)
1
0.75
1
3.7
10
t (weeks)
Figure 11.35
(b) At t = 0 the oxygen is at its normal level in the pond. The level decreases sharply for the first week after the waste is
dumped into the pond. The oxygen level reaches its minimum of approximately one half the normal level by the end
of the first week. Then the level begins to increase.
(c) Eventually, the oxygen level in the pond will once again approach the normal level of one.
(d) The line y = 0.75 is shown in Figure 11.35. After the level has reached its minimum, we can approximate the
intersection of f and y = 0.75 at t ≈ 3.73. Thus, it takes about 3.73 weeks for the level to return to 75% normal.
666
Chapter Eleven /SOLUTIONS
20. (a) To cover its costs of $80,000 and make a profit of $40,000, the printing house must take in $120,000 from sales.
Therefore, if 1000 copies are sold,
120,000
= $120.
Price per copy =
1000
Calculating the price for other projected sales in the same way, we have the result in Table 11.5
Table 11.5
Number of copies sold
1000
2000
4000
6000
Price per copy, $
120
60
30
20
(b) Generalizing the formula we used to calculate the costs in Table 11.5, we have
p=
120,000
.
s
(c) The graph of p = 120000/s is shown in Figure 11.36
p, price ($)
100
50
3000
6000
s, projected sales
Figure 11.36
21. (a) Adding x kg of copper increases both the amount of copper and the total amount of alloy. Originally there are 3 kg
of copper and 12 kg of alloy. Adding x kg of copper results in a total of (3 + x) kg of copper and (12 + x) kg of
alloy. Thus, the new concentration is given by
f (x) =
(b)
3+x
.
12 + x
7
3 + 12
7
1
2
=
=
= 28%. Thus, adding one-half kilogram copper results in an alloy that is 28%
(i) f ( ) =
1
25
2
25
12 + 2
2
copper.
3
(ii) f (0) = 12
= 41 = 25%. This means that adding no copper results in the original alloy of 25% copper.
2
(iii) f (−1) = 11 ≈ 18.2%. This could be interpreted as meaning that the removal of 1 kg copper (corresponding to
x = −1) results in an alloy that is about 18.2% copper.
3+x
(iv) Let y = f (x) =
. Then, multiplying both sides by the denominator we have
12 + x
(12 + x)y = 3 + x
12y + xy = 3 + x
xy − x = 3 − 12y
x(y − 1) = 3 − 12y
3 − 12y
x=
y−1
and so
f −1 (x) =
3 − 12x
x−1
11.4 SOLUTIONS
Using this formula, we have f −1 ( 21 ) =
667
−3
= 6. This means that you must add 6 kg copper in order to obtain
− 21
1
9
, or 50%, copper. (You can check this by finding f (6) = 18
= 21 ).
2
0
3
(v) f −1 (0) =
= −3. Check: f (−3) =
= 0. This means that you must remove 3 kg copper to obtain an
−1
9
alloy that is 0% copper, or pure tin.
(c) The graph of y = f (x) is in Figure 11.37. The axis intercepts are (0, 0.25) and (−3, 0). The y-intercept of (0, 0.25)
indicates that with no copper added the concentration is 0.25, or 25%, which is the original concentration of copper
in the alloy. The x-intercept of (−3, 0) indicates that to make the concentration of copper 0%, we would have to
remove 3 kg of copper. This makes sense, as the alloy has only 3 kg of copper to begin with.
(d) The graph of y = f (x) on a larger domain is in Figure 11.38.
an alloy that is
y
concentration of copper
in alloy
y=1
1
0.5
0.25
−5
−3
−1
0.5
1
3
5
copper
added
x
20
Figure 11.37
40
60
80
100
Figure 11.38
The concentration of copper in the alloy rises with the amount of copper added, x. However, the graph levels
off for large values of x, never quite reaching y = 1 = 100%. This is because as more and more copper is added,
the concentration gets closer and closer to 100%, but the presence of the 9 kg of tin prevents the alloy from ever
becoming 100% pure copper.
22. (a) Originally the total amount of the alloy is 2 kg, one half of which — or equivalently 1 kg — is tin. We have
Total amount of tin
Total amount of alloy
(original amount of tin) + (added tin)
=
(original amount of alloy) + (added tin)
1+x
=
2+x
C(x) =
C(x) is a rational function.
(b) Using our formula, we have
1.5
1 + 0.5
=
= 60%.
2 + 0.5
2.5
This means that if 0.5 kg of tin is added, the concentration of tin in the resulting alloy will be 60%. As for C(−0.5),
we have
1 − 0.5
0.5
C(−0.5) =
=
≈ 33.333%.
2 − 0.5
1.5
A negative x-value corresponds to the removal of tin from the original mixture, so the statement C(−0.5) = 33.333%
would mean that removing 0.5 kg of tin results in an alloy that is 33.333% tin.
(c) To graph y = C(x), let’s see if we can represent the formula as a translation of a power function. We write
C(0.5) =
C(x) =
(x + 2) − 1
x+1
=
x+2
x+2
1
x+2
−
=
x+2
x+2
1
= 1−
x+2
1
+ 1.
=−
x+2
(splitting the numerator)
668
Chapter Eleven /SOLUTIONS
Thus the graph of C will resemble the graph of f (x) = − x1 shifted two units to the left and then one unit up.
Figure 11.39 shows this translation.
y
y
f (x) = − x1
y = C(x) =
x+1
x+2
(−1, 1)
R
(−3, 2)
x
y=1
x
I
(1, −1)
(0, 12 )
x = −2
Figure 11.39: The graph of the rational function y = C(x) is a translation of the graph of the power function
f (x) = − x1 .
For “interesting features”, we start with the intercepts and asymptotes. C(x) has a y-intercept between y = 0
and y = 1, an x-intercept (or zero) at x = −1, a horizontal asymptote of y = 1, and a vertical asymptote of x = −2.
What physical significance do these graphical features have?
First off, since
1
0+1
= = 0.5
C(0) =
0+2
2
we see that the y-intercept is 0.5, or 50%. This means that if you add no tin (i.e. x = 0 kg), then the concentration is
50%, the original concentration of tin in the alloy.
Second, since
0
−1 + 1
= = 0,
C(−1) =
−1 + 2
1
we see that the x-intercept is indeed at x = −1. This means that if you remove 1 kg of tin (i.e. x = −1 kg), then the
concentration will be 0%, as there will be no tin remaining in the alloy.
This fact has a second implication: the graph of C(x) is meaningless for x < −1, as it is impossible to remove
more than 1 kg of tin. Thus, in the context of the problem at hand, the domain of C(x) is x ≥ −1. The graph on
this domain is given by Figure 11.40. Notice that the vertical asymptote of the original graph (at x = −2) no longer
appears, and it has no physical significance.
y
y=1
0.5
x
−1
Figure 11.40: The domain of C(x) is x ≥ −1
The horizontal asymptote of y = 1 is, however, physically meaningful. As x grows large, we see that y approaches 1, or 100%. Since the amount of copper in the alloy is fixed at 1 kg, adding large amounts of tin results in
an alloy that is nearly pure tin. For example, if we add 10 kg of tin then x = 10 and
C(x) =
11
10 + 1
=
= 0.916 . . . ≈ 91.7%.
10 + 2
12
11.4 SOLUTIONS
669
Since the alloy now contains 11 kg of tin out of 12 kg total, it is relatively pure tin—at least, it is 91.7% pure. If
instead we add 98 kg of tin, then x = 98 and
C(x) =
99
98 + 1
=
= 99%.
98 + 2
100
Thus adding 98 kg of tin results in an alloy that is 99% pure. The 1 kg of copper is almost negligible. Therefore,
the horizontal asymptote at y = 1 indicates that as the amount of added tin, x, grows large, the concentration of tin
in the alloy approaches 1, or 100%.
23. The population of Mathville reached 20,000 when
20
4t + 3 = 20
2t + 5
4t + 3
=1
2t + 5
4t + 3 = 2t + 5
2t = 2
t=1
so the population reached 20,000 after 1 year, in 2011.
The function P (t) has a horizontal asymptote given by
lim P (t) = 20
t→∞
4
2
= 40,
that is 40, 000 people. So the population will never reach 50,000.
24. (a) The time to travel the first 10 miles is 10
= 0.25 hours. The time for the remaining 50 miles in 50/V hours so the
40
total journey time is T = 0.25 + 50/V. Thus, the average speed is
Average speed =
60
60
=
T
0.25 +
50
V
(b) If you want to average 60 mph for the trip then you need
=
240V
.
V + 200
240V
= 60.
V + 200
25. (a)
(b)
(c)
(d)
Solving this equation gives V = 200/3 mph, nearly 70 mph.
x
Amount of Alcohol
=
f (x) =
Amount of Liquid
x+5
7
7
f (7) = 7+5
= 12
≈ 58.333%. Also, f (7) is the concentration of alcohol in a solution consisting of 5 gallons of
water and 7 gallons of alcohol.
x
= 0 and so x = 0. The concentration of alcohol is 0% when there is no alcohol in the
f (x) = 0 implies that
x+5
solution, that is, when x = 0.
The horizontal asymptote is given by the ratio of the highest-power terms of the numerator and denominator:
y=
x
= 1 = 100%
x
This means that as the amount of alcohol added, x, grows large, the concentration of alcohol in the solution approaches 100%.
(i)
(ii)
(iii)
(iv)
(b) (i)
26. (a)
C(1) = 5050 means the cost to make 1 unit is $5050.
C(100) = 10,000 means the cost to make 100 units is $10,000.
C(1000) = 55,000 means the cost to make 1000 units is $55,000.
C(10000) = 505,000 means the cost to make 10,000 units is $505,000.
a(1) = C(1)/1 = 5050 means that it costs $5050/unit to make 1 unit.
670
Chapter Eleven /SOLUTIONS
(ii) a(100) = C(100)/100 = 100 means that it costs $100/unit to make 100 units.
(iii) a(1000) = C(1000)/1000 = 55 means that it costs $55/unit to make 1000 units.
(iv) a(10000) = C(10000)/10000 = 50.5 means that it costs $50.50/unit to make 10,000 units.
(c) As the number of units increases, the average cost per unit gets closer to $50/unit, which is the unit (or marginal) cost.
This makes sense because the fixed or initial $5000 expenditure becomes increasingly insignificant as it is averaged
over a large number of units.
27. (a) From Figure 11.41, we see that
Slope of line l =
C(n0 ) − 0
C(n0 )
∆y
=
=
.
∆x
n0 − 0
n0
y
l
C(n)
(n0 , C(n0 ))
5000
n
n0
(0, 0)
Figure 11.41
(b) a(n0 ) = C(n0 )/n0 . Thus, the slope of line l is the same as the average cost of producing n0 units.
28. Line l1 has a smaller slope than line l2 . We know the slope of line l1 represents the average cost of producing n1 units,
and the slope of l2 represents the average cost of producing n2 units. Thus, the average cost of producing n2 units is more
than that of producing n1 units. For these goods, the average cost actually goes up between n1 and n2 units.
29. (a) C(x) = 30000 + 3x
C(x)
30000 + 3x
30000
=
=3+
(b) a(x) =
x
x
x
(c) The graph of y = a(x) is shown in Figure 11.42.
(d) The average cost, a(x), approaches $3 per unit as the number of units grows large. This is because the fixed cost of
$30,000 is averaged over a very large number of goods, so that each good costs only little more than $3 to produce.
(e) The average cost, a(x), grows very large as x → 0, because the fixed cost of $30000 is being divided among a small
number of units.
(f) The value of a−1 (y) tells us how many units the firm must produce to reach an average cost of $y per unit. To find a
formula for a−1 (y), let y = a(x), and solve for x. Then
30000 + 3x
x
yx = 30000 + 3x
y=
yx − 3x = 30000
x(y − 3) = 30000
30000
.
x=
y−3
30000
.
y−3
−1
(g) We want to evaluate a (5), the total number of units required to yield an average cost of $5 per unit.
So, we have a−1 (y) =
a−1 (5) =
30000
30000
=
= 15000.
5−3
2
11.5 SOLUTIONS
671
Thus, the firm must produce at least 15,000 units for the average cost per unit to be $5. The firm must produce at
least 15,000 units to make a profit.
y
y = a(x)
y=3
3
x
50000
Figure 11.42
30. We need to find the coefficient c so that R(x) = f (x) approximates f (x). Notice that R(0) = f (0) = 1 for all values of
c. We can find c by making sure that the function and approximation match at x = 1, so R(1) = f (1) = e. This gives
R(1) =
so
1
= f (1) = e
1+c·1
so c = e−1 − 1 = −0.6321,
1
.
1 − 0.6321x
Figure 11.43 shows f (x) and R(x) on the interval (0, 1). There is reasonable agreement.
R(x) =
y
5
x
−0.5
0.5
1
1.5
Figure 11.43
Solutions for Section 11.5
Exercises
1. The zero of this function is at x = 4. It has vertical asymptotes at x = ±3. Its long-run behavior is: y → 0 as x → ±∞.
See Figure 11.44.
672
Chapter Eleven /SOLUTIONS
y
x = −3
x=3
1
-
(0, 4/9)
y=
x
4
x−4
x2 −9
Figure 11.44
2. The zeros of this function are at x = ±2. It has a vertical asymptote at x = 9. Its long-run behavior is that it looks like
the line y = x. See Figure 11.45.
y
y=
x2 −4
x−9
y=x
(−2, 0)
(2, 0)
R ?
x
x=9
Figure 11.45
3. The zero of this function is at x = −3. It has a vertical asymptote at x = −5. Its long-run behavior is: y → 1 as
x → ±∞. See Figure 11.46.
y
y=
x+3
x+5
(0, 3/5)
R
−3
x = −5
Figure 11.46
y=1
x
11.5 SOLUTIONS
673
4. The zero of this function is at x = −3. It has a vertical asymptote at x = −5. Its long-run behavior is: y → 0 as
x → ±∞. See Figure 11.47.
y
x = −5
(0, 3/25)
x
y=
−3
x+3
(x+5)2
Figure 11.47
5. Since
(x − 2)(x + 2)
x2 − 4
=
,
x3 + 4x2
x2 (x + 4)
the x-intercepts are x = ±2; there is no y-intercept; the horizontal asymptote is y = 0; the vertical asymptotes are
x = 0, x = −4.
g(x) =
6. Since
x(4 − x)
x(4 − x)
=
,
x2 − 6x + 5
(x − 1)(x − 5)
the x-intercepts are x = 0, x = 4; the y-intercept is y = 0; the vertical asymptotes are x = 1, x = 5. Since we can also
write
4x − x2
−x2 + 4x
k(x) = 2
= 2
,
x − 6x + 5
x − 6x + 5
the horizontal asymptote is y = −1
k(x) =
7. The x-intercept is x = 2; the y-intercept is y = −2/(−4) = 1/2; the horizontal asymptote is y = 1; the vertical
asymptote is x = 4.
8. Since
x2 − 9
(x − 3)(x + 3)
=
,
x2 + 9
x2 + 9
the x-intercepts are x = ±3; the y-intercept is y = −9/9 = −1; the horizontal asymptote is y = 1; there are no vertical
asymptotes.
g(x) =
9. (a) See the following table.
x
−5
−4.1
−4.01
−4
−3.99
−3.9
−3
G(x)
10
82
802
Undef
−798
−78
−6
As x approaches −4 from the left the function takes on very large positive values. As x approaches −4 from the
right the function takes on very large negative values.
(b)
x
5
10
100
1000
x
−5
−10
−100
−1000
G(x)
1.111
1.429
1.923
1.992
G(x)
10
3.333
2.083
2.008
For x > −4, as x increases, f (x) approaches 2 from below. For x < −4, as x decreases, f (x) approaches 2
from above.
674
Chapter Eleven /SOLUTIONS
(c) The horizontal asymptote is y = 2. The vertical asymptote is x = −4. See Figure 11.48.
y
10
G(x) =
2x
x+4
2
x
−14
6
−4
−10
Figure 11.48
10. (a) See the following table.
x
−3
−2.1
−2.01
−2
−1.99
−1.9
−1
g(x)
1
100
10,000
Undef
10,000
100
1
As x approaches −2 from the left the function takes on very large positive values. As x approaches −2 from the
right the function takes on very large positive values.
(b)
x
5
10
100
1000
x
−5
−10
−100
−1000
g(x)
0.02
0.007
9.6 · 10−5
10−6
g(x)
0.111
0.016
10−4
10−6
For x > −2, as x increases, f (x) approaches 0 from above. For x < −2 as x decreases, f (x) approaches 0
from above.
(c) The horizontal asymptote is y = 0 (the x-axis). The vertical asymptote is x = −2. See Figure 11.49.
y
8
g(x) =
1
(x+2)2
x
−7
3
−2
−2
Figure 11.49
Problems
11. The graph is the graph of y = 1/x moved up by 2. See Figure 11.50.
11.5 SOLUTIONS
675
y
y = 2 + 1/x
x
Figure 11.50
12. The graph will have vertical asymptotes at x = ±4 and zeros at x = 3 and x = 2. The y-intercept is (0, − 43 ), and for
large positive or negative values of x, we see that y → 2—thus, there is a horizontal asymptote of y = 2. Note that the
graph will intersect the horizontal asymptote if
2x2 − 10x + 12
= 2,
x2 − 16
which implies
2x2 − 10x + 12 = 2x2 − 32
−10x = −44
x = 4.4
Putting all of this information together, we obtain a graph similar to that of Figure 11.51.
y
x
−0.75
23
Figure 11.51
13. (a) To estimate
x
,
5−x
we consider what happens to the function when x is slightly larger than 5. The numerator is positive and the denomx
inator is negative and is approaching 0 as x approaches 5. We suspect that
gets more and more negative as x
5−x
approaches 5 from the right. We can also use either a graph or a table of values as in Table 11.6 to estimate this limit.
We see that
x
= −∞.
lim
x→5+ 5 − x
lim
x→5+
Table 11.6
x
5.1
5.01
5.001
5.0001
f (x)
−51
−501
−5001
−50001
676
Chapter Eleven /SOLUTIONS
(b) To estimate
x
,
5−x
we consider what happens to the function when x is slightly smaller than 5. The numerator is positive and the
x
denominator is positive and is approaching 0 as x approaches 5. We suspect that
gets larger and larger as x
5−x
approaches 5 from the left. We can also use either a graph or a table of values to estimate this limit. We see that
lim
x→5−
lim
x→5−
x
= +∞.
5−x
14. (a) To estimate
5−x
,
(x − 2)2
we consider what happens to the function when x is slightly larger than 2. The numerator is positive and the de5−x
gets larger and larger as x
nominator is positive and is approaching 0 as x approaches 2. We suspect that
(x − 2)2
approaches 2 from the right. We can also use either a graph or a table of values as in Table 11.7 to estimate this limit.
We see that
5−x
lim
= +∞.
2
x→2+ (x − 2)
lim
x→2+
Table 11.7
x
2.1
2.01
2.001
2.0001
f (x)
290
29900
2999000
299990000
(b) To estimate
5−x
,
(x − 2)2
we consider what happens to the function when x is slightly smaller than 2. The numerator is positive and the
5−x
denominator is positive and is approaching 0 as x approaches 2. We suspect that
gets larger and larger as x
(x − 2)2
approaches 2 from the left. We can also use either a graph or a table of values to estimate this limit. We see that
lim
x→2−
lim
x→2−
5−x
= +∞.
(x − 2)2
1
− 1 has a vertical asymptote at x = 5, no x intercept, horizontal asymptote y = −1: (iii)
(x − 5)2
x−2
has vertical asymptotes at x = −1, 3, x intercept at 2, horizontal asymptote y = 0: (i)
y=
(x + 1)(x − 3)
2x + 4
y=
has a vertical asymptote at x = 1, x intercept at x = −2, horizontal asymptote y = 2: (ii)
x−1
x−3+x+1
2x − 2
y=
=
has vertical asymptotes at x = −1, 3, x intercept at 1, horizontal asymptote
(x + 1)(x − 3)
(x + 1)(x − 3)
at y = 0: (iv)
(1 + x)(1 − x)
has vertical asymptote at x = 2, two x intercepts at ±1: (vi)
y=
x−2
1 − 4x
y=
has a vertical asymptote at x = −1, x intercept at x = 41 , horizontal asymptote at y = −2: (v)
2x + 2
15. (a) y = −
(b)
(c)
(d)
(e)
(f)
2
(x)
16. (a) Matches description (v). y = fg(x)
: No vertical asymptotes. Horizontal asymptote at y = 1. x-intercepts at
= xx2 −4
+4
±2.
2
+4
(b) Does not match any of the descriptions. y = fg(x)
: Vertical asymptotes at x = ±2. Horizontal asymptote at
= xx2 −4
(x)
y = 1. No zeros.
11.5 SOLUTIONS
677
= xx+5
(c) Matches description (i). y = h(x)
2 −4 : Vertical asymptotes at x = ±2. Horizontal asymptote at y = 0.
f (x)
x-intercept at −5.
2
(d) Matches description (vii). y = f ( x1 ) = x12 − 4 = 1−4x
: Vertical asymptote at x = 0. Horizontal asymptote at
x2
1
y = −4. x-intercepts at ± 2 .
2
g(x)
+4
= xx+5
: Vertical asymptote at x = −5. No horizontal asymptotes. No x(e) Matches description (ii). y = h(x)
intercepts.
2
2
)
+5
(f) Matches description (ii). y = h(x
= xx+5
: Vertical asymptote at x = −5. No horizontal asymptotes. No xh(x)
intercepts.
1
(g) Does not match any of the descriptions. y = g(x)
= x21+4 : No vertical asymptotes. Horizontal asymptote at y = 0.
No x-intercepts.
(h) Does not match any of the descriptions. y = f (x) · g(x) = (x2 − 4)(x2 + 4): No vertical asymptotes. No horizontal
asymptotes. x-intercepts at ±2.
17. (a) lim f (x) = lim f (x) = 0.
x→∞
x→−∞
(b) The vertical asymptote is x = −2 and we see
lim f (x) = ∞
x→−2+
and
lim f (x) = ∞.
x→−2−
18. (a) lim f (x) = lim f (x) = 2.
x→∞
x→−∞
(b) The vertical asymptote is x = −4 and we see
lim f (x) = −∞
x→−4+
19. (a)
(b)
(c)
(d)
(e)
and
lim f (x) = ∞.
x→−4−
1
If f (n) is large, then f (n)
is small.
1
is large.
If f (n) is small, then f (n)
1
If f (n) = 0, then f (n)
is undefined.
1
If f (n) is positive, then f (n)
is also positive.
1
is negative.
If f (n) is negative, then f (n)
1
1
20. (a) The graph of y = f (x)
will have vertical asymptotes at x = 0 and x = 2. As x → 0 from the left, f (x)
→ −∞,
1
1
and as x → 0 from the right, f (x) → +∞. The reciprocal of 1 is 1, so f (x) will also go through the point (1,1). As
1
1
1
x → 2 from the left, f (x)
→ +∞, and as x → 2 from the right, f (x)
→ −∞. As x → ±∞, f (x)
→ 0 and is
negative.
1
The graph of y = f (x)
is shown in Figure 11.52.
y
x
1
2
Figure 11.52
(b) A formula for f is of the form
f (x) = k(x − 0)(x − 2)
Thus, 1 = k(1)(−1), so k = −1. Thus
The reciprocal
1
f (x)
=
1
− x(x−2)
and
f (x) = −x(x − 2).
is graphed as shown in Figure 11.52.
f (1) = 1.
678
Chapter Eleven /SOLUTIONS
21. (a) The graph of y = −f (−x) + 2 will be the graph of f flipped about both the x-axis and the y-axis and shifted up 2
units . The graph is shown in Figure 11.53.
1
1
will have vertical asymptotes x = −1 and x = 3. As x → +∞, f (x)
→ − 21 and as
(b) The graph of y = f (x)
1
1
1
1
x → −∞, f (x) → 0. At x = 0, f (x) = 2 , and as x → −1 from the left, f (x) → −∞; as x → −1 from the right,
1
1
1
→ +∞; as x → 3 from the left, f (x)
→ +∞; and as x → 3 from the right, f (x)
→ −∞.
f (x)
x = −1
y
(−3, 2)
(1, 2)
(0, 0)
−3
x=3
4
y=4
4
y
x
x
1
y = − 21
Figure 11.53
Figure 11.54
22. (a) The graph shows y = 1/x shifted to the right one and up 2 units. Thus,
y=
1
+2
x−1
is a choice for a formula.
(b) The equation y = 1/(x − 1) + 2 can be written as
y=
2x − 1
.
x−1
(c) We see that the graph has both an x-and y-intercept. When x = 0, y =
then 2x − 1 = 0, so x = 12 . The x-intercept is ( 21 , 0).
−1
−1
= 1, so the y-intercept is (0, 1). If y = 0
23. (a) The graph shows y = 1/x flipped across the x-axis and shifted left 2 units. Therefore
y=−
1
x+2
is a choice for a formula.
(b) The formula y = −1/(x + 2) is already written as a ratio of two linear functions.
(c) The graph has a y-intercept if x = 0. Thus, y = − 21 . Since y cannot be zero if y = −1/(x + 2), there is no
x-intercept. The only intercept is (0, − 12 ).
24. (a) The graph appears to be the graph of y = 1/x shifted down 3 units. Thus,
y=
is a possible formula for the function.
(b) The formula y = (1/x) − 3 can be written as
1
−3
x
−3x + 1
x
after getting a common denominator and combining terms.
(c) The graph has no y-intercept since x = 0 is not in the domain of the function. Any x-intercept(s) will occur when
the numerator of y = (−3x + 1)/x is zero. Then −3x + 1 = 0, so x = 13 . The only intercept is at ( 31 , 0).
y=
11.5 SOLUTIONS
679
1
1
, so p = 1. The graph of y =
has been shifted three units to the right
x
x
and four units up. To find the y-intercept, we need to evaluate f (0):
25. The function f is the transformation of y =
f (0) =
1
11
+4=
.
−3
3
To find the x-intercepts, we need to solve f (x) = 0 for x.
1
+ 4,
x−3
1
,
−4 =
x−3
−4(x − 3) = 1,
0=
Thus,
−4x + 12 = 1,
−4x = −11,
11
so
x=
is the only x-intercept.
4
The graph of f is shown in Figure 11.55.
y
x=3
4
(0,
y=4
11
3 )
x
3
( 11
4 , 0)
Figure 11.55
26. The function g is a transformation of y = 1/x2 , so p = 2. The graph of y = 1/x2 has been shifted 2 units to the right,
flipped over the x-axis and shifted 3 units down. To find the y-intercept, we need to evaluate g(0):
g(0) = −
13
1
−3=− .
4
4
Note that this function has no x-intercepts.
The graph of g is shown in Figure 11.56.
y
x=2
2
−3
x
y = −3
)
(0, − 13
4
Figure 11.56
680
Chapter Eleven /SOLUTIONS
27. First, we can simplify the formula for h(x):
2
1
+
+2
x−1
1−x
1
2
1
=
−
+2=−
+ 2.
x−1
x−1
x−1
h(x) =
1
1
with p = 1. The graph of y = has been shifted one unit to the right, flipped over
x
x
the x-axis and shifted up 2 units. To find the y-intercept, we need to evaluate h(0):
Thus h is a transformation of y =
h(0) = −
1
+ 2 = 3.
−1
To find the x-intercepts, we need to solve h(x) = 0 for x:
1
+2
x−1
1
−2 = −
x−1
−2x + 2 = −1
0=−
−2x = −3,
3
so, x =
is the only x-intercept.
2
The graph of h is shown in Figure 11.57.
y
x=1
(0, 3)
y=2
2
1
( 23 , 0)
x
Figure 11.57
28. (a) The table indicates symmetry about the y-axis. The fact that the function values have the same sign on both sides of
the vertical asymptotes indicates a transformation of y = 1/x2 rather than y = 1/x.
(b) As x takes on large positive or negative values, y → 1. Thus, we try
y=
1
+ 1.
x2
y=
1 + x2
.
x2
This formula works and can be expressed as
29. (a) The table indicates translation of y = 1/x because the values of the function are headed in opposite directions near
the vertical asymptote.
(b) The transformation of y = 1/x given involves a shift to the right by 2 units, so we try
y=
1
.
x−2
11.5 SOLUTIONS
A check of data from the table shows that we must add
y=
1
2
681
to each output of our guess. Thus, a formula would be
1
1
+ .
x−2
2
As a ratio of polynomials, we have
y=
so
1(x − 2)
1(2)
+
(x − 2)(2)
2(x − 2)
y=
x
.
2x − 4
30. (a) The table shows symmetry about the vertical asymptote x = 3. The fact that the function values have the same sign
on both sides of the vertical asymptotes indicates a transformation of y = 1/x2 rather than y = 1/x.
(b) In order to shift the vertical asymptote from x = 0 to x = 3 for the table, we try
y=
1
.
(x − 3)2
checking the x-values from the table in this formula gives y-values that are each 1 less than the y-values of the table.
Therefore, we try
1
y=
+ 1.
(x − 3)2
This formula works. To express the formula as a ratio of polynomials, we take
y=
1(x − 3)2
1
+
,
2
(x − 3)
(x − 3)2
so
y=
x2 − 6x + 10
.
x2 − 6x + 9
31. (a) The table indicates translation of y = 1/x because the values of the function are headed in opposite directions near
the vertical asymptote.
(b) The data points in the table indicate that y → 21 as x → ±∞. The vertical asymptote does not appear to have been
shifted. thus, we might try
1
1
y= + .
x
2
A check of x-values shows that this formula works. To express as a ratio of polynomials, we get a common denominator. Then
1(2)
1(x)
+
x(2)
2(x)
2+x
y=
2x
y=
32. We express the ratio of the volume to the surface area of the box. See Figure 11.58.
The sides of the base are (11 − 2x) and (8.5 − 2x) inches and the depth is x inches, so the volume V (x) =
x(11 − 2x)(8.5 − 2x). The surface area, S(x), is the area of the complete sheet minus the area of the four squares, so
S(x) = 11 · 8.5 − 4x2 . Thus, the ratio we want to maximize is given by
R(x) =
x(11 − 2x)(8.5 − 2x)
V (x)
=
.
S(x)
11 · 8.5 − 4x2
The graph in Figure 11.59 suggests that the maximum of R(x) occurs when x ≈ 1.939 inches. (A good viewing window
is 0 < x < 5 and 0 < y < 1.) So one side is x = 1.939 and therefore the others are 11 − 2(1.939) = 7.123 and
8.5 − 2(1.939) = 4.623 inches.
The dimensions of the box are 7.123 by 4.623 by 1.939 inches.
682
Chapter Eleven /SOLUTIONS
11′′
11 − 2x
6
x
x
6
-
1
x
x
(1.939, 0.813)
x = 4.8
8.5 − 2x
R(x)
8.5′′
?
x
?
Figure 11.58
5
Figure 11.59
• Since the graph has a vertical asymptote at x = 2, let the denominator be (x − 2).
• Since the graph has a zero at x = −1, let the numerator be (x + 1).
• Since the long–range behavior tends toward
x → ±∞, the ratio of the leading terms should be −1.
−1 as
x+1
. You can check that the y–intercept is y = 12 , as it should be.
So a possible formula is y = f (x) = −
x−2
34. We try (x + 3)(x − 1) in the numerator in order to get zeros at x = −3 and x = 1. There is only one vertical asymptote at
x = −2, but in order to have the horizontal asymptote of y = 1, the numerator and denominator must be of same degree.
Thus, try
(x + 3)(x − 1)
y=
.
(x + 2)2
33.
Note that this answer gives the correct y-intercept of (0, − 43 ) and y → 1 as x → ±∞.
(x − 3)(x + 2)
fits the zeros and vertical asymptote of the graph. However, in order to satisfy the y(x + 1)(x − 2)
intercept at (0, −3) and end behavior of y → −1 as x → ±∞, the graph should be “flipped” across the x-axis. Thus try
(x − 3)(x + 2)
y=−
.
(x + 1)(x − 2)
• Since the graph has asymptotes at x = −1 and x = 2, let the denominator be
(x − 2)(x + 1).
• Since the graph has zeros at x = −2 and x = 3, let the numerator be (x + 2)(x − 3).
• Since the long–range behavior tends toward 1 as x → ±∞, the ratio of the leading terms should be 1.
(x + 2)(x − 3)
So a possible formula is y = f (x) =
. You can check that the y-intercept is y = 3, as it should be.
(x − 2)(x + 1)
The graph has vertical asymptotes at x = −1 and x = 1. When x = 0, we have y = 2 and y = 0 at x = 2. The graph of
(x − 2)
satisfies each of the requirements, including y → 0 as x → ±∞.
y=
(x + 1)(x − 1)
x
fits.
The graph of y =
(x + 2)(x − 3)
Factoring the numerator, we have
35. A guess of y =
36.
37.
38.
39.
f (x) =
(x − 9)(x − 2)
(x − 2)
18 − 11x + x2
=
= (x − 9)
.
x−2
x−2
(x − 2)
In this form we see that the graph of y = f (x) is identical to that of y = x − 9, except that the graph of y = f (x)
has no y-value corresponding to x = 2. The line y = x − 9 goes through the point (2, −7) so the graph of y = f (x) will
be the line y = x − 9 with a hole at (2, −7).
11.6 SOLUTIONS
683
40. Factoring the numerator, we have
g(x) =
x2 (x + 5) + (x + 5)
(x + 5)(x2 + 1)
(x + 5)
x3 + 5x2 + x + 5
=
=
= (x2 + 1)
.
x+5
x+5
(x + 5)
(x + 5)
In this form we see that the graph of y = g(x) is identical to that of y = x2 + 1, except that the graph of y = g(x)
has no y-value corresponding to x = −5. The parabola y = x2 + 1 goes through the point (−5, 26) so the graph of
y = g(x) will be the parabola y = x2 + 1 with a hole at (−5, 26).
x−2
41. In order to create a hole in the graph of y = x3 at (2, 8), we use the factor
. Multiplying by this factor is the same
x−2
as multiplying by 1, except at x = 2, where the factor is undefined. So, our function is
(x − 2)
x4 − 2x3
=
.
(x − 2)
x−2
p(x)
−3(x − 2)(x − 3)
42.
f (x) =
=
q(x)
(x − 5)2
We need the factor of −3 in the numerator and the exponent of 2 in the denominator, because we have a horizontal
2
asymptote of y = −3. The ratio of highest term of p(x) to highest term of q(x) will be −3x
= −3.
x2
h(x) = x3
43. The vertical asymptotes indicate a denominator of (x + 2)(x − 3). The horizontal asymptote of y = 0 indicates that the
degree of the numerator is less than the degree of the denominator. To get the point (5,0) we need (x − 5) as a factor in
the numerator. Therefore, try
(x − 5)
g(x) =
.
(x + 2)(x − 3)
44. The description of h agrees with the description of g from part (b) except h has a horizontal asymptote of y = 1.
Therefore, the degree of numerator and denominator must be the same. In fact, the highest-powered terms should be the
same. Note that we can accomplish this without adding a zero by changing the function of (b) to
h(x) =
(x − 5)2
.
(x + 2)(x − 3)
Solutions for Section 11.6
Exercises
1. The function is exponential, because p(x) = (5x )2 = 52x = (52 )x = 25x .
2. The function fits neither, because the variable in the exponent is squared.
3. The function fits neither form. If the expression in the parentheses expanded, then m(x) = 3(9x2 + 6x + 1) = 27x2 +
18x + 3.
4. The function is exponential, because n(x) = 3 · 23x+1 = 3 · 23x · 21 = 6 · 8x .
5. The function fits an exponential, because r(x) = 2 · 3−2x = 2(3−2 )x = 2( 91 )x .
6. The function is a power function, because s(x) =
7. Since larger powers grow faster in the long-run,
A - (i)
B - (iv)
C - (ii)
D - (iii)
4
5x−3
= 45 x3 .
684
Chapter Eleven /SOLUTIONS
8. Larger powers of x give smaller values for 0 < x < 1.
A - (iii)
B - (ii)
C - (iv)
D - (i)
9. (a)
Table 11.8
x
f (x)
g(x)
−3
1/27
−27
−2
1/9
−8
−1
1/3
−1
0
1
0
1
3
1
2
9
8
3
27
27
(b) As x → −∞, f (x) → 0. For f , large negative values of x result in small f (x) values because a large negative power
of 3 is very close to zero. For g, large negative values of x result in large negative values of g(x), because the cube
of a large negative number is a larger negative number. Therefore, as x → −∞, g(x) → −∞.
As x → ∞, f (x) → ∞ and g(x) → ∞. For f (x), large x-values result in large powers of 3; for g(x), large x
values yield the cubes of large x-values. f and g both climb fast, but f climbs faster than g (for x > 3).
10. As x → ∞, we know x3 dominates x2 . Multiplying by a positive constant a or b, does not change the outcome, so
y = ax3 dominates.
11. Since 0.99x is a decreasing exponential function, y = 7 · 0.99x → 0 as x → ∞, so y = 6x35 dominates.
12. As x → ∞, increasing exponential functions dominate power functions, so y = 4ex dominates.
13. As x → ∞, the higher power dominates, so x1.1 dominates x1.08 . The coefficients 1000 and 50 do not change this, so
y = 50x1.1 dominates.
Problems
14. Table 11.9 shows that 3−x approaches zero faster than x−3 as x → ∞.
Table 11.9
x
2
10
100
3−x
1/9
0.000017
1.94 × 10−48
x−3
1/8
0.001
10−6
15. The function y = e−x will approach zero faster. To see this, note that a doubling of x in the cubic function y = x−3
will cause the y-value to decrease by a factor of 2−3 = 81 ; while a doubling of x in the exponential function y = e−x
will cause the y-value to be squared. For small values of y, squaring decreases faster than multiplying by 2−3 . To see this
numerically, look in Table 11.10.
Table 11.10
x
1
10
x−3
1.0
0.001
e−x
0.368
0.0000454
11.6 SOLUTIONS
685
16. Neither. The formula for an exponential function is y = bx ; the formula for a power function is y = xp , where b and p
are constants. The function y = xx is a variable raised to a variable, and thus it fits neither description.
17. (a) Let f (x) = ax + b. Then f (1) = a + b = 18 and f (3) = 3a + b = 1458. Solving simultaneous equations gives us
a = 720, b = −702. Thus f (x) = 720x − 702.
(b) Let f (x) = a · bx , then
ab3
1458
f (3)
=
= b2 =
= 81.
f (1)
ab
18
Thus,
b2 = 81
b=9
(since b must be positive)
Using f (1) = 18 gives
a(9)1 = 18
a = 2.
Therefore, if f is an exponential function, a formula for f would be
f (x) = 2(9)x .
(c) If f is a power function, let f (x) = kxp , then
f (3)
k(3)p
=
= (3)p
f (1)
k(1)p
and
f (3)
1458
=
= 81.
f (1)
18
Thus,
3p = 81
Solving for k, gives
Thus, a formula for f is
18 = k(14 )
k = 18.
so
f (x) = 18x4 .
18. (a) If f is linear,
m=
and
p = 4.
so
128 − 16
= 112,
2−1
16 = 112(1) + b,
b = −96.
so
Thus,
(b) If f is exponential, then
f (x) = 112x − 96.
a(b)2
128
=
= b,
16
a(b)
so
b=8
and
16 = a(8),
so
Therefore
f (x) = 2(8)x .
a = 2.
686
Chapter Eleven /SOLUTIONS
(c) If f is a power function, f (x) = k(x)p . Then
k(2)p
f (2)
128
=
= (2)p =
= 8,
f (1)
k(1)p
16
so p = 3. Using f (1) = 16 to solve for k, we have
16 = k(13 ),
so
k = 16.
Thus,
f (x) = 16x3 .
19. (a) If f is linear, then the formula for f (x) is of the form
f (x) = mx + b,
where
m=
Thus, f (x) =
63
x
4
189
48 − 43
∆y
189
63
=
= 4 =
=
∆x
2 − (−1)
3
12
4
+ b. Since f (2) = 48, we have
48 =
48 −
63
(2) + b
4
63
=b
2
33
b=
2
Thus, if f is a linear function,
33
63
x+
.
4
2
(b) If f is exponential, then the formula for f (x) is of the form
f (x) =
f (x) = abx ,
b > 0, b 6= 1.
Taking the ratio of f (2) to f (−1), we have
ab2
f (2)
= −1 = b3 ,
f (−1)
ab
and
f (2)
4
48
= 3 = 48 · = 64.
f (−1)
3
4
Thus
b3 = 64,
and
b = 4.
To solve for a, note that
f (2) = a(4)2 = 48,
which gives a = 3. Thus, an exponential model for f is f (x) = 3 · 4x .
(c) If f is a power function, then the formula for f (x) is of the form
f (x) = kxp ,
k and p constants.
Taking the ratio of f (2) to f (−1), we have
f (2)
k · 2p
2p
=
=
= (−2)p .
f (−1)
k · (−1)p
(−1)p
11.6 SOLUTIONS
Since we know from part (b) that
f (2)
f (−1)
687
= 64, we have
(−2)p = 64.
Thus, p = 6. To solve for k, note that
f (2) = k · 26 = 48,
which gives
64k = 48
48
3
k=
= .
64
4
Thus, f (x) = 43 x6 is a power function which satisfies the given data.
20. Explanations should include the recognition that Table 11.18 indicates a constant rate of change (linear behavior), and
that Table 11.19 is the only candidate for the logarithmic function (since power functions do not have a zero at x = 1 and
the change in function values is not linear). Table 11.20 must be the quadratic power function, since the function values
are symmetric about the vertical axis, and Table 11.17 is left as the cubic power function.
Using the data, we find:
Table 11.17 : j(x) = .3x3
Table 11.18 : k(x) = 1.2x − 3
Table 11.19 : m(x) = log x
Table 11.20 : z(x) = .4x2
21. Note:
5
7
>
9
16
>
3
8
>
3
,
11
and we know that for x > 1, the higher the exponent, the more steeply the graph climbs, so
A is kx5/7 ,
B is kx9/16 ,
C is kx3/8 ,
D is kx3/11 .
22. (a) For 0 < x < 1, we know that x3 < x2 and we know that for x > 0, x2 < 2x2 . Therefore, f (x) is graph C, g(x) is
graph A, h(x) is graph B.
(b) Yes, B = g(x) = h(x) and A = x3 = 2x2 , so
x3 = 2x2
3
x − 2x2 = 0
x2 (x − 2) = 0
So x = 2 is the only solution for x > 0.
(c) No, since 2x2 > x2 for x > 0.
23. If f (x) = mx1/3 goes through (1, 2), then m(1)1/3 = 2, so m = 2 and f (x) = 2x1/3 . Using x = 8 in f (x) = 2x1/3
gives t = 4. If g(x) = kx4/3 goes through (8, 4), then k = 41 . Thus, m = 2, t = 4, and k = 41 .
24. (a) We are given t(v) = v −2 = 1/v 2 and r(v) = 40v −3 = 40/v 3 therefore
1
40
= 3.
v2
v
Multiplying by v 3 we get v = 40.
(b) We found in part (a) that v = 40. By graphing or substituting values of x between 0 and 40, we see that r(x) > t(x)
for 0 < x < 40.
(c) For values of x > 40 we see by graphing or substituting values that t(x) > r(x).
25. Since x8 is much larger than x2 + 5 for large x (either positive or negative), the ratio tends to zero. Thus, y → 0 as
x → ∞ or x → −∞.
26. Since the denominator has highest power of t6 , which dominates t2 , the ratio tends to 0. Thus, y → 0 as t → ∞ or
t → −∞.
688
Chapter Eleven /SOLUTIONS
27. For large positive t, the value of 5t is much larger than the value of 2t . Thus, y → 0 as t → ∞.
For large negative t, the values of 2t and 5t go to zero, so
y→
7
0+7
=
0+9
9
as t → −∞.
28. For large positive t, the value of 3−t → 0 and 4t → ∞. Thus, y → 0 as t → ∞.
For large negative t, the value of 3−t → ∞ and 4t → 0. Thus,
y→
Very large positive number
0+7
as t → −∞.
So y → ∞ as t → −∞.
29. Multiplying out the numerator gives a polynomial with highest term x3 , which dominates the x2 in the denominator. Thus,
y → x as x → ∞ or x → −∞. So y → ∞ as x → ∞, and y → −∞ as x → −∞.
30. For large positive x, the value of 2x is much larger than the value of x2 . Thus, y → ∞ as x → ∞. For large negative x,
the value of 2x → 0, but x2 → ∞. Thus, y → 0 as x → −∞.
√
31. Since x dominates ln x, the value of y is very small for large positive x. Thus, y → 0 as x → ∞. Since negative
x-values are not in the domain of this functions, we do not consider x → −∞.
32. Since et and t2 both dominate ln |t|, we have y → ∞ as t → ∞.
For large negative t, the value of et → 0, but t2 is large and dominates ln |t|. Thus, y → ∞ as t → −∞,
33. As x → ∞, the value of ex → ∞ and e−x → 0. As x → −∞, the value of e−x → ∞ and ex → 0. Thus, y → ∞ as
x → ∞ and y → −∞ as x → −∞.
34. As x → ∞, the value of ex → ∞ and e−x → 0. As x → −∞, the value of e−x → ∞ and ex → 0. Thus,
As x → ∞,
As x → −∞,
y=
y=
ex − e−x
ex − 0
→ x
= 1.
x
−x
e +e
e +0
ex − e−x
0 − e−x
→
= −1.
x
−x
e +e
0 + e−x
35. Since ex dominates x100 , for large positive x, the value of y is very large. Thus, y → ∞ as x → ∞.
For large negative x, the value of ex is very small, but x100 is very large. Thus, y → 0 as x → −∞.
36. Since e3t dominates e2t , the value of y is very small for large positive t. Thus, y → 0 as t → ∞.
For large negative t, the value of e2t → 0, so y → 0 as t → −∞.
37. The trigonometric function should oscillate, or in other words, the function values should move periodically back and
forth between two extremes. It seems f (x) best displays this behavior. The graph in Figure 11.60 shows the points for f
from Table 11.21 in the problem. One possible curve has been dashed in. We can recognize the curve as having the same
shape as the sine function. The amplitude is 2 (the curve only varies 2 units up or down from the central value of 4). It is
raised 4 units from the x-axis. Also, the period is 4 because the curve makes one full cycle in the space of 4 units, which
π
2π
= . A formula for the curve in Figure 11.60 could be
tells us that the frequency is
4
2
f (x) = 2 sin
π
x + 4.
2
(Note: Answers are not unique!)
The exponential function should take the form y = a · bx . Since neither a nor b can be zero, the function cannot pass
through the point (0, 0). Therefore, g(x) cannot be exponential. Try h as the exponential function. Figure 11.61 shows
the points plotted from h(x) with the curve dashed in.
11.6 SOLUTIONS
689
6
1
4
2
x
−2
−1
1
x
2
−2
Figure 11.60: f (x) best fits a trigonometric function
Rewriting h(0) = 0.33 as h(0) =
1
3
−1
1
2
Figure 11.61: h(x) could be an exponential function
gives a = 31 . Thus, using any other point, say (1, 0.17), we have
0.17 =
1
b
3
b ≈ 0.50 =
We find a possible formula to be
1
.
2
1 1 x
.
3 2
The power function is left for g. A power function takes the form y = k · xp . Solving for k and p, we find
h(x) =
5
g(x) = − x3 .
2
We see that the data for g satisfy this formula.
38. (a)
No. of years elapsed
Start-of-year balance
End-of-year deposit
0
$1000.00
$1000
End-of-year interest
$60.00
1
$2060.00
$1000
$123.60
2
$3183.60
$1000
$191.02
3
$4374.62
$1000
$262.48
4
$5637.10
$1000
$338.23
5
$6975.33
$1000
$418.52
(b) The balance is not growing at a linear rate because the change (increase) in balance is increasing each year. If the
growth were linear, the increase would be the same amount each year.
The balance is not growing exponentially, either, because the ratio of successive balance amounts is not constant.
For example,
3183.60
Balance in year 2
=
≈ 1.55
Balance in year 1
2060.00
and
Balance in year 3
4374.62
=
≈ 1.37.
Balance in year 2
3183.60
Thus, neither a linear nor an exponential function represents the growth of the balance in this situation.
690
Chapter Eleven /SOLUTIONS
39. (a) If pn (r) represents the balance after n years where r is the rate of interest, we have the pattern in Table 11.11.
Table 11.11
In year n =
balance
0
1000
1
1000 + 1000 · r + 1000
=1000(1 + r) + 1000
2
1000(1 + r) + 1000 + (1000(1 + r) + 1000) ·r + 1000
|
{z
} |
{z
=(1000(1 + r) + 1000)(1 + r) + 1000
=1000(1 + r)2 + 1000(1 + r) + 1000
3
|
2
{z
}
}
1000(1 + r) + 1000(1 + r) + 1000 +[(1000)(1 + r)2 + 1000(1 + r) + 1000] · r + 1000
|
{z
}
=(1000(1 + r)2 + 1000(1 + r) + 1000)(1 + r) + 1000
=1000(1 + r)3 + 1000(1 + r)2 + 1000(1 + r) + 1000
.
..
in year n
1000(1 + r)n + 1000(1 + r)n−1 + · · · + 1000(1 + r) + 1000
Thus,
p5 (r) = 1000(1 + r)5 + 1000(1 + r)4 + 1000(1 + r)3 + 1000(1 + r)2
+1000(1 + r) + 1000
and
p10 (r) = 1000(1 + r)10 + 1000(1 + r)9 + 1000(1 + r)8 + 1000(1 + r)7
+1000(1 + r)6 + 1000(1 + r)5 + 1000(1 + r)4 + 1000(1 + r)3 +
1000(1 + r)2 + 1000(1 + r) + 1000.
(b) We can enter p5 (r) as is on our calculator and solve for where p5 (r) = 10,000. Alternatively, let x = (1 + r), and
solve
1000x5 + 1000x4 + 1000x3 + 1000x2 + 1000x + 1000 = 10, 000
or equivalently solve for x such that
x5 + x4 + x3 + x2 + x − 9 = 0.
A graphical solution shows x ≈ 1.20279. Since x = 1 + r, r ≈ 20.279% interest.
40. The function f (d) = b · dp/q , with p < q, because f (d) increases more and more slowly as d gets larger, and g(d) =
a · dp/q , with p > q, because g(d) increases more and more quickly as d gets larger.
Solutions for Section 11.7
Exercises
1. f (x) = axp for some constants a and c. Since f (1) = 1 = a(1)p , it follows that a = 1. Also, f (2) = 2p = c. Solving
for p we have p = ln c/ ln 2. Thus, f (x) = xln c/ ln 2
2. Regression on a calculator returns the exponential function h(x) = 2.35(1.44)x .
3. Regression on a calculator returns the power function g(x) = 2x1.2 .
691
11.7 SOLUTIONS
(i) Calculator result: y = 6.222t0.504 . Answers may vary.
(ii) Calculator result: y = 0.066t2 − 1.515t + 25.571. Answers may vary.
(b) For the time period 1970-2000, the quadratic function is the better fit. The power function is approximately the square
root function so it is concave down, but the catch values increased rapidly toward the end of this period. Notice that
the power function goes through (0, 0), meaning that the predicted value of the 1965 catch is zero—not likely to be
a realistic prediction. The quadratic function is shifted and stretched so is the better fit. See Figure 11.62. However,
outside of the interval 1970-2000, there is no reason to suppose that either function is a good fit. Our results hold only
for this time period.
4. (a)
y
60
50
Power
y = 6.222t0.504
40
?
30
I
20
Quadratic
10
y = 0.066t2 − 1.515t + 25.571
5
10
15
20
25
30
35
t
Figure 11.62
5. (a) Regression on a calculator returns the power function f (x) = 201.353x2.111 , where f (x) represents the total dry
weight (in grams) of a tree having an x cm diameter at breast height.
(b) Using our regression function, we obtain f (20) = 201.535(20)2.111 = 112,313.62 gm.
(c) Solving f (x) = 201.353x2.111 = 100,000 for x we get
100,000
201.353
100,000 1/2.111
x=
= 18.930 cm.
201.353
x2.111 =
6. (a) Performing regression with a power function on the calculator returns the function Q(b) = 221.425b0.685 .
(b) The estimated brain weight of the Erythrocebus is Q(7800) = 221.425(7800)0.685 ≈ 102,639 mg.
7. The slope of this line is
6−0
4−0
8. The slope of this line is
2−0
0−(−3)
e2 eln(x
2/3
)
= 32 . The vertical intercept is 0. Thus ln y =
3
2
ln x, and y = e(3/2) ln x = eln(x
= 32 . The vertical intercept is 2. So ln y = 2 +
2
3
2
2
y2 −y1
x2 −x1
= 32 . The vertical intercept is 0, thus y = 32 x.
= 1 and the vertical intercept is 2, thus ln y = x + 2, so y = ex+2 .
10. The slope of this line is
2−1
0−(−1)
11. The slope of this line is
2−0
= 0.4. The vertical intercept is 0. Thus ln y = 0.4x, and y =
5−0
1.7−0
= −1.7. The vertical intercept is 0. Thus ln y = −1.7x. So
−1−0
12. The slope of this line is
)
3
= x2 .
2
ln x, and y = e2+ 3 ln x = e2 e 3 ln x =
= e2 x 3 .
9. The slope of this line is m =
3/2
e0.4x .
y = e−1.7x .
692
Chapter Eleven /SOLUTIONS
Problems
13. (a) The function y = −83.039 + 61.514x gives a superb fit, with correlation coefficient r = 0.99997.
(b) When the power function is plotted for 2 ≤ x ≤ 2.05, it resembles a line. This is true for most of the functions we
have studied. If you zoom in close enough on any given point, the function begins to resemble a line. However, for
other values of x (say, x = 3, 4, 5 . . .), the fit no longer holds.
14. (a) The FM band appears linear, because the FM frequency always increases by 4 as the distance increases by 10. See
Figure 11.63.
frequency
160
140
120
100
80
60
millimeters
10 20 30 40 50
Figure 11.63
(b) The AM band is increasing at an increasing rate. These data could therefore represent an exponential relation.
(c) We recall that any linear function has a formula f (x) = b + mx. Since the rate of change, m, is the change in
4
= 0.4. So
frequency, 4, compared to the change in length, 10, then m = 10
y = b + 0.4x.
But the table tells us that f (5) = b + 0.4(5) = b + 2 = 88. Therefore, b = 86, and
y = 86 + 0.4x.
We could have also used a calculator or computer to determine the coefficients for the linear regression.
(d) Since the data for the AM band appear exponential, we wish to plot the natural log of the frequency against the length.
Table 11.12 gives the values of the AM station numbers, y, and the natural log, ln y, of those station numbers as a
function of their location on the dial, x.
Table 11.12
x
5
15
25
35
45
55
y
53
65
80
100
130
160
3.97
4.17
4.38
4.61
4.87
5.08
ln y
The (x, ln y) data are very close to linear. Regression gives the coefficients for the linear equation ln y = b+mx,
yielding
ln y = 3.839 + 0.023x.
Solving for y gives:
eln y = e3.839+0.023x
y = e3.839+0.023x
y=e
Since e3.839 ≈ 46.5,
3.839 0.023x
e
(since eln y = y).
(since ax+y = ax ay )
y = 46.5e0.023x .
11.7 SOLUTIONS
693
15. By plotting x against t0.4 we see a straight line with slope≈ 3.50. Alternatively, by calculating x/t0.4 for each of the data
points, except (0, 0), we find a common value of approximately≈ 3.50. These give an estimate of a ≈ 3.50. Figure 11.64
shows the function x = 3.50t0.4 and the data set from Table 11.32, which seems to model the situation well.
Distance
(in cms)
30
20
10
Time
100
200
300 (in minutes)
Figure 11.64: The wetting front as a function
of time and the function x = 3.49t0.4
16. (a) A computer or calculator gives
N = −14t4 + 433t3 − 2255t2 + 5634t − 4397.
(b) The graph of the data and the quartic in Figure 11.65 shows a good fit between 1980 and 1996.
N , AIDS deaths
N AIDS deaths
(thousands)
400
(thousands)
N=
−14t4
16
+
433t3
N = −14t4 + 433t3 − 2255t2
+5634t − 4397
2255t2
−
+5634t − 4397
400
t (years)
10
20
Figure 11.65
30
t, (years)
Figure 11.66
(c) Figure 11.65 shows that the quartic model fits the 1980-1996 data well. However, this model predicts that in 2000, the
number of deaths decreases. See Figure 11.66. Since N is the total number of deaths since 1980, this is impossible.
Therefore the quartic is definitely not a good model for t > 20.
17. (a)
Recliner price ($)
399
499
599
699
799
Demand (recliners)
62
55
47
40
34
Revenue ($)
24,738
27,445
28,153
27,960
27,166
(b) Using quadratic regression, we obtain the following formula for revenue, R, as a function of the selling price, p:
R(p) = −0.0565p2 + 72.9981p + 4749.85.
(c) By using a graphing calculator to zoom in, we see that the price which maximizes revenue is about $646. The revenue
generated at this price is about $28,349.
694
Chapter Eleven /SOLUTIONS
18. (a) The function appears to be decreasing and concave up.
(b) Using a calculator and starting with t = 1 for 2006, a power regression function is obtained: P (t) = 2.976t−0.002 .
(c) Substituting t = 7, we get P (7) = 2.976 · 7−0.002 = 2.964. In 2012, we extrapolate the Armenian population to be
approximately 2,964,000.
19. (a) Use a calculator or computer to find an exponential regression function starting with t = 0 for 1985, C(t) =
841.368(1.333)t . Answers may vary.
(b) From the equation, the growth factor is 1.333, so cellular subscriptions are increasing by 33.3% per year.
(c) Although the number of cell subscriptions may continue to increase, we eventually expect slower growth. The graph
would become concave down.
20. (a) The curve does not fit well because the projected maximum, at the vertex, is about two years late.
(b) Use a calculator to find v = −0.145h3 + 1.850h2 − 1.313h + 10.692. Answers may vary. Yes, this equation is a
better fit. See Figure 11.67.
v, users (millions)
50
40
30
20
10
5
10
h, years since 1990
Figure 11.67
21. (a) See Figure 11.68.
Population (thousands)
2000
1500
1000
500
Years
20 40 60 80 100 120 since 1650
Figure 11.68
(b) Using a calculator or computer, we get P (t) = 56.108(1.031)t . Answers may vary.
(c) The 56.108 represents a population of 56,108 people in 1650. Note that this is more than the actual population of
50,400. The growth factor of 1.031 means the rate of growth is approximately 3.1% per year.
(d) We find P (100) = 1194.308, which is slightly higher than the given data value of 1,170.8.
(e) The estimated population, P (150) = 5510.118, is higher than the given census population.
22. (a) P (t) = 3.956(1.030)t . Answers may vary.
(b) The value, P (10) = 3.956 · 1.03010 = 5.314 for 1800, is an interpolation, while in the other problem, 1800 was
outside the data set of 1650-1790 and was a less accurate extrapolation.
(c) P (210) = 3.956 · 1.030220 = 2639.038. This is more than 2.6 billion, and far from being correct, since the US
population was about 310 million in 2010.
11.7 SOLUTIONS
695
23. (a) The formula is N = 1148.55e0.3617t . See Figure 11.69.
N
2,500,000,000
2,000,000,000
1,500,000,000
1,000,000,000
500,000,000
5
10
15
20
25
30
35
40
t
Figure 11.69
(b) The doubling time is given by ln 2/0.3617 ≈ 1.916. This is consistent with Moore’s Law, which states that the
number of transistors doubles about once every two years. Dr. Gordon E. Moore is Chairman Emeritus of Intel
Corporation According to the Intel Corporation, “Gordon Moore made his famous observation in 1965, just four
years after the first planar integrated circuit was discovered. The press called it ‘Moore’s Law’ and the name has
stuck. In his original paper, Moore observed an exponential growth in the number of transistors per integrated circuit
and predicted that this trend would continue.”
24. (a)
(b)
(c)
(d)
Calculator result:y = 2.183t1.271 . Answers may vary.
Calculator result: y = 18.916 · 1.060t . Answers may vary.
Calculator result: y = 0.553t2 − 22.884t + 226.956. Answers may vary.
The power function is a poor fit because it does not rise fast enough to fit values from 1990 to 2000; its projection
is likely too low for 2010. The exponential function fits the data better and although its values are low for the period
1990 to 2000; it probably gives the best projection for 2010. The quadratic function fits the data best of all by passing
close to most points, but its 2010 projection is likely too high. See Figure 11.70.
y , fish (1000s metric tons)
1400
1200
Exponential
y = 18.916(1.060)t
1000
Quadratic
800
-
y = 0.553t2 −
22.884t + 226.956
600
400
Power
y = 2.132t1.271
200
10
20
30
40
50
60
70
y , years since 1935
Figure 11.70
25. (a) Using t = 5, 10, . . . , 50, yields y = 0.310t2 − 12.177t + 144.517. Answers may vary.
(b) Using t (
= 55, 60, 65, 70, yields y = 3.01t2 − 348.43t + 10,955.75. Answers may vary.
y = 0.310t2 − 12.177t + 144.517 5 ≤ t ≤ 50
(c) f (t) =
y = 3.01t2 − 348.43t + 10,955.75 55 ≤ t ≤ 70
696
Chapter Eleven /SOLUTIONS
y , fish (1000s metric tons)
1400
1200
1000
6
800
y = 3.01t2 − 348.43t + 10,955.75
600
400
y = 0.310t2 − 12.177t + 144.517
50
60
200
10
20
30
40
70
t, years since 1935
Figure 11.71
26. (a) Table 11.13 shows the transformed data, where y = ln N . Figure 11.72 shows that the transformed data lie close to
a straight line.
y = ln N
Table 11.13
t
y = ln N
1
2
10
t
y = ln N
5.069
9
11.408
6.433
10
11.708
3
7.664
11
11.972
4
8.637
12
12.203
5
9.442
13
12.405
6
10.115
14
12.587
7
10.624
15
12.740
8
11.039
16
12.837
5
4
8
12
t (years
16 since 1980)
Figure 11.72: Domestic deaths from AIDS,
1981–96 (ln N against t)
(b) We now use linear regression to estimate a line to fit the data points (t, ln N ). The formula provided by a calculator
(and rounded) is
y = 6.445 + 0.47t.
(c) To find the formula for N in terms of t, we substitute ln N for y and solve for N :
ln N = 6.445 + 0.47t
N = e6.445+0.47t
= e6.445
and since e6.445 ≈ 630, we have
e0.47t ,
N ≈ 630e0.47t .
11.7 SOLUTIONS
697
27. (a) We take the log of both sides and make the substitution y = ln N to obtain
y = ln N = ln (atp )
= ln a + ln tp
= ln a + p ln t.
We make the substitution b = ln a, so the equation becomes
y = b + p ln t.
Notice that this substitution does not result in y being a linear function of t. However, if we make a second substitution, x = ln t, we have
y = b + px.
So y = ln N is a linear function of x = ln t.
(b) If a power function fits the original data, the points lie on a line. If the points do not lie on a line, a power function
does not fit the data well.
28. (a) We transform the data using the substitutions x = ln t and y = ln N . The transformed data in Table 11.14 are plotted
in Figure 11.73.
y = ln N
Table 11.14
x = ln t
y = ln N
x = ln t
y = ln N
0
5.069
2.20
11.408
0.69
6.433
2.30
11.708
1.10
7.664
2.40
11.972
1.39
8.637
2.48
12.203
1.61
9.442
2.56
12.405
1.79
10.115
2.64
12.587
1.95
10.624
2.71
12.740
2.08
11.039
2.77
12.837
10
5
1
2
3
x = ln t
Figure 11.73: Domestic deaths from AIDS, 1981–96
(ln N against ln t)
(b) Using regression to fit a line to these data gives
y = 4.670 + 3.005x.
(c) Now transform the equation back to the original variables, t and N . Since our original substitutions were y = ln N
and x = ln t, this equation becomes
ln N = 4.670 + 3.005 ln t.
Raise e to the power of both sides
eln N = e4.670+3.005 ln t
N = (e4.670 )(e3.005 ln t ).
Since e4.670 ≈ 107 and e3.005 ln t = eln t
3.005
= t3.005 , we have
N ≈ 107t3.005 ,
which is the formula of a power function.
29. (a) Quadratic is the only choice that increases and then decreases to match the data.
(b) Using ages of x = 20, 30, . . . , 80, a quadratic function is y = −34.136x2 + 3497.733x − 39,949.714. Answers
may vary.
(c) The value of the function at 37 is y = −34.136 · 372 + 3497.733 · 37 − 39,949.714 = $42,734.
(d) The value of the function for age 10 is y = −34.136 · 102 + 3497.733 · 10 − 39,949.714 = −$8386. Answers may
vary. Not reasonable, as income is positive. In addition, 10-year olds do not usually work.
698
Chapter Eleven /SOLUTIONS
30. (a) Formulas for the best fit curves are given by he = 16.2e−0.0032t , ha = 12.3e−0.0076t , and hb = 14.7e−0.0067t .
Since 1/0.0032 = 312.5, we have 0.0032 = 1/312.5. Thus the formula for he can be rewritten as
he = 16.2e−t/312.5 .
Similarly 0.0076 = 1/131.6 and 0.0067 = 1/149.3. Thus we have
ha = 12.3e−t/131.6
and
hb = 14.7e−t/149.3 .
Figures 11.74–11.76 show the plots with the corresponding exponentials.
he
ha
15
15
10
10
5
5
100
200
t
300
100
Figure 11.74
200
t
300
Figure 11.75
hb
15
10
5
100
200
t
300
Figure 11.76
(b) The value of h0 tells us the initial height of the foam, and the constant τ , called the time constant, tells us how
long it takes the beer to drop by a factor of 1/e ≈ 36.8%. In other words, every τ seconds 63.2% of the foam
disappears. For Erdinger Weissbier, the foam is initially 16.2 cm high and goes down by 63.2% every 312.5 seconds.
For Augustinerbr¨au M¨unchen, the foam is initially 12.3 cm high and goes down by 63.2% every 131.6 seconds. For
Budweiser Budvar, the foam is initially 14.7 cm high and goes down by 63.2% every 149.3 seconds.
31. (a) Using a computer or calculator, we find that t = 8966.1H −2.3 . See Figure 11.77.
(b) Using a computer or calculator, we find that r = 0.0124H − 0.1248. See Figure 11.78.
r
t
0.25
40
0.2
30
0.15
20
0.1
10
0.05
5
10
15
20
Figure 11.77
25
30
H
5
10
15
20
Figure 11.78
25
30
H
11.7 SOLUTIONS
699
(c) From part (a), we see that t → ∞ as H → 0, and since r = 1/t, we have r → 0 as H → 0. From part (b), we can
solve for the value of H making r = 0 as follows:
0.0124H − 0.1248 = 0
0.0124H = 0.1248
H ≈ 10.1.
Thus, the first model predicts that the development rate will fall to r = 0 only at H = 0◦ C (the freezing point of
water), whereas the second model predicts that r will reach 0 at around 10◦ C (or about 50◦ F). The latter prediction
seems far more reasonable: certainly weevil eggs (or any other eggs) would not grow at temperatures near freezing.
32. (a) Because the data gives a decreasing function, we expect p to be negative.
(b) The point (0, 244) must be omitted because a power function with a negative p is not defined for c = 0. A computer
or calculator gives
l = 720c−0.722 .
(Different programs may give slightly different results.)
(c) If l = kcp , we want a formula using natural logs such that
ln l = ln(kcp )
ln l = ln k + ln cp
ln l = ln k + p ln c.
Since y = ln l and x = ln c,
y = ln k + px.
Letting b = ln k, we have
y = b + px.
(d) We must omit the point (0, 244) because ln 0 is undefined. The data is in Table 11.15.
Table 11.15
x = ln c
y = ln l
−0.223
5.347
1.758
5.323
3.136
5.118
3.970
4.828
4.673
4.673
5.011
4.277
5.333
3.555
5.660
2.485
5.808
1.569
6.011
0.833
6.205
0.182
y = ln l
5
5
x = ln c
Figure 11.79
(e) The plotted points in Figure 11.79 don’t look linear, so the power function won’t give a good fit.
33. By plotting P against D3/2 we see a straight line with slope≈ 0.2. Alternatively, by calculating P/D3/2 for each of
the planets, we find a common value of approximately 0.2. We see that Kepler’s model fits the data reasonably well,
with k ≈ 0.2. Kepler’s equation is P = 0.2d3/2 . Figure 11.80 shows the function P = 0.2D3/2 and the data set from
Table 11.44, which models the situation well.
700
Chapter Eleven /SOLUTIONS
Period
90000
60000
30000
Distance
1000
3000
6000
Figure 11.80: The period as a function of
distance and the function P = 0.2D3/2
Solutions for Chapter 11 Review
Exercises
1. This function represents proportionality to a power.
y=
Thus k = 1/6 and p = −7.
1
3
2x7
=
1
=
6x7
1
x(−7) .
6
2. This function represents proportionality to a power.
y=
6
= −3x5 = (−3)x(5) .
−2/x5
Thus k = −3 and p = 5.
3. This function does not represent proportionality to a power because we cannot get it into the form y = kxp . Instead, we
have a function of the form y = kpx , with constant p.
4. This function does not represent proportionality to a power because we cannot get it into the form y = kxp .
5. While y = 6x3 is a power function, when we add two to it, it can no longer be written in the form y = kxp , so this is not
a power function.
6. Dividing both sides by three, we get y = 3x2 , so it is a power function.
7. Expanding the right side, we get y − 9 = x2 − 9. Adding nine to both sides, we see that this is the power function, y = x2 .
8. Expanding the right side, we get y = 4x2 − 16 + 16 = 4x2 . Thus, this is a power function, y = 4x2 .
9. Since the graph is symmetric about the y-axis, the power function is even.
10. Since the graph is steeper near the origin and less steep away from the origin, the power function has a power between 0
and 1.
11. Since the graph is symmetric about the origin, the power function has an odd power.
12. Since the graph is symmetric about the y-axis, the power function has an even power.
13. Since the graph is symmetric about the origin, the power function has an odd power.
14. Since the graph is steeper near the origin and less steep away from the origin, the power function is fractional.
SOLUTIONS to Review Problems For Chapter Eleven
15. The values are
We have
701
√
3
k=2 7
11
.
p=
15
√
√
√
√ √
5
5
3
3
r(x) = 2 7x x2 = 2 7 3 x x2
√
1/5
3
= 2 7 · x1/3 x2
√
3
= 2 7 · x1/3 · x2/5
√
2
1
3
= 2 7 · x3+5
√
3
= 2 7 · x11/15 .
16. By the ratio method,
f (5)
f (3)
k(5)p
k(3)p
5 p
3
p
3
5
3
=
5
3
=
5
= −1.
=
Solving for k, we have
k(3)−1 = 5
k = 15.
17. By multiplying out the expression (x2 − 4)(x2 − 2x − 3) and then simplifying the result, we see that
y = x4 − 2x3 − 7x2 + 8x + 12
which is a fourth-degree polynomial.
18. Since the leading term of the polynomial is 16x3 , the value of y goes to infinity as x → ∞. The graph resembles
y = 16x3 .
19. Since the leading term of the polynomial is 4x4 , the value of y goes to infinity as x → ∞. The graph resembles y = 4x4 .
√
√
√
20. Since 5x2 /x3/2 = 5x1/2 = 5 x, this function can be written
√ as y = 5 x + 2. Since the leading term is 5 x, the value
of y goes to infinity as x → ∞. The graph resembles y = 5 x.
21. Since 2x2 /x−7 = 2x9 , we can rewrite this polynomial as y = 2x9 − 7x5 + 3x3 + 2. Since the leading term of the
polynomial is 2x9 , the value of y goes to infinity as x → ∞. The graph resembles y = 2x9 .
22. This problem cannot be solved algebraically. Note that we cannot use the quadratic formula, as this is a 5th degree
polynomial and not a 2nd degree polynomial. A graph of the function is shown in Figure 11.81 for −1 ≤ x ≤ 1,
−10 ≤ y ≤ 10. From the graph, we approximate the zero to be at x ≈ −0.143.
y
zoom in here to find
x ≈ −0.143
5
R
−5
Figure 11.81
x
702
Chapter Eleven /SOLUTIONS
23. By using the quadratic formula, we find that y = 0 if
x=
3±
√
9 − 4(2)(−3)
3 ± 33
=
.
4
4
p
24. This is a rational function, as we can put it in the form of one polynomial divided by another:
f (x) =
5
x2 − 5
x2
−
=
.
x−3
x−3
x−3
25. This is not a rational function, as we cannot put it in the form of one polynomial divided by another, since ex is an
exponential function, not a polynomial.
26. Since y = 7/x−4 = 7x4 and since x4 dominates x3 , we see that 7/x−4 dominates.
27. Since y = 4/e−x = 4ex and since increasing exponential functions dominate power functions, y = 4/e−x dominates.
28. (a) Since the long-run behavior of a rational function is given by the ratio of the leading terms, we have
lim
x→∞
(b) We have
lim
x→−∞
x(x2 − 4)
1
x3 − 4x
x3
= lim
= lim
= .
3
3
x→∞ 5x + 5
x→∞ 5x3
5 + 5x
5
3x(x − 1)(x − 2)
3x3 − 9x2 + 6x
3x3
= lim
= lim
= 0.
x→−∞
x→−∞ −6x4
5 − 6x4
−6x4 + 5
29. (a) Since the long-run behavior of r(x) = p(x)/q(x) is given by the ratio of the leading terms of p and q, we have
lim
x→∞
(b) We have
lim
x→−∞
2x + 1
2x
= lim
= 2.
x→∞ x
x−5
5x
2 + 5x
5
= lim
= .
x→−∞ 6x
6x + 3
6
30. One way to approach this problem is to consider the graphical interpretations of even and odd functions. Recall that if a
function is even, its graph is symmetric about the y-axis. If a function is odd, its graph is symmetric about the origin.
(a) The function f (x) = x2 + 3 is y = x2 shifted three units up. Note that y = x2 is symmetric about the y-axis. An
upward shift will not affect the symmetry. Therefore, f is even.
(b) We consider g(x) = x3 + 3 as an upward shift (by three units) of y = x3 . However, although y = x3 is symmetric
about the origin, the upward-shifted function will not have that symmetry. Therefore, g(x) = x3 + 3 is neither even
nor odd.
(c) The function y = x1 is symmetric about the origin (odd). The function h(x) = x5 is merely a vertical stretch of
y = x1 . This would not affect the symmetry of the function. Therefore, h(x) = x5 is odd.
(d) If y = |x|, the graph is symmetric about the y-axis. However, a shift to the right by four units would make the
resulting function symmetric to the line x = 4. The graph of j(x) = |x − 4| is neither even nor odd.
(e) The function k(x) = log x is neither even nor odd. Since k is not defined for x ≤ 0, clearly neither type of symmetry
would apply for k(x) = log x.
(f) If we take l(x) = log(x2 ), we have now included x < 0 into the domain of l. For x < 0, the graph looks similar to
y = log x flipped about the y-axis. Thus, l(x) = log(x2 ) is symmetric about the y-axis. —It is even.
(g) Clearly y = 2x is neither even nor odd. Likewise, m(x) = 2x + 2 is neither even nor odd.
(h) We have already seen that y = cos x is even. Note that the graph of n(x) = cos x + 2 is the cosine function shifted
up two units. The graph will still be symmetric about the y-axis. Thus, n(x) = cos x + 2 is even.
Problems
31. All power function go through the origin (0, 0). The graph does not, so it does not represent a power function.
SOLUTIONS to Review Problems For Chapter Eleven
703
32. (a) x1/n is concave down: its values increase quickly at first and then more slowly as x gets larger. The function xn , on
the other hand, is concave up. Its values increase at an increasing rate as x gets larger. Thus
f (x) = x1/n
and
g(x) = xn .
(b) Since the point A is the intersection of f (x) and g(x), we want the solution of the equation xn = x1/n . Raising both
sides to the power of n we get
(xn )·n = (x1/n )·n
or in other words
2
xn = x
Since x 6= 0 at the point A, we can divide both sides by x, giving
2
xn
−1
= 1.
Since n2 − 1 is just some integer we rewrite the equation as
xp = 1
where p = n2 − 1. If p is even, x = ±1, if p is odd x = 1. By looking at the graph we can tell that we are not
interested in the situation when x = −1. When x = 1, the quantities xn and x1/n both equal 1. Thus the coordinates
of point A are (1, 1).
33. Graph (i) looks periodic with amplitude of 2 and period of 2π, so it best corresponds to function J,
y = 2 sin(0.5x).
Graph (ii) appears to decrease exponentially with a y-intercept < 10, so it best corresponds to function L,
y = 2e−0.2x .
Graph (iii) looks like a rational function with two vertical asymptotes, no zeros, a horizontal asymptote at y = 0 and
a negative y-intercept, so it best corresponds to function O,
y=
1
.
x2 − 4
Graph (iv) looks like a logarithmic function with a negative vertical asymptote and y-intercept at (0, 0), so it best
corresponds to function H,
y = ln(x + 1).
34. Graph (i) looks trigonometric with period π and amplitude 2, so it is (B).
Graph (ii) looks like a logarithm which has been shifted right, so it is (L).
Graph (iii) looks cubic, shifted right and down, so it is (G).
Graph (iv) is a line with positive slope and negative y-intercept, so it is (H).
Graph (v) looks like a rational function with two vertical asymptotes and no zeros, so it is (U).
Graph (vi) looks like an exponential decay function flipped over the x-axis, so it is (S).
Graph (vii) looks trigonometric with amplitude less than 1 and period π, so it is (A).
Graph (viii) looks rational with two asymptotes and one positive zero, so it is (E).
35. The x intercepts are at x = −4, −2, 2, so let y = k(x + 4)(x + 2)(x − 2). The y-intercept is at (0, 24), so substituting
x = 0, y = 24:
24 = k(4)(2)(−2)
24 = k(−16)
24
3
k=
=
.
−16
−2
Therefore, y = − 23 (x + 4)(x + 2)(x − 2) is a possible formula.
704
Chapter Eleven /SOLUTIONS
36. Since the graph opens upward, we expect an even degree polynomial with a positive leading coefficient. Since the graph
crosses the x-axis at −1 and −3, there are factors of (x + 1) and (x + 3), giving y = a(x + 1)(x + 3). Since the graph
crosses the y-axis at 12, we can find a by substituting x = 0:
12 = a(0 + 1)(0 + 3)
12 = 3a
4 = a.
Thus, a possible polynomial is y = 4(x + 1)(x + 3).
37. The graph appears to have x intercepts at x = − 12 , 3, 4, so let
y = k(x +
1
)(x − 3)(x − 4).
2
The y-intercept is at (0, 3), so substituting x = 0, y = 3:
which gives
or
1
3 = k( )(−3)(−4),
2
3 = 6k,
1
k= .
2
Therefore, y = 21 (x + 12 )(x − 3)(x − 4) is a possible formula for f .
38. The shape of the graph suggests an odd degree polynomial with a positive leading term. Since the graph crosses the x-axis
at 0, −2, and 2, there are factors of x, (x + 2) and (x − 2), giving y = ax(x + 2)(x − 2). To find a we use the fact that
at x = 1, y = −6. Substituting:
−6 = a(1)(1 + 2)(1 − 2)
−6 = −3a
2 = a.
Thus, a possible polynomial is y = 2x(x + 2)(x − 2).
39. The graph has x intercepts at x = −3, 0, 2, so the equation is of the form
y = kx(x + 3)(x − 2).
The sign of k must be negative. To find its value, use the point x = 1, y = 4:
4 = k · 1(1 + 3)(1 − 2)
4 = −4k
k = −1.
So the equation is y = −x(x + 3)(x − 2).
40. The shape of the graph suggests an odd degree polynomial with a positive leading term. Since the graph crosses the x-axis
at −2, there is a factor of (x + 2), and since it touches the x-axis at 3, there should be a factor of (x − 3)2 , giving
y = a(x + 2)(x − 3)2 . Since the graph crosses the y-axis at 126, we can find a by substituting x = 0:
126 = a(0 + 2)(0 − 3)2
126 = 18a
7 = a.
Thus, a possible polynomial is y = 7(x + 2)(x − 3)2 .
SOLUTIONS to Review Problems For Chapter Eleven
705
41. We use the position of the “bounce” on the x-axis to indicate a multiple zero at that point. Since there is not a sign change
at those points, the zero occurs an even number of times.
We have
y = k(x + 3)x2 ,
and using the point (−1, 2) gives
2 = k(−1 + 3)(−1)2 = 2k,
so
k = 1.
2
Thus, y = x (x + 3) is a possible formula.
42. We use the position of the “bounce” on the x-axis to indicate a multiple zero at that point. Since there is not a sign change
at those points, the zero occurs an even number of times. Letting
y = k(x + 2)2 (x)(x − 2)2
represent (c), we use the point (1, −3) to get
−3 = f (1) = k(3)2 (1)(−1)2 ,
so
−3 = 9k,
1
k=− .
3
Thus, a possible is
1
y = − (x + 2)2 (x)(x − 2)2 .
3
43. This one is tricky. However, we can view it as a translation of another function. Consider the graph in Figure 11.82. A
formula for the graph in Figure 11.82 could be of the form y = k(x + 3)(x + 2)(x + 1). Since y = 6 if x = 0,
6 = k(0 + 3)(0 + 2)(0 + 1), therefore 6 = 6k, which yields k = 1. Note that the graph of y is a vertical shift (by 4) of
the graph in Figure 11.82, giving y = (x + 3)(x + 2)(x + 1) + 4 as a possible formula for the function.
y
6
x
−3
−2
−1
Figure 11.82
44. We know from the figure that f (−3) = 0, f (0) = 0, f (1) = 2, f (3) = 0. We also know that f does not change sign at
x = −3 or at x = 0. Therefore, a possible formula for f (x) is
f (x) = k(x + 3)2 (x − 3)x2 , k 6= 0.
Using f (1) = 2, we have
1
so −32k = 2. Thus, k = − 16
, and
f (1) = k(1 + 3)2 (1 − 3)(1)2 = −32k,
f (x) = −
is a possible formula for f .
1
(x + 3)2 (x − 3)x2
16
706
Chapter Eleven /SOLUTIONS
45. (a) The graph indicates the graph of y = 1/x2 has been shifted to the right by 2 and down 1. Thus,
y=
1
−1
(x − 2)2
is a possible formula for it.
(b) The equation y = 1/(x − 2)2 − 1 can be written as
y=
−x2 + 4x − 3
x2 − 4x + 4
by obtaining a common denominator and combining terms.
(c) The graph has x-intercepts when y = 0 so the numerator of y = (−x2 + 4x − 3)/(x2 − 4x + 4) must equal zero.
Then
−x2 + 4x − 3 = 0
−(x2 − 4x + 3) = 0
−(x − 3)(x − 1) = 0,
so either x = 3 or x = 1. The x-intercepts are (1, 0) and (3, 0). Setting x = 0, we find y = − 43 , so (0, − 43 ) is the
y-intercept.
46. (a) The graph shows the graph of y = 1/x2 shifted up 2 units. Therefore, a formula is
y=
1
+ 2.
x2
y=
2x2 + 1
.
x2
(b) The equation y = (1/x2 ) + 2 can be written as
(c) We see that the graph has no intercepts on either axis. Algebraically this is seen by the fact that x = 0 is not in the
domain of the function, and there are no real solutions to 2x2 + 1 = 0.
47. (a) The graph appears to be y = 1/x2 shifted 3 units to the right and flipped across the x-axis. Thus,
y=−
1
(x − 3)2
is a possible formula for it.
(b) The equation y = −1/(x − 3)2 can be written as
y=
−1
.
x2 − 6x + 9
(c) Since y can not equal zero if y = −1/(x2 − 6x + 9), the graph has no x-intercept. The y-intercept occurs when
−1
1
1
x = 0, so y = (−3)
2 = − 9 . The y-intercept is at (0, − 9 ).
48. (a) If g is a power function, g(x) = kxp . Taking the ratio of g(4) to g(2), we have
k(4)p
g(4)
=
= 2p
g(2)
k(2)p
and
g(4)
96
=
= 4 = 22 .
g(2)
24
Therefore, p = 2. Then, using g(2) = 24 to find k, we have
g(2) = k · 22 = 24
k = 6.
Thus, if g is a power function, its formula is g(x) = 6x2 .
SOLUTIONS to Review Problems For Chapter Eleven
707
(b) If g is linear, then g(x) = b + mx. We use the two points (2, 24) and (4, 96) to solve for m:
m=
Solving for b, we have
96 − 24
72
∆y
=
=
= 36.
∆x
4−2
2
24 = b + 36(2)
b = −48.
Thus, if g is linear, its formula is g(x) = −48 + 36x.
(c) If g is exponential, then g(x) = abx . Taking ratios, we have
g(4)
ab4
= 2 = b2 .
g(2)
ab
We also know that
g(4)
96
=
= 4.
g(2)
24
Putting these two statements together gives us
b2 = 4
so
b = 2.
Having determined the value of b, we now have
g(x) = a(2x ).
To solve for a, we use the fact that g(2) = 24
g(2) = a(2)2 = 24
a = 6.
x
This gives us the exponential formula g(x) = 6(2 ).
49. (a) In factored form, f (x) = x2 +5x+6 = (x+2)(x+3). Thus, f has zeros at x = −2 and x = −3. For g(x) = x2 +1
there are no real zeros.
(x)
(b) If r(x) = fg(x)
, the zeros of r are where the numerator is zero (assuming g is not also zero at those points). Thus, r
has zeros x = −2 and x = −3. There is no vertical asymptote since g(x) is positive for all x. As x → ±∞, r will
2
behave like y = xx2 = 1. Thus, r(x) → 1 as x → ±∞. The graph of y = r(x) is shown in Figure 11.83.
6
4
2
x
−10
−5
5
10
Figure 11.83: The rational function r(x) =
x2 + 5x + 6
x2 + 1
(c) In fact s does not have a zero near the origin—it does not have a zero anywhere. If s(x) = 0 then g(x) = 0, which
is never true. The function does have two vertical asymptotes, at the zeros of f , which are x = −2 and x = 3. As
x → ±∞, s(x) → 1.
708
Chapter Eleven /SOLUTIONS
50. Notice that f (x) = (x − 3)2 has one zero (at x = 3), g(x) = (x − 2)(x + 2) has 2 zeros (at x = 2 and x = −2),
h(x) = x + 1 has one zero (at x = −1), and j(x) has no zeros.
x2 − 4
has 2 zeros, no vertical asymptote, and a horizontal asymptote at y = 1.
x2 + 1
2
r(x) = (x − 3) (x + 1) has 2 zeros, no vertical asymptote, and no horizontal asymptote.
h(x)/f (x) fits this description, but is not among the functions above.
x+1
(x − 3)2
and q(x) = 2
each have 1 zero, 2 vertical asymptotes, and a horizontal asymptote. p has a
p(x) = 2
x −4
x −4
horizontal asymptote at y = 1, and q has a horizontal asymptote at y = 0.
x2 + 1
has no zeros, 1 vertical asymptote, and a horizontal asymptote at y = 1.
v(x) =
(x − 3)2
1
1
t(x) =
=
has no zeros, 1 vertical asymptote, and a horizontal asymptote at y = 0.
h(x)
x+1
(a) s(x) =
(b)
(c)
(d)
(e)
(f)
51. (a)
(b)
(c)
(d)
False. For example, f (x) = x2 + x is not even.
False. For example, f (x) = x2 is not invertible.
True.
False. For example, if f (x) = x2 , then
f (x) → ∞
x→∞
as
f (x) → ∞
as
x → −∞.
52. (a) On the standard viewing screen, the graph of f is shown in Figure 11.84:
10
−10
x
−5
10
−10
4
Figure 11.84: f (x) = x − 17x2 + 36x − 20
(b) No. The graph of f is very steep near x = −5, but that does not mean it has a vertical asymptote. Since f is a
polynomial function, it is defined for all values of x.
(c) The function has 3 zeros. A good screen to see the zeros is −6 ≤ x ≤ 3, −3 ≤ y ≤ 3.
(d) Using a graphing calculator or a computer, we find that f has zeros at x = −5, x = 1, and a double zero at x = 2.
Thus, f (x) = (x + 5)(x − 1)(x − 2)2 .
(e) The function has 3 turning points, two of which are visible in the standard viewing window (see Figure 11.84), and
one of which is in the third quadrant, but off the bottom of the screen. It is not possible to see all the turning points in
the same window. To see the left-most turning point, a good window is −6 ≤ x ≤ 6, −210 ≤ y ≤ 50, but the other
turning points are invisible on this scale. To see the other turning points, a good window is 0 ≤ x ≤ 3, −1 ≤ y ≤ 2,
but the left-most turning point is far too low to see on this window.
p(x)
(x + 3)(x − 2)
53.
f (x) =
=
q(x)
(x + 5)(x − 7)
54. A denominator of (x + 1) will give the vertical asymptote at x = −1. The numerator will have a highest-powered term
of 1 · x1 to give a horizontal asymptote of y = 1. If there is a zero at x = −3, try
f (x) =
Note, this agrees with the y-intercept at y = 3.
(x + 3)
.
(x + 1)
SOLUTIONS to Review Problems For Chapter Eleven
709
55. The function f has a vertical asymptote at x = 1 and a zero at x = −1. A possible formula for f (x) is
f (x) =
x+1
.
x−1
56. The function g has a double zero at x = −1 and a vertical asymptote at x = 1. A possible formula for g is
g(x) =
(x + 1)2
.
(x − 1)2
57. We have f (x) = k(x + 3)(x − 2)(x − 5). Solving for k,
k(0 + 3)(0 − 2)(0 − 5) = −6
30k = −6
1
k=− .
5
58.
Thus, f (x) = (−1/5)(x + 3)(x − 2)(x − 5).
p(x) = k(x + 3)(x − 2)(x − 5)(x − 6)2
7 = p(0) = k(3)(−2)(−5)(−6)2
= k(1080)
7
k=
1080
7
(x + 3)(x − 2)(x − 5)(x − 6)2
p(x) =
1080
59. Let h(x) = j(x) + 7 where j(x) has zeros at x = −5, −1, 4. Since j has zeros at these x-values, h has “sevens”
there—that is, the value of h equals 7 at these x-values. A formula for j(x) = k(x + 5)(x + 1)(x − 4), so h(x) =
k(x + 5)(x + 1)(x − 4) + 7. Solving for k, we have
h(0) = 3
k(5)(1)(−4) + 7 = 3
−20k = −4
1
k= .
5
Thus, h(x) = (1/5)(x + 5)(x + 1)(x − 4) + 7.
60. Using the hint, we have w(x) = p(x) + v(x). If w(x) equals v(x) at x = −4, 1, 3, then p(x) = 0 at those x-values, so
p(x) = k(x + 4)(x − 1)(x − 3). We have
w(x) = k(x + 4)(x − 1)(x − 3) + 2x + 5
|
{z
p(x)
}
w(0) = k(4)(−1)(−3) + 2(0) + 5
| {z }
v(x)
2 = 12k + 5
1
k=− .
4
Thus, p(x) = (−1/4)(x + 4)(x − 1)(x − 3) and w(x) = (−1/4)(x + 4)(x − 1)(x − 3) + 2x + 5.
61. The map distance is directly proportional to the actual distance (mileage), because as the actual distance, x, increases, the
map distance, d, also increases.
Substituting the values given into the general formula d = kx, we have 0.5 = k(5), so k = 0.1, and the formula is
d = 0.1x.
When d = 3.25, we have 3.25 = 0.1(x) so x = 32.5. Therefore, towns which are separated by 3.25 inches on the
map are 32.5 miles apart.
710
Chapter Eleven /SOLUTIONS
62. Since the frequency, f , is inversely proportional to the length, L, we have
f =k
1
,
L
so
1
L=k .
f
Thus, the length of the string is inversely proportional to the frequency.
63. (a) We are given that if d is the radius of the earth,
k
,
d2
180 =
so
k
.
180
On a planet whose radius is three times the radius of the earth, the person’s weight is
d2 =
k
k
= 2.
(3d)2
9d
w=
Therefore,
w=
k
k
9( 180
)
=
180
= 20 lbs.
9
If the radius is one-third of the earth’s,
k
w=
=
( 31 d)2
k
1 2
d
9
=
9k
9k
= k
d2
180
= 9(180) = 1620 lbs.
(b) Let x be the fraction of the earth’s radius, d. Since 1 ton = 2000 lb, we have
2000 =
k
k
= 2 2.
(xd)2
x d
We can use the information from part (a) to substitute
d2 =
so that
2000 =
Thus
x2 =
k
180
k
180
= 2 .
k
x
x2 ( 180
)
180
= 0.09
2000
and
x = 0.3
The radius is
3
10
(discarding negative x).
of the earth’s radius.
64. (a) Kepler’s Law states that the square of the period, P , is proportional to the cube of the distance, d. Thus, we have
P 2 = kd3 .
Solving for P gives
P =
√
kd3 =
√
kd3/2 = k1 d3/2 .
For the earth, P = 365 and d = 93,000,000. Thus,
365 = k1 (93,000,000)3/2 ,
SOLUTIONS to Review Problems For Chapter Eleven
so
k1 =
365
= 4.1 · 10−10 .
(93,000,000)3/2
This gives
P =
365
d3/2
· d3/2 = 365
= 365
3/2
(93,000,000)
(93,000,000)3/2
or
(b) For Jupiter, d = 483,000,000, so we have
711
d
93,000,000
3/2
,
P = 4.1 · 10−10 d3/2 .
P = 365
483,000,000
93,000,000
3/2
which gives
P ≈ 4320 earth days,
or almost 12 earth years.
65. (a) The graph of the function on the suggested window is shown in Figure 11.85. At x = 0 (when Smallsville was
founded), the population was 5 hundred people.
(b) The x-intercept for x > 0 will show when the population was zero. This occurs at x ≈ 8.44. Thus, Smallsville
became a ghost town in June of 1908.
(c) There are two peaks on the graph on 0 ≤ x ≤ 10, but the first occurs before x = 5 (i.e.,before 1905). The second
peak occurs at x ≈ 7.18. The population at that point is ≈ 7.9 hundred. So the maximum population was ≈ 790 in
February of 1907.
Population
12
10
8
6
4
2
0
−2
1 2 3 4 5 6 7 8 9 10
Years since founded
Figure 11.85
66. (a) The graph of C(x) = (x − 1)3 + 1 is the graph of y = x3 shifted right one unit and up one unit. The graph is shown
in Figure 11.86.
(b) The price is $1000 per unit, since R(1) = 1 means selling 1000 units yields $1,000,000.
(c)
Profit = R(x) − C(x)
= x − [(x − 1)3 + 1]
= x − (x − 1)3 − 1.
712
Chapter Eleven /SOLUTIONS
y
y
C(x)
0
2
x
1
2
1
x
1
2
Figure 11.86
Figure 11.87
The graph of R(x) − C(x) is shown in Figure 11.87. Profit is negative for x < 1 and for x > 2. Profit = 0 at
x = 1 and x = 2. Thus, the firm will break even with either 1000 or 2000 units, make a profit for 1000 < x < 2000
units, and lose money for any number of units between 0 and 1000 or greater than 2000.
67. (a) The length x of a fish proportional to L implies that x = aL, where a is a positive constant. The weight y proportional
to its volume, and therefore to L3 , implies y = bL3 , where b is a positive constant. Eliminating L between these two
equations gives y = (b/a3 )x3 = kx3 where k = b/a3 .
(b) The ratio x3 /y should be approximately constant. The successive values of x3 /y are 113.2, 113.1, 114.4, 116.1,
115.9, 114.1, 114.6, 115.7, 114.7, 113.9, and 113.7, with an average of 114.5. Since y = kx3 , the ratio x3 /y =
1/k = 114.5. Thus, k ≈ 1/114.5 = 0.0087.
(c) Figure 11.88 shows the function y = 0.0087x3 and the data in Table 11.47. This function is a reasonable model.
y
700
600
500
400
300
200
100
x
10
20
30
40
50
Figure 11.88: The weight of plaice as a
function of its length and the function
y = 0.0087x3
68. (a) We have T = kR2 D4 .
(b) If R doubles, the thrust increases by a factor of 4. To see why, suppose
T1 = kR12 D4
and
T2 = kR22 D4 .
Now suppose R2 = 2R1 . Then
T2 = k(2R1 )2 D4 = k22 R12 D4 = 4kR12 D4 .
So
Thus, the thrust has been multiplied by 4.
T2
4kR12 D4
= 4.
=
T1
kR12 D4
SOLUTIONS to Review Problems For Chapter Eleven
713
(c) If D doubles, the thrust increases by a factor of 24 = 16. To see why, suppose
T1 = kR2 D14
and let D2 = 2D1 . Then
and
T2 = kR2 D24
T2 = kR2 (2D1 )4 = kR2 24 D14 = 16kR2 D14 ,
so
16kR2 D14
T2
=
= 16.
T1
kR2 D14
Thus, the thrust has been multiplied by 16.
(d) Suppose R1 and D1 are the original values of the speed and diameter, and R2 and D2 are the new values. We are
told that D2 = 1.5D1 , and we want to find the relation between R1 and R2 . Since T remains constant,
T = kR12 D14 = kR22 D24 .
Substituting D2 = 1.5D1
kR12 D14 = kR22 (1.5D1 )4
kR12 D14 = kR22 (1.5)4 D14
Canceling k and D14 gives
R12 = (1.5)4 R22 .
We are interested in the ratio R2 /R1 , so we solve::
1
R22
=
R12
(1.5)4
R2 2
1
=
R1
(1.5)4
R2
1
= 0.444 = 44.4%.
=
R1
(1.5)2
Thus, the speed should be reduced to 44.4% of its previous value.
69. (a)
13
14
15
12
+
+
+
≈ 2.71666 . . . .
2
6
24
120
This is accurate to 2 decimal places, since e ≈ 2.718.
(b) p(5) ≈ 91.417. This is not at all close to e5 ≈ 148.4.
(c) See Figure 11.89. The two graphs are difficult to tell apart for −2 ≤ x ≤ 2, but for x much less than −2 or much
greater than 2, the fit gets worse and worse.
p(1) = 1 + 1 +
y
f (x)
54
p(x)
−4
f (x)
x
4
6
p(x)
Figure 11.89
714
Chapter Eleven /SOLUTIONS
70. (a) See Figure 11.90.
r(%)
Exponential
16
r = 15.597e−0.323v
12
Power
r = 10.53v−0.641
8
?
4
1
2
3
4
5
v(MeV)
Figure 11.90
(b) The value of r tends to go down as v increases. This means that the telescope is better able to distinguish between
high-energy gamma ray photons than low-energy ones.
(c) The first curve is a power function given by r = 10.53v −0.641 . The second is an exponential function given by
r = 15.597e−0.323v . The power function appears to give a better fit.
(d) The power function predicts that r → ∞ as v → 0, and so is most consistent with the prediction that the telescope
gets rapidly worse and worse at low energies. In contrast, exponential function predicts that r gets close to 15.6% as
E gets close to 0.
CHECK YOUR UNDERSTANDING
1. False. The quadratic function y = 3x2 + 5 is not of the form y = kxn , so it is not a power function.
2. False. This is an exponential function.
3. True. Evaluate g(−1) = (−1)p . Since p is even, g(−1) = 1, so the point (−1, 1) is on the graph.
4. True. Evaluate g(−x) = (−x)p . Since p is even, g(−x) = g(x), so the graph of g is symmetric about the y-axis.
5. True. All positive even power functions have an upward opening U shape.
6. False. Since f (0) = 1/0, we see that f (x) is undefined at x = 0.
7. False. The y-axis is also an asymptote.
8. False. The function values of f (x) = x−1 approach +∞ as x approaches zero from the right side, but the function values
approach −∞ as x approaches zero from the left side.
9. True. The x-axis is an asymptote for f (x) = x−1 , so the values approach zero.
10. True. In the long run, an increasing exponential function grows faster than any power function.
11. True. The function g(x) = ln x grows slower than h(x) = xa , for every a greater than 0.
12. False. We have 2x < x2 at x = 3.
13. False. As x grows very large the exponential decay function g approaches the x-axis faster than any power function with
a negative power.
14. False. This is an exponential function.
15. True. It is of the form y = kxn , where k = 3 and n = 1.
16. True. A quadratic function is a polynomial function of degree 2.
17. False. For example, the polynomial x2 + x3 has degree 3 because the degree is the highest power, not the first power, in
the formula for the polynomial.
18. True. In the long run the highest-power term eventually dominates.
19. False. A zero of a polynomial is any x value for which p(x) = 0.
CHECK YOUR UNDERSTANDING
715
20. True. The graph of a polynomial p intersects the x-axis where p(x) = 0, and these values of x are the zeros of p.
21. True. The graph crosses the y-axis at the point (0, p(0)).
22. True. Since g is a polynomial of degree 4 with positive leading coefficient, it will eventually dominate any polynomial of
degree 3.
23. False. Any polynomial of positive even degree has a graph which in the long run looks like a U, so it does not pass the
horizontal line test.
24. False. Some odd degree polynomials have an inverse, but not all of them do. As a counterexample, consider p(x) =
x(x − 1)(x + 1), a polynomial of degree 3 that has no inverse because p(x) = 0 has more than one solution.
25. True. We can write p(x) = (x − a) · C(x). Evaluating at x = a we get p(a) = (a − a) · C(a) = 0 · C(a) = 0.
26. True. If a polynomial p(x) has n + 1 zeros, r1 , r2 , . . . , rn+1 , then the product (x − r1 )(x − r2 ) · · · (x − rn+1 ) is a factor
of p(x) of degree n + 1, so p(x) must have degree n + 1 or higher. This is impossible if p(x) has degree n.
27. True. There is a multiple zero at x = −2 because the graph is tangent to the x-axis at x = −2.
28. False. There is a single zero at x = 0 because the graph crosses the x-axis at x = 0. The fact that the graph does not
appear to be tangent to the x-axis at x = 0 leads us to conclude that there is not a multiple zero at x = 0.
29. True. This is the definition of a rational function.
30. True. This is the quotient of two polynomials, f (x) = 1 and g(x) = x.
31. True. The highest-power term in a polynomial determines its long-run behavior, so by considering the ratio of the highestpower terms in the numerator and denominator, we are determining the long-run behavior of the rational function.
32. False. The ratio of the highest degree terms in the numerator and denominator is x/x = 1 so for large positive x-values,
y approaches 1.
33. True. The ratio of the highest degree terms in the numerator and denominator is 2x/x2 = 2/x, so for large positive
x-values, y approaches 0.
34. False. The ratio of the highest degree terms in the numerator and denominator is x3 /4x3 = 1/4 so the asymptote is
y = 1/4.
35. False. The ratio of the highest degree terms in the numerator and denominator is −4x2 /x2 = −4. So for large positive
x-values, y approaches −4.
36. False. The ratio of the highest terms in the numerator and denominator is 5x/x = 5. So for large positive x-values, y
approaches 5.
37. False. The ratio of the highest degree terms in the numerator and denominator is 3x4 /x2 = 3x2 . So for large positive
x-values, y behaves like y = 3x2 .
38. True. The ratio of the highest terms in the numerator and denominator is x3 /(−x2 ) = −x. Since the x-values are negative,
y approaches positive infinity.
39. True. A zero numerator forces the fraction value to be zero when the fraction is defined. It is undefined when the denominator is zero.
40. True. This is the definition of the zero of a function.
41. True. At x = −4, we have f (−4) = (−4 + 4)/(−4 − 3) = 0/(−7) = 0, so x = −4 is a zero.
42. False. At x = −2, we have f (−2) = (−2 + 2)/((−2)2 − 4) = 0/0, so the function is not defined at x = −2, and
x = −2 is not a zero of the function.
43. False. The numerator is never zero so g has no zeros.
44. False. The rational function r has a zero, not an asymptote, at each of the zeros of p(x).
45. False. If p(x) has no zeros, then r(x) has no zeros. For example, if p(x) is a nonzero constant or p(x) = x2 + 1, then
r(x) has no zeros.
46. False. If the asymptote is horizontal it is possible. For example, the graph of f (x) = (x − 1)(x + 4)/((x + 2)(x − 2))
crosses the horizontal asymptote y = 1 where x = 0.
47. False. It has a zero at w = 1, since g(1) = 0/(−55) = 0.
716
Chapter Eleven /SOLUTIONS
Solutions to Skills for Chapter 11
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
11.
12.
13.
14.
15.
16.
17.
18.
3
4
3·7+4·5
21 + 20
41
+ =
=
=
5
7
35
35
35
7
2
7 · 15 − 2 · 10
105 − 20
85
17
−
=
=
=
=
10
15
150
150
150
30
1 · 3 − 2(2x)
2
3 − 4x
1
− =
=
2x
3
6x
6x
6y + 63y
69y
69
6
9
6 · y + 9 · 7y
=
=
=
+ =
7y
y
7y 2
7y 2
7y 2
7y
−2(1 − 2y)
−2
4
−2z + 4yz
−2 + 4y
+ =
=
=
yz
z
yz 2
yz
yz
−2(z − 2)
4
−2z + 4
−2z
+ =
=
y
y
y
y
2
3
2 − 3x
− =
x2
x
x2
3
20
60
15
3
4
=
=
= ·
7
4
7
28
7
20
5
6
15
3
x
x2
6
3
x
6
x2
5 1
1
·
=
6 15
18
3 6
18
= · 2 = 3
x x
x
=
=
3 x2
x
·
=
x 6
2
14
13
14
13
2 · 14 + 13
41
+
=
+
=
=
x−1
2x − 2
x−1
2(x − 1)
2(x − 1)
2(x − 1)
xy(4zy 3 − 3wx)
4z
3w
4zxy 4 − 3wx2 y
4y3z − 3wx
−
=
=
=
x2 y
xy 4
x3 y 5
x3 y 5
x2 y 4
10
3
10
3
7
+
=
−
=
y−2
2−y
y−2
y−2
y−2
8(y + 4)
32
8y + 32
8y
+
=
=
y−4
y−4
y−4
y−4
8y
32
8y
32
8y − 32
8(y − 4)
+
=
−
=
=
=8
y−4
4−y
y−4
y−4
y−4
y−4
9
8
9
8
−
=
−
3x2 − x − 4
x+1
(x + 1)(3x − 4)
x+1
8 − 9(3x − 4)
=
(x + 1)(3x − 4)
−27x + 44
=
(x + 1)(3x − 4)
15
7
15(x + 5) + 7(x − 3)
+
=
(x − 3)2 (x + 5)
(x − 3)(x + 5)2
(x − 3)2 (x + 5)2
15x + 75 + 7x − 21
=
(x − 3)2 (x + 5)2
22x + 54
=
(x − 3)2 (x + 5)2
2(11x + 27)
=
(x − 3)2 (x + 5)2
SOLUTIONS TO SKILLS FOR CHAPTER 11
717
19. The common denominator is (x − 4)(x + 4) = x2 − 16. Therefore,
2
3(x + 4)
2(x − 4)
3
−
=
−
x−4
x+4
(x − 4)(x + 4)
(x + 4)(x − 4)
3(x + 4) − 2(x − 4)
3x + 12 − 2x + 8
=
x2 − 16
x2 − 16
x + 20
.
= 2
x − 16
20. If we rewrite the second fraction −
1
1
as
, the common denominator becomes x − 1. Therefore,
1−x
x−1
x2
1
x2
1
x2 + 1
−
=
+
=
.
x−1
1−x
x−1
x−1
x−1
21. The second denominator 4r 2 +6r = 2r(2r+3), while the first denominator is 2r+3. Therefore the common denominator
is 2r(2r + 3). We have:
3
1
3
1
+ 2
=
+
2r + 3
4r + 6r
2r + 3
2r(2r + 3)
3
1 · 2r
+
=
2r(2r + 3)
2r(2r + 3)
2r + 3
1
=
=
.
2r(2r + 3)
2r
22. The common denominator is u + a. Therefore,
u+a+
(u + a)(u + a)
(u + a)2 + u
u
u
=
+
=
.
u+a
u+a
u+a
u+a
√
23. The common denominator is ( x)3 .
√
( x)2
1
x−1
1
1
√ − √ 3 = √ 3 − √ 3 = √ 3
x
( x)
( x)
( x)
( x)
It is fine to leave the answer in the form
(
x−1
√
,
x)3
or we can rationalize the denominator:
√
√
√
x(x − 1)
x−1
x−1
x x− x
.
√ 3 = √ =
√ √
=
x2
( x)
x x
x x x
24. The common denominator is e2x . Thus,
1
1
1
ex
1 + ex
+ x = 2x + 2x =
.
2x
e
e
e
e
e2x
25. If we factor the number and denominator of the second fraction, we can cancel some terms with the first,
2(4x + 1)
a + b 8x + 2
a+b
4x + 1
· 2
=
·
=
.
2
b − a2
2
(b + a)(b − a)
b−a
26. The common denominator is 4M . Therefore,
0.07
3
(0.07)(4)
(3M 2 )M
.28 + 3M 3
+ M2 =
+
=
.
M
4
4M
4M
4M
718
Chapter Eleven /SOLUTIONS
27. Each of the denominators are different and therefore the common denominator is r1 r2 r3 . Accordingly,
1
1
r2 r3 + r1 r3 + r1 r2
1
+
+
=
.
r1
r2
r3
r1 r2 r3
8(y − 4)
32
8y − 32
8y
−
=
=
=8
y−4
y−4
y−4
y−4
2a + 3
29.
(a + 3)(a − 3)
30. We change this division example to a multiplication problem by writing the reciprocal of the second faction. Therefore,
28.
x3
x−4
x2
x(x − 4)(x + 2)
x3
x2 − 2x − 8
x2
=
·
=
= x(x + 2).
− 2x − 8
x−4
x2
x−4
31. First we find a common denominator for the two fractions in the numerator. Thus,
1
(x+h)2
h
−
1
x2
=
x2 − (x + h)2 1
·
x2 (x + h)2
h
h(−2x − h) 1
x2 − x2 − 2xh − h2 1
· = 2
·
x2 (x + h)2
h
x (x + h)2 h
−2x − h
.
= 2
x (x + h)2
=
32. Recall that the terms a−2 and b−2 can be written as
1
1
and 2 respectively. Therefore,
a2
b
2
2
1
b +a
+ b12
b2 + a 2
1
1
a−2 + b−2
a2
a2 b2
=
=
=
· 2
= 2 2.
a 2 + b2
a 2 + b2
a 2 + b2
a 2 b2
a + b2
a b
33. We expand within the first brackets first. Therefore,
[4 − (x2 + 2xh + h2 )] − [4 − x2 ]
[4 − (x + h)2 ] − [4 − x2 ]
=
h
h
[4 − x2 − 2xh − h2 ] − 4 + x2
−2xh − h2
=
=
h
h
= −2x − h.
34.
b−1 (b − b−1 )
=
b+1
35.
1
b
b − 1b
b+1
=
1 (b2 − 1)
(b + 1)(b − 1)
b−1
=
=
.
b2 b + 1
b2 (b + 1)
b2
1 − a12
(a − 1)(a + 1)
a2 − 1
a
a−1
1
1 − a−2
=
=
·
=
=
=1− .
1
−1
1+a
a2
a+1
a(a + 1)
a
a
1+ a
36. We simplify the second complex fraction first. Thus,
p−
p
q
q
+
q
p
= p−
=
q
p2 +q 2
qp
= p−q·
qp
p2 + q 2
p(p2 + q 2 ) − q 2 p
p3
= 2
.
2
2
p +q
p + q2
SOLUTIONS TO SKILLS FOR CHAPTER 11
719
5
3
3x − 5
− 2
3x − 5
x4 y 2
x2 y
xy
x y
x2 y
37.
=
=
·
=
2
(3x − 5)(2x + 1)
x2 y
(3x − 5)(2x + 1)
2x + 1
6x − 7x − 5
y4 y2
x4 y 2
38. Cancellation is employed here to simplify. Therefore,
3x2 − (ln x)(6x)
1
x
(3x2 )2
=
3x − (ln x)(6x)
9x4
=
1 − (ln x)(2)
1 − 2 ln x
=
.
3x3
3x3
39. We cancel the common factor x3 + 1 in both numerator and denominator. Therefore,
2x(x3 + 1) − x2 (2)(3x2 )
2x(x3 + 1)2 − x2 (2)(x3 + 1)(3x2 )
=
3
2
2
[(x + 1) ]
(x3 + 1)3
=
2x4 + 2x − 6x4
2x − 4x4
=
.
(x3 + 1)3
(x3 + 1)3
40. Write
1
(2x
2
− 1)−1/2 (2) − (2x − 1)1/2 (2x)
=
(x2 )2
1
(2x−1)1/2
−
2x(2x−1)1/2
1
(x2 )2
.
Next a common denominator for the top two fractions is (2x − 1)1/2 . Therefore we obtain,
1
(2x−1)1/2
−
2x(2x−1)
(2x−1)1/2
x4
=
1 − 4x2 + 2x 1
−4x2 + 2x + 1
√
· 4 =
.
1/2
x
(2x − 1)
x4 2x − 1
41. Dividing 2x3 into each term in the numerator yields:
26x
1
13
1
26x + 1
=
+ 3 = 2 + 3.
2x3
2x3
2x
x
2x
√
42. Dividing 3 x into both terms in the numerator yields:
√
√
3
1
1
x+3
x
√
= √ + √ = +√ .
3
3 x
3 x
3 x
x
43.
6l2 + 3l − 4
6l2
3l
4
2
1
4
= 4 + 4 − 4 = 2 + 3 − 4
4
3l
3l
3l
3l
l
l
3l
44. The denominator p2 + 11 is divided into each of the two terms of the numerator. Thus,
7+p
7
p
= 2
+ 2
.
p2 + 11
p + 11
p + 11
45.
1
x
3
−
2x
1
2
=
x
3
2x
−
1
2
2x
=
x 1
1 1
1
1
·
− ·
= −
3 2x
2 2x
6
4x
46. In this example, dividing the denominator into each term of the numerator involves the same base t. Therefore we subtract
exponents.
t−1/2
t1/2
1
t−1/2 + t1/2
1
=
+ 2 = t−1/2−2 + t1/2−2 = t−5/2 + t−3/2 = 5/2 + 3/2
2
2
t
t
t
t
t
720
Chapter Eleven /SOLUTIONS
47. We write the numerator x − 2 as x + 5 − 7. Therefore,
x−2
(x + 5) − 7
7
=
=1−
.
x+5
x+5
x+5
48. The numerator q − 1 = q − 4 + 3. Thus,
(q − 4) + 3
3
q−1
=
=1+
.
q−4
q−4
q−4
49. Dividing the denominator R into each term in the numerator yields,
R+1
R
1
1
=
+
=1+ .
R
R
R
R
50. Rewrite 3 + 2u = 2u + 3 = (2u + 1) + 2. Thus,
(2u + 1) + 2
3 + 2u
2u + 3
2
=
=
=1+
.
2u + 1
2u + 1
2u + 1
2u + 1
51. Dividing by cos x yields:
52.
cos x
sin x
sin x
cos x + sin x
=
+
=1+
.
cos x
cos x
cos x
cos x
1
ex
1
1
1 + ex
= x + x = x +1 =1+ x
ex
e
e
e
e
53. False
54. True
55. False
56. False
57. True
58. True, factor x−1/3 on the left and rewrite it with positive exponent in the denominator.