Solutions for Chapter 1 Review
96. Looking at
3.12 3.74≈ 3.74
4.49≈ 4.49 5.39≈ 5.39
6.47≈ 6.47 7.76≈ 0.83.
Thus𝑔 is an exponential function, and so 𝑓 and 𝑘 are the power functions. Each is of the form 𝑎𝑥2or𝑎𝑥3, and since 𝑘(1.0) = 9.01 we see that for 𝑘, the constant coefficient is 9.01. Trial and error gives
𝑘(𝑥) = 9.01𝑥2,
since𝑘(2.2) = 43.61 ≈ 9.01(4.84) = 9.01(2.2)2. Thus𝑓(𝑥) = 𝑎𝑥3and we find𝑎 by noting that 𝑓(9) = 7.29 = 𝑎(93) so 𝑎 =7.29
93 = 0.01 and𝑓(𝑥) = 0.01𝑥3.
97. (a) See Figure 1.155.
(b) The graph is made of straight line segments, rising from the𝑥-axis at the origin to height 𝑎 at 𝑥 = 1, 𝑏 at 𝑥 = 2, and 𝑐 at 𝑥 = 3 and then returning to the 𝑥-axis at 𝑥 = 4. See Figure 1.156.
−1 0 1 2 3 4 5
1 2 3 4
𝑥 𝑦
Figure 1.155
−1 0 1 2 3 4 5
𝑏 𝑐 𝑎
𝑥 𝑦
Figure 1.156
98. (a) Reading the graph of 𝜃 against 𝑡 shows that 𝜃 ≈ 5.2 when 𝑡 = 1.5. Since the coordinates of 𝑃 are 𝑥 = 5 cos 𝜃, 𝑦 = 5 sin 𝜃, when 𝑡 = 1.5 the coordinates are
(𝑥, 𝑦) ≈ (5 cos 5.2, 5 sin 5.2) = (2.3, −4.4).
(b) As𝑡 increases from 0 to 5, the angle 𝜃 increases from 0 to about 6.3 and then decreases to 0 again. Since 6.3 ≈ 2𝜋, this means that𝑃 starts on the 𝑥-axis at the point (5, 0), moves counterclockwise the whole way around the circle (at which time𝜃 ≈ 2𝜋), and then moves back clockwise to its starting point.
99. (a) III (b) IV (c) I (d) II
100. The functions 𝑦(𝑥) = sin 𝑥 and 𝑧𝑘(𝑥) = 𝑘𝑒−𝑥for𝑘 = 1, 2, 4, 6, 8, 10 are shown in Figure 1.157. The values of 𝑓(𝑘) for 𝑘 = 1, 2, 4, 6, 8, 10 are given in Table 1.19. These values can be obtained using either tracing or a numerical root finder on a calculator or computer.
From Figure 1.157 it is clear that the smallest solution of sin𝑥 = 𝑘𝑒−𝑥for𝑘 = 1, 2, 4, 6 occurs on the first period of the sine curve. For small changes in𝑘, there are correspondingly small changes in the intersection point. For 𝑘 = 8 and𝑘 = 10, the solution jumps to the second period because sin 𝑥 < 0 between 𝜋 and 2𝜋, but 𝑘𝑒−𝑥is uniformly positive.
Somewhere in the interval 6≤ 𝑘 ≤ 8, 𝑓(𝑘) has a discontinuity.
2 4
6 8
10 12
−1 0 1 2
𝑥
+ 𝑦(𝑥) = sin 𝑥
𝑧𝑘(𝑥) = 𝑘𝑒−𝑥
Figure 1.157
Table 1.19 𝑘 𝑓(𝑘) 1 0.588 2 0.921 4 1.401 6 1.824 8 6.298 10 6.302 101. For any values of 𝑘, the function is continuous on any interval that does not contain 𝑥 = 2.
Since 5𝑥3− 10𝑥2= 5𝑥2(𝑥 − 2), we can cancel (𝑥 − 2) provided 𝑥 ≠ 2, giving 𝑓(𝑥) = 5𝑥3− 10𝑥2
𝑥 − 2 = 5𝑥2 𝑥 ≠ 2.
Thus, if we pick𝑘 = 5(2)2= 20, the function is continuous.
102. At 𝑥 = 0, the curve 𝑦 = 𝑘 cos 𝑥 has 𝑦 = 𝑘 cos 0 = 𝑘. At 𝑥 = 0, the curve 𝑦 = 𝑒𝑥− 𝑘 has 𝑦 = 𝑒0− 𝑘 = 1 − 𝑘. If 𝑗(𝑥) is continuous, we need
𝑘 = 1 − 𝑘, so 𝑘 = 1 2 . 103. Two possible graphs are shown in Figures 1.158 and 1.159.
time velocity
Figure 1.158: Velocity of the car
time distance
Figure 1.159: Distance
The distance moved by the car is continuous. (Figure 1.159 has no breaks in it.) In actual fact, the velocity of the car is also continuous; however, in this case, it is well-approximated by the function in Figure 1.158, which is not continuous on any interval containing the moment of impact.
104. Figure 1.160 suggests that as 𝑡 → ∞, 3 − 2𝑒−𝑡 → 3. We shall use limit properties to confirm this. We now calculate the limit in stages using the properties to justify each step:
lim𝑡→∞(3 − 2𝑒−𝑡) = lim
𝑡→∞(3) + lim
𝑡→∞(−2𝑒−𝑡) (Property 2)
= lim
𝑡→∞(3) − 2 lim
𝑡→∞(𝑒−𝑡) (Property 1)
= 3 − 2(0) = 3 (Property 6 and because𝑒−𝑡→ 0)
𝑦 = 3 − 2𝑒−𝑡 𝑦 = 3 𝑡 𝑦
Figure 1.160
105. (a) To find lim
𝑥→4+𝑓(𝑥), we observe how 𝑓(𝑥) behaves as 𝑥 approaches 4 from the right. If 𝑥 > 4, then 𝑓(𝑥) = 9 − 𝑥; so as 𝑥 approaches 4 from the right, 𝑓(𝑥) approaches 9 − 4 = 5. So lim
𝑥→4+𝑓(𝑥) = 5.
(b) When𝑥 < 4, we have 𝑓(𝑥) = 2𝑥 − 3; so as 𝑥 approaches 4 from the left, 𝑓(𝑥) approaches 2(4) − 3 = 5. So
𝑥→4lim−𝑓(𝑥) = 5.
(c) From (a) and (b), we know lim
𝑥→4+𝑓(𝑥) = lim
𝑥→4−𝑓(𝑥) = 5, so 𝑓(𝑥) approaches 5 as 𝑥 approaches 4 from either direction.
Therefore, lim
𝑥→4𝑓(𝑥) = 5.
106. (a) See Figure 1.161.
(b) For any value of𝑘, the function is continuous at every point except 𝑥 = 2. We choose 𝑘 to make the function continuous at𝑥 = 2.
Since (𝑥 − 2)2+ 3 takes on the value (2 − 2)2+ 3 = 3 at 𝑥 = 2, we choose 𝑘 so that 𝑘𝑥 = 3 at 𝑥 = 2, so 2𝑘 = 3 and𝑘 = 3∕2.
(c) See Figure 1.162.
2 4
7 𝑓(𝑥), 𝑘 = 1
𝑥 Figure 1.161
2 4
7 𝑓(𝑥), 𝑘 = 1.5
𝑥
Figure 1.162
107. (a) Figure 1.163 shows a possible graph of 𝑔(𝑥), yours may be different.
−2 2
𝑥 = 3
𝑦 = 5
𝑥 𝑦
Figure 1.163
(b) In order for𝑔 to approach the horizontal asymptote at 5 from above it is necessary that 𝑔 eventually become concave up. It is therefore not possible for𝑔 to be concave down for all 𝑥 < −2.
108. This statement is true; since the graph of 𝑔(𝑡) must approach the vertical asymptote at 𝑡 = 5 on the right or the left (or both),𝑔(𝑡) cannot approach a finite real number as 𝑡 approaches 5 from both sides. So lim
𝑡→5𝑔(𝑡) does not exist.
109. This statement is false. Since the graph of 𝑔(𝑡) has horizontal asymptotes at both 𝑦 = −1 and at 𝑦 = 2, the graph must approach one of these lines as𝑡 approaches ∞ and the other as 𝑡 approaches −∞. So lim
𝑡→−∞𝑔(𝑡) must exist, and must be equal to −1 or 2.
110. We cannot determine whether this statement is true or false. We know that the graph of 𝑔(𝑡) has a vertical asymptote at 𝑡 = 5, but we cannot be sure from the information given whether 𝑔(𝑡) approaches ∞ or −∞ as 𝑡 approaches 5. To say that lim𝑡→5𝑔(𝑡) = ∞, we would need to know that 𝑔(𝑡) approaches ∞ as 𝑡 approaches 5 from both sides.
111. We cannot determine whether this statement is true or false. We know that the graph of 𝑔(𝑡) has horizontal asymptotes at𝑦 = −1 and at 𝑦 = 2, and the graph must approach each of these. However, it could be that 𝑔(𝑡) approaches 2 as 𝑡 approaches ∞ and −1 as𝑡 approaches −∞, or it could be the other way around.
112. (a) The values of 𝑓(𝑥) as 𝑥 gets closer to 5 are getting closer to 3.5, which suggests lim
𝑥→5𝑓(𝑥) = 3.5.
(b) 𝑓(𝑥) is continuous at 𝑥 = 5 if the values of 𝑓(𝑥) approach 𝑓(5) as 𝑥 approaches 5. The values approach 3.5, but 𝑓(5) = 8, so the function is not continuous.
113. (a) The values of 𝑓(𝑥) as 𝑥 gets closer to 5 are getting closer to 3.5, which suggests lim
𝑥→5𝑓(𝑥) = 3.5.
(b) 𝑓(𝑥) is continuous at 𝑥 = 5 if the values of 𝑓(𝑥) approach 𝑓(5) as 𝑥 approaches 5. The values approach 3.5, and 𝑓(5) = 3.5, so the function is continuous.
114. We have,
117.
119. We cannot use the limit properties to calculate this limit since lim
𝑥→9
124. We have 125. The limit appears to be 1; a graph and table of values is shown below.
0.02 0.04 0.06 0.08 0.1
126. There are many possible correct answers.
(a) If𝑓(𝑥) = 2𝑥 and 𝑔(𝑥) = −𝑥 then 𝑓(𝑥) + 𝑔(𝑥) = 𝑥 so lim𝑥→∞𝑓(𝑥) + 𝑔(𝑥) = ∞.
(b) Property 3 of the limit properties states
lim𝑥→𝑐(𝑓(𝑥)𝑔(𝑥)) =(
lim𝑥→𝑐𝑓(𝑥)) (
lim𝑥→𝑐𝑔(𝑥))
provided the limits on the right hand side exist. In this case, lim
𝑥→0
1
𝑥 does not exist, so we cannot invoke the property.
128. From Table 1.20, it appears the limit is 1. This is confirmed by Figure 1.164. An appropriate window is −0.0033 < 𝑥 <
0.0033, 0.99 < 𝑦 < 1.01.
129. From Table 1.21, it appears the limit is −1. This is confirmed by Figure 1.165. An appropriate window is −0.099 < 𝑥 <
0.099, −1.01 < 𝑦 < −0.99.
130. From Table 1.22, it appears the limit is 0. This is confirmed by Figure 1.166. An appropriate window is −0.005 < 𝑥 <
131. From Table 1.23, it appears the limit is 0. This is confirmed by Figure 1.167. An appropriate window is −0.0033 < 𝑥 <
0.0033, −0.01 < 𝑦 < 0.01.
132. From Table 1.24, it appears the limit is 2. This is confirmed by Figure 1.168. An appropriate window is −0.0865 < 𝑥 <
0.0865, 1.99 < 𝑦 < 2.01.
133. From Table 1.25, it appears the limit is 3. This is confirmed by Figure 1.169. An appropriate window is −0.047 < 𝑥 <
0.047, 2.99 < 𝑦 < 3.01.
134. From Table 1.26, it appears the limit is 1. This is confirmed by Figure 1.170. An appropriate window is −0.0198 < 𝑥 <
0.0198, 0.99 < 𝑦 < 1.01.
Table 1.26
𝑥 𝑓(𝑥)
0.1 1.0517 0.01 1.0050 0.001 1.0005 0.0001 1.0001
𝑥 𝑓(𝑥)
−0.0001 1.0000
−0.001 0.9995
−0.01 0.9950
−0.1 0.9516
−0.0198 0.0198
0.99 1.01
Figure 1.170
135. From Table 1.27, it appears the limit is 2. This is confirmed by Figure 1.171. An appropriate window is −0.0049 < 𝑥 <
0.0049, 1.99 < 𝑦 < 2.01.
Table 1.27
𝑥 𝑓(𝑥)
0.1 2.2140 0.01 2.0201 0.001 2.0020 0.0001 2.0002
𝑥 𝑓(𝑥)
−0.0001 1.9998
−0.001 1.9980
−0.01 1.9801
−0.1 1.8127
−0.0049 0.0049
1.99 2.01
Figure 1.171