Exercises
1. Yes, becausex − 2 is not zero on this interval.
2. No, becausex − 2 = 0 at x = 2.
3. Yes, because2x − 5 is positive for 3 ≤ x ≤ 4.
4. Yes, because the denominator is never zero.
5. Yes, because2x + x2/3is defined for allx.
6. No, because2x + x−1is undefined atx = 0.
7. No, becausecos(π/2) = 0.
8. No, becausesin 0 = 0.
9. No, becauseex− 1 = 0 at x = 0.
10. Yes, becausecos θ is not zero on this interval.
11. We have thatf (0) = −1 < 0 and f(1) = 1 > 0 and that f is continuous. Thus, by the Intermediate Value Theorem applied tok = 0, there is a number c in [0, 1] such that f (c) = k = 0.
12. We have thatf (0) = 1 > 0 and f (1) = e − 3 < 0 and that f is continuous. Thus, by the Intermediate Value Theorem applied tok = 0, there is a number c in [0, 1] such that f (c) = k = 0.
13. We have thatf (0) = −1 < 0 and f(1) = 1 − cos 1 > 0 and that f is continuous. Thus, by the Intermediate Value Theorem applied tok = 0, there is a number c in [0, 1] such that f (c) = k = 0.
14. Sincef is not continuous at x = 0, we consider instead the smaller interval [0.01, 1]. We have that f (0.01) = 20.01− 100 < 0 and f (1) = 2 − 1/1 = 1 > 0 and that f is continuous. Thus, by the Intermediate Value Theorem applied to k = 0, there is a number c in [0.01, 1], and hence in [0, 1], such that f (c) = k = 0.
15. (a) Atx = 1, on the line y = x, we have y = 1. At x = 1, on the parabola y = x2, we havey = 1. Thus, f (x) is continuous. See Figure 1.68.
(b) Atx = 3, on the line y = x, we have y = 3. At x = 3, on the parabola y = x2, we havey = 9. Thus, g(x) is not continuous. See Figure 1.69.
1 f (x)
x y
Figure 1.68
3 g(x)
x y
Figure 1.69
Problems
16. (a) Even if the car stops to refuel, the amount of fuel in the tank changes smoothly, so the fuel in the tank is a continuous function; the quantity of fuel cannot suddenly change from one value to another.
(b) Whenever a student joins or leaves the class the number jumps up or down immediately by 1 so this is not a continuous function, unless the enrollment does not change at all.
(c) Whenever the oldest person dies the value of the function jumps down to the age of the next oldest person, so this is not a continuous function.
17. Two possible graphs are shown in Figures 1.70 and 1.71.
time velocity
Figure 1.70: Velocity of the car
time distance
Figure 1.71: Distance
The distance moved by the car is continuous. (Figure 1.71 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 1.70, which is not continuous on any interval containing the moment of impact.
18. The voltagef (t) is graphed in Figure 1.72.
7 6
12
time (seconds)
Figure 1.72: Voltage change from 6V to 12V
Using formulas, the voltage,f (t), is represented by
f (t) = (
6, 0 < t ≤ 7 12, 7 < t
Although a real physical voltage is continuous, the voltage in this circuit is well-approximated by the functionf (t), which is not continuous on any interval around7 seconds.
19. The value ofy on the line y = kx at x = 3 is y = 3k. To make f (x) continuous, we need 3k = 5 so k = 5
3. See Figure 1.73.
3 5
■
y = (5/3)x
✻
y = 5 f (x)
x y
Figure 1.73
20. For any value ofk, the function is continuous at every point except x = 2. We choose k to make the function continuous atx = 2.
Since3x2takes the value3(22) = 12 at x = 2, we choose k so that the graph of kx goes through the point (2, 12).
Thusk = 6.
21. If the graphs ofy = t + k and y = kt meet at t = 5, we have 5 + k = 5k
k = 5/4.
See Figure 1.74.
5 10
5 4
y = (5/4) + t y = (5/4)t g(t)
t
Figure 1.74
22. Atx = π, the curve y = k cos x has y = k cos π = −k. At x = π, the line y = 12 − x has y = 12 − π. If h(x) is continuous, we need
−k = 12 − π k = π − 12.
23. (a) See Figure 1.75.
(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 and k = 3/2.
(c) See Figure 1.76.
2 4
7 f (x), k = 1
x y
Figure 1.75
2 4
7 f (x), k = 1.5
x y
Figure 1.76
24. For any value ofk, the function is continuous at every point except x = 1. We choose k to make the function continuous atx = 1.
Sincex + 3 takes the value 1 + 3 = 4 at x = 1, we choose k so that the graph of kx goes through the point (1, 4).
Thusk = 4.
25. For any value ofk, the function is continuous at every point except x = 1. We choose k to make the function continuous atx = 1.
Sincekx takes the value k · 1 = k at x = 1, we choose k so that the graph of 2kx + 3 goes through the point (1, k).
This gives
2k · 1 + 3 = k k = −3.
26. For any value ofk, the function is continuous at every point except x = π. We choose k to make the function continuous atx = π.
Sincek sin x takes the value k sin π = 0 at x = π, we cannot choose k so that the graph of x + 4 goes through the point(π, 0). Thus, this function is discontinuous for all values of k.
27. For any value ofk, the function is continuous at every point except x = 2. We choose k to make the function continuous atx = 2.
Sincex + 1 takes the value 2 + 1 = 3 at x = 2, we choose k so that the graph of ekxgoes through the point(2, 3).
This gives
e2k= 3 2k = ln 3
k = ln 3 2
28. For any value ofk, the function is continuous at every point except x = 1. We choose k to make the function continuous atx = 1.
Sincesin(kx) takes the value sin k at x = 1, we choose k so that the graph of 0.5x goes through the point (1, sin k).
This gives
sin k = 0.5
k = sin−10.5 = π 6. Other solutions are possible.
29. For any value ofk, the function is continuous at every point except x = 2. We choose k to make the function continuous atx = 2.
Sinceln(kx + 1) takes the value ln(2k + 1) at x = 2, we choose k so that the graph of x + 4 goes through the point (2, ln(2k + 1)). This gives
ln(2k + 1) = 2 + 4 = 6 2k + 1 = e6
k =e6− 1 2 .
30. (a) The initial value is whent = 0, and we see that P (0) = ek·0= e0= 1000.
(b) Since the function is continuous, att = 12, we have ekt= 100 and we solve for k:
e12k= 100 12k = ln 100
k =ln 100
12 = 0.384.
(c) The population is increasing exponentially for 12 months and then becoming constant.
31. Forx > 0, we have |x| = x, so f(x) = 1. For x < 0, we have |x| = −x, so f(x) = −1. Thus, the function is given by
f (x) =
(1 x > 0
0 x = 0
−1 x < 1 , sof is not continuous on any interval containing x = 0.
32. The graph ofg suggests that g is not continuous on any interval containing θ = 0, since g(0) = 1/2.
33. The drug first increases linearly for half a second, at the end of which time there is0.6 ml in the body. Thus, for 0 ≤ t ≤ 0.5, the function is linear with slope 0.6/0.5 = 1.2:
Q = 1.2t for 0 ≤ t ≤ 0.5.
Att = 0.5, we have Q = 0.6. For t > 0.5, the quantity decays exponentially at a continuous rate of 0.002, so Q has the form
Q = Ae−0.002t 0.5 < t.
We chooseA so that Q = 0.6 when t = 0.5:
0.6 = Ae−0.002(0.5)= Ae−0.001 A = 0.6e0.001.
Thus
Q =
(1.2t 0 ≤ t ≤ 0.5
0.6e0.001e−.002t 0.5 < t.
34.
0.2 0.4 0.6
0.8 1
0.2 0.4 0.6
0.8 1
0.2 0.4 0.6
0.8 1
35. Since polynomials are continuous, and sincep(5) < 0 and p(10) > 0 and p(12) < 0, there are two zeros, one between x = 5 and x = 10, and another between x = 10 and x = 12. Thus, p(x) is a cubic with at least two zeros.
Ifp(x) has only two zeros, one would be a double zero (corresponding to a repeated factor). However, since a polynomial does not change sign at a repeated zero,p(x) cannot have a double zero and have the signs it does.
Thus,p(x) has three zeros. The third zero can be greater than 12 or less than 5. See Figures 1.77 and 1.78.
p(x)
x y
Figure 1.77
p(x) x y
Figure 1.78
36. (a) The graphs ofy = exandy = 4 − x2cross twice in Figure 1.79. This tells us that the equationex= 4 − x2has two solutions.
Sincey = exincreases for allx and y = 4 − x2increases forx < 0 and decreases for x > 0, these are only the two crossing points.
−2 2
y = ex y = x2− 4
x y
Figure 1.79
(b) Values off (x) are in Table 1.5. One solution is between x = −2 and x = −1; the second solution is between x = 1 andx = 2.
Table 1.5
x −4 −3 −2 −1 0 1 2 3 4
f (x) 12.0 5.0 0.1 −2.6 −3 −0.3 7.4 25.1 66.6
37. (a) Figure 1.80 shows a possible graph off (x), yours may be different.
3 5
2 9
x
Figure 1.80
(b) In order forf to approach the horizontal asymptote at 9 from above it is necessary that f eventually become concave up. It is therefore not possible forf to be concave down for all x > 6.
38. (a) Sincef (x) is not continuous at x = 1, it does not satisfy the conditions of the Intermediate Value Theorem.
(b) We see thatf (0) = e0 = 1 and f (2) = 4 + (2 − 1)2 = 5. Since exis increasing betweenx = 0 and x = 1, and since4 + (x − 1)2is increasing betweenx = 1 and x = 2, any value of k between e1 = e and 4 + (1 − 1)2= 4, such ask = 3, is a value such that f (x) = k has no solution.
Strengthen Your Understanding
39. The Intermediate Value theorem only makes this guarantee for a continuous function, not for any function.
40. The Intermediate Value Theorem guarantees that for at least one value ofx between 0 and 2, we have f (x) = 5, but it does not tell us which value(s) ofx give f (x) = 5.
41. We want a function which has a value at every point but where the graph has a break atx = 15. One possibility is
f (x) =
(1 x ≥ 15
−1 x < 15
42. One example isf (x) = 1/x, which is not continuous at x = 0. The Intermediate Value Theorem does not apply on an interval that contains a point where a function is not continuous.
43. Letf (x) = (
1 x ≤ 2
x x > 2. Thenf (x) is continuous at every point in [0, 3] except at x = 2. Other answers are possible.
44. Letf (x) = (
x x ≤ 3
2x x > 3. Thenf (x) is increasing for all x but f (x) is not continuous at x = 3. Other answers are possible.
45. False. For example, letf (x) = (
1 x ≤ 3
2 x > 3 , then f (x) is defined at x = 3 but it is not continuous at x = 3. (Other examples are possible.)
46. False. A counterexample is graphed in Figure 1.81, in whichf (5) < 0.
5 10
100
f (x)
x
Figure 1.81
47. False. A counterexample is graphed in Figure 1.82.
a b
f (b)
f (a) f (x)
x
Figure 1.82