1. from the Graph of G, State the Intervals on Which G Is Continuous

97909_02_ch02_p120-129.qk:97909_02_ch02_p120-129 9/21/10 9:20 AM Page 127

SECTION 2.5 CONTINUITY 127 a root must lie between 1.2 and 1.3. A calculator gives, by trial and error, f ͑1.22͒ ෇ Ϫ0.007008 Ͻ 0 and f ͑1.23͒ ෇ 0.056068 Ͼ 0 so a root lies in the interval ͑1.22, 1.23͒ .

We can use a graphing calculator or computer to illustrate the use of the Intermediate Value Theorem in Example 10. Figure 10 shows the graph of f in the viewing rectangle ͓Ϫ1, 3͔ by ͓Ϫ3, 3͔ and you can see that the graph crosses the x -axis between 1 and 2. Fig- ure11 shows the result of zooming in to the viewing rectangle ͓1.2, 1.3͔ by ͓Ϫ0.2, 0.2͔ .

3 0.2

_1 3 1.2 1.3

_3 _0.2 FIGURE 10 FIGURE 11

In fact, the Intermediate Value Theorem plays a role in the very way these graphing de- vices work. A computer calculates a ﬁnite number of points on the graph and turns on the pixels that contain these calculated points. It assumes that the function is continuous and takes on all the intermediate values between two consecutive points. The computer therefore connects the pixels by turning on the intermediate pixels.

2.5 Exercises In Class problem sheet number: 4; Wednesday,January 18 2017;

1. Write an equation that expresses the fact that a function f 4.1.From From the the graph graph of t of , stateg, state the theintervals intervals on which on which t is g is continuous. is continuous at the number 4. continuous. y 2. If f is continuous on ͑Ϫϱ, ϱ͒ , what can you say about its graph?

3. (a) From the graph of f , state the numbers at which f is discontinuous and explain why. _4 _2 2468 x (b) For each of the numbers stated in part (a), determine whether f is continuous from the right, or from the left, or neither. y 5–8SolutionSketch the 1: graph of a function f that is continuous except for the stated discontinuity. 5.gDiscontinuous,is continuous but on thecontinuous following from intervals: the right, at 2

6.[−Discontinuities4, −2), (−2, 2at) ,Ϫ1[2, and 4), 4,( 4,but 6 )continuous, (6, 8). from the left at Ϫ1 and from the right at 4 0 x _4 _2 246 7. Removable discontinuity at 3, jump discontinuity at 5 8. Neither left nor right continuous at Ϫ2 , continuous only from the left at 2

; Graphing calculator or computer required 1. Homework Hints available at stewartcalculus.com x2 − 4 Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic2. rights, Use some continuity third party content may to be evaluate suppressed from the the eBook limit and/or lim eChapter(s).arctan . Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rightsx →restrictions2 require it.3x2 − 6x Solution 2:

Inverse trigonometric functions are continuous at every number in their domain (a useful fact to remember). Therefore, arctan is continuous. Now if you write 2 2 f (g(x)) = x −4 g(x) = x −4 f (x) = (x) f arctan 3x2−6x i.e., 3x2−6x and arctan . As is continuous function thus lim f (g(x)) = f (lim g(x)). x→2 x→2 x2 − 4 (x−2)(x + 2) 4 2 lim g(x) = lim = lim = = . x→2 x→2 3x2 − 6x x→2 3x(x− 2) 6 3 2 =⇒ lim f (g(x)) = arctan( ) ≈ 0.5880 x→2 3

1 3. Find the values of a and b that make f continuous everywhere.

 2 x −4 if x < 2  x−2 f (x) = ax2 − bx + 3 if 2 ≤ x < 3  2x − a + b if x ≥ 3

Solution 3:

2 x − 4 (x− 2)(x + 2) At x = 2: lim = lim = lim (x + 2) = 4 and x→2− x − 2 x→2− (x− 2) x→2− lim ax2 − bx + 3 = 4a − 2b + 3. Thus for f (x) to be continuous at x = 2, x→2+ 4a − 2b + 3 = 4 =⇒ 4a − 2b = 1 —– (1)

At x = 3: lim (ax2 − bx + 3) = 9a − 3b + 3 and lim (2x − a + b) = 6 − a + b. Thus for f (x) x→3− x→3+ to be continuous at x = 3 we must have 9a − 3b + 3 = 6 − a + b =⇒ 10a − 4b = 3 —– (2)

Solving (1) and (2)

4a − 2b = 1 × −2 Substituting a = 1/2 to eq. (1) 10a − 4b = 3 4(1/2) − 2b = 1 =⇒ b = 1/2 .

2a = 1 =⇒ a = 1/2 Thus a = b = 1/2 for f (x) to be continuous on (−∞, ∞).

4. A Tibetan monk leaves the monastery at 7:00 AM and takes his usual path to the top of the mountain, arriving at 7:00 PM. The following morning, he starts at 7:00 AM at the top and takes the same path back, arriving at the monastery at 7:00 PM. Use the Intermediate Value Theorem to show that there is a point on the path that the monk will cross at exactly the same time of day on both days.

Solution 4:

Let d be the distance between the monastery and the top of the hill. Deﬁne x1(t) to be the distance traveled by the monk in t hours on day one and similarly deﬁne x2(t) to be the distance traveled by the monk in t hours on day two.

In the above setup x1(0) = 0, x1(12) = d, x2(0) = d and x2(12) = 0. Next consider the function (x1 − x2)(t), which is continuous on [−d, d]. Also, note that (x1 − x2)(0) = −d and (x1 − x2)(12) = d.

Therefore, due to the Intermediate Value Theorem there must be a time t0; 0 < t0 < 12, where (x1 − x2)(t0) = 0 =⇒ x1(t0) = x2(t0).