# Math 112 Solutions For Section 2.5, p. 154

The following solutions do not include the graphs. You will need to include the appropriate graphs in your solutions you hand in.

1.
(a) Define . (b) Let . (f) Redefine .

2.
(a) as and as .

(c) as and as .

3.
(a) as . (c) as .

4.
(a) has an infinite discontinuity at 3 (b) is continuous at -3. (c) continuous.

5.
(b) continuous except at (c) continuous except at (e) .

6.
(b) not continuous at 1. (c) yes. (e) infinite discontinuity at 0. (f) yes. (g) yes.

7.
Let be the number of liters of fuel in the tank and the distance traveled. Let . Clearly, and if is when , then . If is continuous, there is a point such that , which means that .

14.
Correction: In line 2, remove ".

(a) Correction: Change B" to C". Solution: Let . Then .

(b) Let be continuous on an interval containing , and let be given. Then since is continuous at , there is such that if and is in the domain of , then . Thus if , then , so is bounded below by and above by over .

(c) If is bounded on , there exists such that for . If is bounded on , then these exists such that for . L:et . Then for , and is bounded on .

(d) Theorem. If is continuous on , then is bounded on .

Proof. Let is bounded on . Since is defined at , and is non-empty. If , then , so that is bounded above. Let be the least upper bound of . Then ; suppose . Then given any there exists such that and . That means that for , so that is bounded beyond , and is not the least upper bound of after all. This contradiction forces , and thus is bounded on .

15.
(a) Since is continuous on , is bounded on . That is, is bounded above, so has a least upper bound; call it .

(b) If , then . If is always less than , then is continuous on , and hence bounded on .

(c) Let and . Because is the least upper bound of , is not an upper bound of .

(d) There is some such that . Then .

(e) Since is arbitrary, must not be bounded. Hence for some , and achieves a maximum on .

(f) Let . Then is continuous on , so by (a)-(e) achieves a maximum at some point . That is, for allk . Hence for all , and achieves a minimum at .

16.
Suppose to the contrary there are two numbers, and , such that for every . Then means . Hence is unique.

17.
Zero Theorem. If is continuous on , , and , then there is some such that .

Proof. Let . If , we are done. If , let and . If , let and . Then is a subinterval of , is continuous on , , and . Let . If , we are done. If , let and ; if , let and . Then is a subinterval of . Continue in this fashion. We thus either find a point such that , and are done, or we create a sequence of intervals such that . By the Nested Interval Theorem, there is a point such that for all . Since is the midpoint of , the length of is , which approaches 0 as . Thus is unique. Now suppose . Let . Since is continuous at , there is some such that for . But for some , making , so that . But , a contradiction. Hence it cannot be that . Similarly, it cannot be that , so .

18.
Let be continuous on and let be a value between and . First suppose that . Then let . Then is continuous on , , and . Hence by the Zero Theorem there is a point such that , and . If , let ; the same result follows.

24.
Given any interval containing a point , there will be values of differing by 1 at points in that interval. Thus given any , if then it will be impossible to force by choosing close to .

31.
(a) , so has a zero between 0 and -1.

(b) , so has a zero between -2 and -1.

(a) , so has a zero between 1 and 2.

32.
(b) Given on [0, 1], we find and , so , being continuous, will have a zero between 0 and 1. Four bisections yield the interval [.6875, .75].