The following solutions do not include the graphs. You will need to include the appropriate graphs in your solutions you hand in.
(c) as and as .
(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 .
(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 .
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 .
(b) , so has a zero between -2 and -1.
(a) , so has a zero between 1 and 2.