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].