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Unformatted text preview: he function
f fails to be closed.
We can define f x x for 0 x 1.
4. Give an example of a function f that is continuous on a bounded set H such that the range f H of the function
f fails to be bounded.
We can define f x 1 for 0 x 1.
x
5. Prove that if a set H of real numbers is unbounded above and f x x for every number x in H, then f is a
continuous function on H and f fails to have a maximum.
Since the range of f is the set H which is assumed to be unbounded, the function f must be
unbounded above.
6. Prove that if H is a set of real numbers and a number a is close to H but does not belong to H, and if we
define
1
fx
x a 
for every x H then f is a continuous function on H but f has no maximum. Solution: We can see that f has no maximum by showing that f is unbounded above. Given any
positive number q the inequality
fx q
says that
1
q
x a 
which holds when
x a  1
q
But since a is close to the set H we know that there do indeed exist members x of H for which the
inequality
x a  1
q
holds. Therefore there are members x in H for which f x q and we have shown that f fails to be
bounded above. Exercises on Continuity of Functions on Intervals
1. Given that S is a set of positive numbers and that f x x for all x S, prove that f is a oneone continuous
function on S. Prove that S is an interval if and only if the set f S is an interval. Hint: Once you have shown that f is strictly increasing and continuous on S, the fact that S is an
interval if and only if f S is an interval will follow from the Bolzano intermediate value theorem and
this theorem.
2. Prove that there are three real numbers x satisfying the equation
x 3 4x 2 0. Solution: First look at a sketch of the graph y x 3
199 4x 2 4 2
2 1x 1 2 0
2 4 We now define f x x 3 4x 2 for every real number x and observe that f 2 0 and f 1 0.
Therefore Bolzano’s intermediate theorem guarantees that the equation x 3 4x 2 0 has a solution
between 2 and 1. Since f 1 0 and f 0 0 we know that there is a solution of the equation between
1 and 0. Finally, from the fact that f 0 0 and f 3 0 we know that there is a solution of the equation
between 0 and 3.
3. Is it true that if a set S of real numbers is not an interval then there must exist a oneone continuous function
on S whose inverse function fails to be continuous?
Not at all. In fact, we know that if S is closed and bounded then every oneone continuous function
S must have a continuous inverse function. Look also at the case in which S is the union of two
open intervals that do not intersect with each other. If f is a oneone continuous function on S then
the range of f is also the union of two open intervals that do not intersect with each other and the
inverse function of f will be continuous.
4. Is it true that if a set S of real numbers is not an interval and is not closed then there must exist a oneone
continuous function on S whose inverse function fails to be continuous? Hint: Look at the cas...
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 Fall '08
 STAFF
 Math, Calculus

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