324 CHAPTER 9 CALCULUS
For example, f(x) : = l/x is continuous on (0,5), but achieves neither maximum nor
minimum values on this interval.
On the other hand, the IVT, while "obvious" (see Problem 9.2.22 for hints about
its proof), has many immediate applications. Here is one simple example. The crux
move, defining a new function, is a typical tactic in problems of this kind.
Example 9.2.4 Let f : [0, 1] ---+ [0, 1] be continuous. Prove that f has a fixed point;
i.e., there exists x E [0, 1] such that f(x ) = x.
Solution: Let g(x) := f(x) - x. Note that g is continuous (since it is the difference
of two continuous functions), and that g(O) = f(O) 2: 0 and g( 1) = f( 1) - 1 :S O. By
the IVT, there exists u E [0, 1] such that g( u) = O. But this implies that f( u ) = u. •Uniform Continuity
Continuous functions on a closed interval (i.e., the domain is a closed interval) possess
another important property, that of uniform continuity. Informally, this means that
the amount of "wiggle" in the graph is constrained in the same way throughout the
domain. More precisely,
A function f : A ---+ B is uniformly continuous on A if, for each e > 0,
there exists 8> 0 such that if XI,X 2 E A satisfy IXI -x 21 < 8, then
If(xt} -f(X 2 ) I < e.
The important thing in this definition is that the value of 8 depends only on e
and not on the x-value. For each positive e, there is a single 8 that works everywhere
on the domain. Because it is rather difficult to prove that all continuous functions on
closed intervals are uniformly continuous, the concept of uniform continuity is not
often introduced in elementary calculus classes. But it is such a useful idea that we
will accept it, for now, on faith.^6
Example 9.2.5 The function f(x) = x^2 is uniformly continuous on [-3,3]. As long
as I XI -x 21 < 8, we are guaranteed that If(xt} -f(X 2 ) I < 68. It is easy to see why:
For any X l , X 2 E [-3,3], the largest possible value for IXI + x 21 is 6, and thenIf(xt} -f(X 2 ) I = IXI +x 21 ·lxl -x 21 :S^61 xl -x 21 ·
Consequently, if we want to be sure that the function values are within e, we need only
require that the x-values be within e/ 6.
Example 9.2.6 The function f(x) = l/x defined on (0,00) is not uniformly continu
ous. For x-values close to 0, the function changes too fast. Given an e, no single 8
will do, if the x-values are sufficiently close to O. Note, however, that on any closed
interval, f(x) is uniformly continuous. For example, verify that if we are restricting
our attention to x E [2, 1000], then the " 8 response" to the "e challenge" is 8 = 4e.
In other words, if we are challenged to constrain the f-values to be within e of each
other, we need only choose x-values within 4e of one another.(^6) Consult any of the texts mentioned on page 357.