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Wednesday, 27 March 2013

The Intermediate Value Theorem


Your teacher probably told you that you can draw the graph of a continuous function without lifting your pencil off the paper. This is made precise by the following result:
Intermediate Value Theorem. Let f (x) be a continuous function on the interval [ab]. If d $ \in$ [f (a), f (b)], then there is a c $ \in$ [ab] such that f (c) = d.



In the case where f (a) > f (b), [f (a), f (b)] is meant to be the same as [f (b), f (a)]. Another way to state the Intermediate Value Theorem is to say that the image of a closed interval under a continuous function is a closed interval. We will present an outline of the proof of the Intermediate Value Theorem on the next page.


Here is a classical consequence of the Intermediate Value Theorem:
Example. Every polynomial of odd degree has at least one real root.
We want to show that if P(x) = anxn + an - 1xn - 1 + ... a1x + a0 is a polynomial with n odd and an $ \not=$ 0, then there is a real number c, such that P(c) = 0.
First let me remind you that it follows from the results in previous pages that every polynomial is continuous on the real line. There you also learned that

$\displaystyle \lim_{n\to\infty}^{}$$\displaystyle {\frac{P(x)}{a_n x^n}}$ = 1 and $\displaystyle \lim_{n\to-\infty}^{}$$\displaystyle {\frac{P(x)}{a_n x^n}}$ = 1.

Consequently for | x| large enough, P(x) and anxn have the same sign. But anxn has opposite signs for positive x and negative x. Thus it follows that if an > 0, there are real numbers x0 < x1 such that P(x0) < 0 and P(x1) > 0. Similarly if an < 0, we can find x0 < x1 such that P(x0) > 0 and P(x1) < 0. In either case, it now follows directly from the Intermediate Value Theorem that (for d = 0) there is a real number c $ \in$ [x0x1] with P(c) = 0.


The natural question arises whether every function which satisfies the conclusion of the Intermediate Value Theorem must be continuous. Unfortunately, the answer is no and counterexamples are quite messy. The easiest counterexample is the function

f (x) = $\displaystyle \left\{\vphantom{
\begin{array}{cl}\sin\left(\frac{1}{x}\right)&\mbox{,
if } x\not=0  0&\mbox{, if }x=0\end{array}}\right.$$\displaystyle \begin{array}{cl}\sin\left(\frac{1}{x}\right)&\mbox{,
if } x\not=0  0&\mbox{, if }x=0\end{array}$
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