# Theorem on continuous function crossing x-axis

1. Aug 16, 2014

### dimitri151

I think this is a theorem, and I'm telling myself that I've proved it. Just a shout out for any possible counter-examples:
If a function f(x) is continuous on some interval and has non-zero derivatives at its root(s) (where f(x')=0 ) then there is some interval around the roots where there are no other roots, and f(x)><0 for x><x' as f'(x')><0.

It just says a function crossing the x-axis comes from below and goes above or comes from above and goes below when the derivative at the point is not zero. Any counter-examples?

2. Aug 16, 2014

### gopher_p

There is a more general theorem that can be proven:

If $f$ is differentiable at $a$ and $f'(a)\neq 0$, then there is an open interval $I$ with $a\in I$ such that for all $x\in I\setminus\{a\}$, $f(x)\neq f(a)$.

3. Aug 16, 2014

### jbunniii

Here's a simple proof for the result mentioned by gopher_p.

Suppose that $f'(a) \neq 0$. By definition, this means that
$$\lim_{h \rightarrow 0}\frac{f(a+h) - f(a)}{h} = f'(a) \neq 0$$
Therefore if we set $\epsilon = |f'(a)|/2$ then there is a $\delta > 0$ such that
$$\left|\frac{f(a+h) - f(a)}{h}\right| > \frac{|f'(a)|}{2} > 0$$
for all $h$ satisfying $0 < |h| < \delta$. Therefore, for all such $h$, we have
$$|f(a+h) - f(a)| > |h|\frac{|f'(a)|}{2} > 0$$
and as a result, $f(a+h) \neq f(a)$ for all $h \in (-\delta, \delta) \setminus \{0\}$.

Last edited: Aug 17, 2014
4. Aug 16, 2014

### WWGD

if you don't mind the heavy machinery; using a tank to kill a fly and f' is continuous, just use the inverse function theorem.