# Prove there's $x_0$ such that $f'(x_0)=0$

Let $f:\mathbb{R}\rightarrow \mathbb{R}$ differentiable at $\mathbb{R}$ and:
$$\lim_{x\rightarrow \infty}\left( f(x)-f(-x) \right) = 0$$

Show there's $x_0$ such that $f'(x_0) = 0$.

I tried to use the limit definition, but couldn't get much further.

Thanks.

• do you mean as $x \to \infty$ – Anurag A Jun 18 '14 at 18:36
• If $f$ takes two distinct values $a<b$, then there is an $N$ so that both $f(N)$ and $f(-N)$ are either strictly bounded above by $b$ or strictly bounded below by $a$. Use the Intermediate Value Theorem to show $f(c)=f(d)$ for some $c\ne d$. Wrap things up with Rolle's Theorem. – David Mitra Jun 18 '14 at 18:47
• Isn't the needed result a direct consequence of Rolle's theorem? – Mathguy Jun 19 '14 at 16:41

If $f$ is a constant function, the result is easy.

Here is an outline of one way to prove such an $x_0$ exists if $f$ is not a constant function:

1. Choose $a, b$ with $f(a)< f(b)$ for some $a,b$.

2. Show that there is an $N>\max\{|a|,|b|\}$ so that either $\max\{f(N), f(-N)\}<f(b)$ or $\min\{f(N), f(-N)\}>f(a)$ (choose $N$ so that $f(N)$ is within $|f(b)-f(a)|/2$ of $f(-N)$; drawing a picture will help here).

3. Use the result of 2. and the Intermediate Value Theorem to show $f(c)=f(d)$ for some $c\ne d$.

4. Wrap things up with Rolle's Theorem.

Alternatively:

If $f'(x)$ is never $0$, then by Darboux's Theorem, $f'$ is either positive everywhere or negative everywhere). In this case $f$ is either strictly increasing or strictly decreasing. But then, for $x>1$, $|f(x)-f(-x)|> |f(1)-f(-1)|>0$; whence $\lim\limits_{x\rightarrow\infty} \bigl(f(x)-f(-x)\bigr)\ne0$.

• Not that I mind upvotes. But a few seconds after I post? Please read answers before upvoting (especially mine, which most likely will contain errors). – David Mitra Jun 18 '14 at 19:16
• I upovted automatically for your efforts. Anyway, I'm confused by (2) and particularly by this sentence: "choose N so that f(N) is within |f(b)−f(a)|/2 of f(−N)". Can you make it clearer to me? – AnnieOK Jun 18 '14 at 19:26
• You choose $N$ large enough so that by your limit condition, $f(N)$ and $f(-N)$ will either both be strictly smaller than $f(b)$ or both strictly greater than $f(a)$. Choosing $\epsilon=|f(b)-f(a)|/2$, per the definition of limit, will give you this $N$ ("$|f(N)-f(-N)|<\epsilon$" means "$f(N)$ is within $\epsilon$ of $f(-N)$". Then, if one of $f(N)$, $f(-N)$ is greater than or equal to $f(b)$, both of them will be strictly greater than $f(a)$. – David Mitra Jun 18 '14 at 19:33
• The upshot is, you first show that there are points $a<b<c$ where both $f(a)$ and $f(c)$ are strictly less than $f(b)$ or both $f(a)$ and $f(c)$ are strictly greater than $f(b)$. You can then choose points in $(a,b)$ and $(b,c)$ where $f$ takes the same value. (The $a$, $b$, and $c$ in this comment aren't the same as those used above.) – David Mitra Jun 18 '14 at 19:41
• I get it now. Drawing made it easier to understand. Thanks! – AnnieOK Jun 18 '14 at 20:06