Every bounded function has an inflection point? Hello from a first time user! I'm working through a problem set that's mostly about using the first and second derivatives to sketch curves, and a question occurred to me: 
Let $f(x)$ be a function that is twice differentiable everywhere and whose domain is $ \Bbb R$. If $f(x)$ is bounded, then must $f$ have at least one inflection point?
The answer is clearly yes, but I can't think of how to prove it. (I thought of using the converse of the MVT, but Google revealed that the converse is untrue.) In the case where $f''(x)$ is always positive, for example, the three possibilities for $f(x)$ are nicely illustrated by considering the three branches of $g(x)$ (disregarding the vertical asymptotes), $g(x)=\left |\frac {1}{(x-a)(x-b)}\right |$, where $a < b$. 
$(-\infty, a)$: $g'(x)$ is bound below, unbounded above, and $g(x)$ is unbounded above
$(a, b)$ :   $g'(x)$ is unbounded below and above, and $g(x)$ is unbounded above
$(b, \infty)$: $g'(x)$ is unbounded below, bound above, and $g(x)$ is unbounded above.
In all three cases, $g''(x) > 0$ implies that $g(x)$ is unbounded.
Is all this right? How can we prove that the answer to the above question about $f(x)$ is true?

If $f''>0$ identically and $f$ is bounded, then $f'\leq 0$ identically (for otherwise, $f(\infty)=\infty$). Likewise, $f'\geq 0$ since otherwise $f(-\infty)=-\infty$. Hence $$f'=0$$ identically, so $f$ is constant. The result follows similarly in the case $f''<0$.
