TITLE: Does there exist a continuously differentiable function with the following properties?
QUESTION [8 upvotes]: Does there exist a continuously differentiable function 

$f: [1,5] \rightarrow \mathbb{R}$, such that $f(1) \lt 0, f(5) \gt 3$ and $f'(x) \leq e^{-f(x)}$?

Now do I just integrate it to get $f(x) = -e^{-f(x)}$? This is only true at $f(x) = 0$ though, what the!!

REPLY [16 votes]: Suppose such a function exists. Since $f$ is continuously differentiable then chain rule shows that 
$f'(x)\le e^{-f(x)}$
so 
$\frac{d}{dx}(e^{f(x)})=e^{f(x)}f'(x)\le1$
By the mean value theorem we have that:
$e^{3}-1<e^{f(5)}-e^{f(1)}=e^{f(c)}f'(c)(5-1)\le4$
which implies
$e^{3}<5$ which is not true. Thus, no such function exists.