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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!!

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You cannot integrate $e^{-f(x)}$ with respect to $x$ to get $-e^{-f(x)} + C$! If you differentiate the latter you will see that it is not at all the same as what you started with. –  user21820 May 18 at 2:43

1 Answer 1

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.

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+1 nice answer!! –  Integrals May 19 at 2:15

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