6

Does there exist a continuously differentiable function $f:[1,5]\rightarrow\mathbb{R}$ such that $f(1)<0 , f(5)>3$ and $f'(x)\le e^{-f(x)}$ ?

My Attempt : If such $f$ exists the mean value theorem states: $\exists c \in(1,5) :$

$$f(5)-f(1)=f'(c)(5-1)$$

Now

$$f'(c)(5-1)\le 4e^{-f(c)}\implies f(5)-f(1)\le 4e^{-f(c)}$$

and

$$f(5)-f(1)>3$$

Therefore

$$3<4e^{-f(c)} \implies -f(c)>\ln(3/4)$$

$$\implies f(c)<-\ln(3/4)$$

This dosen't result in the nice contradiction I was searching for, does such a function exist?

Fermat
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uqtredd1
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2 Answers2

5

Assume there is such $f$. Let $g(x)=e^{f(x)}$. Then $g(1)<1$, $g(5)>e^3>2^3=8$, hence $g'(x)>\frac74>1$ for some $x\in(1,5)$. But $g'(x)=f'(x)\cdot\underbrace{e^{f(x)}}_{>0}\le e^{-f(x)}e^{f(x)}=1.$

Hagen von Eitzen
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Every solution $f$ is such that $f'(x)\mathrm e^{f(x)}\leqslant1$ hence $\mathrm e^{f(x)}-\mathrm e^{f(1)}\leqslant x-1$ for every $x\geqslant1$, that is, $$ f(x)\leqslant\log(x-1+\mathrm e^{f(1)}). $$ If $f(1)\lt0$, one gets $f(x)\lt\log x$ for every $x\geqslant1$. Since $\log5\lt1.61$, $f(5)\gt1.61$ is impossible.

Did
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