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Let $f$ be continuous on $[a,b]$, differentiable on $(a,b)$ and positive for all $x \in(a,b).$ Prove that there exists $c\in(a,b)$ such that $$\frac{f'(c)}{f(c)} = \frac{1}{a-c}+\frac{1}{b-c}.$$

This seems like just an application of the mean value theorem, but it doesn't seem to work out when I try.

My first attempt was to find an explicit equation for $f(x)$ since $$f'(x) = f(x)\left( \frac{1}{a-x} + \frac{1}{b-x} \right)\Rightarrow f(x) = e^{-\ln((a-x)(b-x)} \left( \frac{1}{a-x} + \frac{1}{b-x} \right)$$

But applying the mean value theorem doesn't quite work here because $f(a)$ and $f(b)$ are not defined so $f(x)$ isn't continuous on $[a,b].$

Any help would be appreciated.

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Consider the function $g:[a,b]\to \mathbb{R}$ with $g(x)=(x-a)(x-b)f(x)$. Then $g$ is continuous on $[a,b]$ and differentiable on $(a,b)$ and $g(a)=g(b)=0$. From Rolle's theorem there exists some $c\in (a,b)$ such that $g'(c)=0$ or

$$(c-b)f(c)+(c-a)f(c)+(c-a)(c-b)f'(c)=0$$

and this is equivalent with:

$$\frac{f'(c)}{f(c)} = \frac{1}{a-c}+\frac{1}{b-c}$$

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