Trying to apply Rolle's Theorem

Question

Let $f$ be a differentiable function in the interval $[a,b]$. Prove that there exists $c\in ]a,b[$ such that $$f'(c)=f(c)\dfrac{(a+b-2c)}{(c-a)(c-b)}$$

My ideas are: maybe we need to find a function having a derivate similar to $\dfrac{(a+b-2x)}{(x-a)(x-b)}f(x)$?

Answer 1

Consider the differentiable function $$g(x)=f(x)(x-a)(x-b)$$ then $g(a)=g(b)=0$ and by Rolle's Theorem there is a $c\in (a,b)$ such that $g'(c)=0$: $$g'(c)=f'(c)(c-a)(c-b)+f(c)(c-b)+f(c)(c-a)=0$$ that is $$f'(c)=\frac{f(c)(a+b-2c)}{(c-a)(c-b)}.$$

Answer 2

Take $F(x)=(x-a)(x-b)f(x)=\left(x^2-(a+b)x+ab\right)f(x)$

$F$ is differentiable in $[a,b]$ and $F(a)=F(b)=0$

So the Rolle theorem tells us that $\exists c\in ]a,b[,F’(c)=0$

Now $F’(x)=(x-a)(x-b)f’(x)+\left(2x-(a+b)\right)f(x)$

And we have $(c-a)(c-b)f’(c)=\left(-2c+a+b\right)f(c)$