Prove that $a<c<b$, $\frac{f(b)-f(a)-(b-a)f'(a)}{g(b)-g(a)-(b-a)g'(a)}=\frac{f"(c)}{g"(c)}$

Question

Prove that if $a<c<b$,and $f'(x)$ is differentiable on $ (a,b)$,then $$\frac{f(b)-f(a)-(b-a)f'(a)}{g(b)-g(a)-(b-a)g'(a)}=\frac{f''(c)}{g''(c)}$$

I tried to construct an auxiliary function $h(x)=(f(b)-f(a)-(b-a)f'(a))g'(x)-f'(x)(g(b)-g(a)-(b-a)g'(a))$ I tried to use Rolle's theorem but I am not getting $h(a)=h(b)$

Answer 1

No, you want to apply Cauchy's Mean Value Theorem to a pair of functions. Here's one: $F(x) = f(x)+(b-x)f'(x)$. Probably there is a suitably sneaky way to get it out of Rolle's Theorem. Perhaps my function will help you do that.