If $f(x)$ is continuous on $[a,b]$ and differentiable on $(a,b)$ then $(\frac{f'(c)}{3c^2}) = (\frac{f(b)-f(a)}{b^3 -a^3})$ where$c ∈ (a,b)$.

Question

If $f(x)$ is continuous on $[a,b]$ and differentiable in $(a,b)$, then $(\frac{f'(c)}{3c^2}) = (\frac{f(b)-f(a)}{b^3 -a^3})$ for some $c ∈ (a,b)$.

I have tried in this manner:

Let us assume $H(x)=(b^3 -a^3) f(x)-(f(b)-f(a))x^3$ which is continuous on $[a,b]$ and differentiable on $(a,b)$. $H(a)=b^3 f(a)-a^3 f(b)=H(b)$ which is Rolle's theorem. I am stuck in this step. Could anybody please help me?

Answer 1

let $$g(x) = f(x) - kx^3, k \text{ to be fixed.} $$ requiring $g(a) = g(b)$ gives you $$k = \frac{f(b) - f(a)}{b^3 - a^3} $$ now apply rolle's theorem that tells you that there ia $c$ such that $g'(c) = 0.$