If $f(a)/a=f(b)/b$ then there exists $c$ in $(a,b)$ such that $f'(c)=f(c)/c$. [closed]

Suppose that $f:[a,b]\to\mathbb{R}$ be continuous in $[a,b]$ and differentiable in $(a,b)$ if $f(a)/a=f(b)/b$ then there exists '$c$ 'in (a,b) such that $f'(c)=f(c)/c$.

Instead of required proof I got the result $f'(c)=f(a)/a=f(b)/b$ which can be proved using Lagrange's mean value theorem and componendo and divido rules... Can any body help me to prove required result?

closed as off-topic by Nosrati, Adrian Keister, Namaste, Taroccoesbrocco, max_zornAug 14 '18 at 5:37

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Nosrati, Adrian Keister, Namaste, Taroccoesbrocco, max_zorn
If this question can be reworded to fit the rules in the help center, please edit the question.

We may need more condition on $a,b$: they have same sign. If not, there is a counter example: take $f(x)=1-x^2$ and $a=-1,b=1$. Then $f’(c)=f(c)/c$ leads to $c^2=-1$.
If we add the condition, it follows from the Rolle's theorem applying to $g(x)=f(x)/x$.