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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?

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

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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 $g(x)=f(x)/x$.

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