$f$ is continuous and differentiable in $(-1,1)$, $f(1) = 0$, $f(x)>0 (x∈(-1,1))$. Show $∃c∈(-1,1); mf(c)f'(-c) = nf'(c)f(-c)$ in which $m∈N, n∈N$ https://ift.tt/eA8V8J

My problem is the following: "Suppose that $f\colon[-1,1]\to\mathbb{R}$ is continuous and differentiable on $(-1,1)$, $f(1) = 0$ and $f(x)>0$ for $x\in(-1,1)$. Show that there exists $c\in(-1,1)$ such that $mf(c)f'(-c) = nf'(c)f(-c)$, in which $m\in N$ and $n\in N$."

I can prove if when $M=N$: there exists $c\in(-1,1)$ for which $f(c)f'(-c) = f'(c)f(-c)$ using $g(x) = f(x)f(-x)$, which satisfies $g(1) = g(-1) = 0$ and Rolle's theorem, but I can't prove it when $m ≠ n$. Can someone please tell me how to prove it?

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