Suppose that $f$ is continuous on $[0,1]$ and differentiable on (0,1); $c\in (0,1)$. Show that there exist $y,z\in (0,1)$ such that $$2zf(1)+(c^2-1)f’(z)=f(y).$$

Thanks in advance!



1) First take $c=\frac{\sqrt{3}}{2}$, $f(x)=(x-\frac{1}{2})^2$, and show that the hypothesis $y\in )0,1($ cannot be satisfied. I suppose that you want $y\in [0,1]$.

2) a) Put $u=c^2(f(1)-f(0))+f(0)$. Show that we have $f(0)<u<f(1)$ (if $f(1)>f(0)$) or $f(1)<u<f(0)$ (if $f(1)<f(0)$), and deduce in all cases (the case $f(0)=f(1)$ is left to you) there exists an $y$ such that $u=f(y)$ .

b) Put $g(x)=(c^2-1)f(x)+x^2f(1)$, compute $g(1)-g(0)$ and finish the proof.


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