# An exercise of mean value theorem

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 $$2zf(1)+(c^2-1)f’(z)=f(y).$$

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 $$f(0) (if $$f(1)>f(0)$$) or $$f(1) (if $$f(1)), and deduce in all cases (the case $$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.