-1
$\begingroup$

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!

$\endgroup$
1
$\begingroup$

Hints:

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.