7
$\begingroup$

The function $ f(x) $ is continuous on $[0,1]$ and is differentiable on $(0,1)$,$c∈(0,1)$,prove: $∃ξ, η∈(0,1)$ $$2ηf(1)+(c^2-1)f'(η)=f(ξ)$$

I tried to use the Lagrange mean value theorem and the Rolle mean value theorem on $[0,1]$, but failed.

$\endgroup$
7
$\begingroup$

Let $g(\eta)= \eta^2 f(1)+(c^2-1) f(\eta)$.

Then $g(1)-g(0)=c^2 f(1)+(1-c^2) f(0)$ so by Rolle there exists $\eta$ such that: $$g'(\eta)=2 \eta f(1)+(c^2-1) f'(\eta)=\frac{g(1)-g(0)}{1-0}=c^2 f(1)+(1-c^2) f(0)$$ More over by the mean value theorem as $f$ is continuous and $c^2 f(1)+(1-c^2) f(0) \in [f(0),f(1)]$ there exists $\xi \in [0,1]$ such that $$f(\xi)=c^2 f(1)+(1-c^2) f(0)$$ And so: $$2 \eta f(1)+(c^2-1) f'(\eta)=c^2 f(1)+(1-c^2) f(0)=f(\xi)$$

$\endgroup$
  • $\begingroup$ Shouldn't you have $g'(\eta)=2\eta\,f(1)+(c^2-1)\,f'(\eta)$? $\endgroup$ – Batominovski Sep 10 '18 at 15:51
  • $\begingroup$ @Batominovski Indeed thank you :-) $\endgroup$ – Delta-u Sep 10 '18 at 15:51
  • $\begingroup$ You fixed wrong things. It is already correct that $g(1)-g(0)=c^2\,f(1)+(1-c^2)\,f(0)$. $\endgroup$ – Batominovski Sep 10 '18 at 15:53
  • $\begingroup$ Thank you again :-) $\endgroup$ – Delta-u Sep 10 '18 at 15:55
  • $\begingroup$ @Delta-u I think the $η$ in $g'(η)$ is not the $η$ in $η^2*f(1)η+(c^2-1)*f(η)$ $\endgroup$ – King.Max Sep 10 '18 at 16:14

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.