# A problem about the differential mean value theorem $2ηf(1)+(c^2-1)f'(η)=f(ξ)$

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.

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