Let $f:[0,4] \rightarrow \mathbb{R}$ is differentiable. Then show that $f'(c)=\frac{1}{6}(f'(1)+2f'(2)+3f'(3))$ for some $c \in (0,4)$.

Question

Let $f:[0,4] \to \mathbb{R}$ be a differentiable function. Show that there exists a $c \in [0,4]$ s.t $f'(c)=\frac{1}{6}\left(f'(1)+2f'(2)+3f'(3)\right)$.

I don't understand where to start this problem. I only took that $f'(\alpha)=\min\{f'(1),f'(2),f'(3)\}$ and $f'(\beta)=\max\{f'(1),f'(2),f'(3)\}$, where $\alpha, \beta \in \{1,2,3\}$.

Answer 1

First observe that

$$ \begin{align} f'(\alpha)&=\frac{1}{6}\left( f'(\alpha)+ 2f'(\alpha)+ 3f'(\alpha) \right)\\ &\leq\frac{1}{6}\left( f'(1)+ 2f'(2)+ 3f'(3) \right)\\ &\leq\frac{1}{6}\left( f'(\beta)+ 2f'(\beta)+ 3f'(\beta) \right)=f'(\beta) \end{align} $$

Hence $\frac{1}{6}\left( f'(1)+ 2f'(2)+ 3f'(3) \right)\in\left[f'(\alpha),f'(\beta)\right]$.

As Christophe Leuridan noted, from Darboux theorem it follows that there is $c\in\left[\min(\alpha,\beta),\max(\alpha,\beta)\right]\subset[1,4]$ s.t $f'(c)=\frac{1}{6}\left( f'(1)+ 2f'(2)+ 3f'(3) \right)\\$.