For some $A \in \left(-\frac{1}{2},\frac{1}{2} \right)$ and $c \in (0,1), \; f(1)=f(0)+f'\left(\frac{1}{2}\right)+Af''(c)$

Suppose $$f(x)$$ is continuous on $$[0,1]$$ and twice differentiable on $$(0,1)$$.

Show that there exists $$A \in \left(-\frac{1}{2},\frac{1}{2} \right)$$ and $$c \in (0,1)$$ such that

$$$$f(1)=f(0)+f'\left(\frac{1}{2}\right)+Af''(c)$$$$

By MVT, there exists $$c \in (0,1)$$ such that

$$$$f'(c)=\frac{f(1)-f(0)}{1-0} = f(1)-f(0)$$$$

Comparing with the required statement I see that it suffices to show, for some $$A \in \left(-\frac{1}{2},\frac{1}{2} \right)$$, $$$$f'(c)=f'\left(\frac{1}{2}\right)+Af''(c)$$$$ Rearranging, $$$$\frac{f'(c)-f'\left(\frac{1}{2}\right)}{A}=f''(c)$$$$

This looks like an application of MVT again, but I'm not very sure how to proceed, as $$c$$ appears on both LHS and RHS.

Any help would be appreciated!

$$f'(c)-f'(\frac 1 2) =(c-\frac 1 2)f''(d)$$ for some $$d$$ between $$c$$ and $$\frac 1 2$$ Note that $$|c-\frac 1 2| \leq \frac 1 2$$. Hence the required equation holds with $$A=c-\frac 1 2$$ and $$c$$ changed to $$d$$.