Show there exists a number $c \in (a,b)$ such that $f(b) = f(a) + (b-a)f'(a) + \frac{(b-a)^2}{2}f''(c)$. Full question:
Suppose that $f'$ exists and is continuous on $[a, b]$ and $f''$ exists on $(a, b)$, where $a < b$.
Show that there exists a number $c \in (a, b)$ such that
$$
f(b) = f(a) + (b-a)f'(a) + \frac{(b-a)^2}{2}f''(c)
$$
Manipulating the above equation gives me
\begin{align*}
\frac{f(b) - f(a)}{b - a} &= f'(a) + \frac{(b-a)}{2}f''(c) \\
f'(c) &= f'(a) + \frac{(b-a)}{2}f''(c)
\end{align*}
The question suggests the use of the Mean Value Theorem. Using this for both $f$ and $f'$ gives me
\begin{align*}
f'(c) &= \frac{f(b) - f(a)}{b - a} \\
f''(c) &= \frac{f'(b) - f'(a)}{b - a}
\end{align*}
But I am stuck here. I tried formulating the equation of the tangent line through $(a, f(a))$ for $f$ as well as through $(a, f'(a))$ for $f'$ but to no avail. Is there some sort of manipulation for this which can help?

I think you should look at the Taylor expansion of f near a to the first order and use the Lagrange form of the remainder (the  remainder is of second order), then substitute b.
