Consider a function $g : [a, b] \to \mathbb{R}$ defined by
\begin{equation}
g(x) : = f(x) - \frac{f(b) - f(a)}{b - a} (x - a)
\end{equation}
Now
\begin{equation}
g(a)
= f(a) - \frac{f(b) - f(a)}{b - a} (a - a)
= f(a)
\end{equation}
and
\begin{equation}
g(b)
= f(b) - \frac{f(b) - f(a)}{b - a} (b - a)
= f(b) - f(b) + f(a)
= f(a)
\end{equation}
That is, $g(a) = g(b)$. Furthermore, $g$ is a sum of $f$ and a first-degree polynomial, whence results
guarantee that $g$ is differentiable on $(a, b)$. Result
R1074: Rolle's theorem now guarantees that there exists $x \in (a, b)$ such that
\begin{equation}
0
= g'(x)
= f'(x) - \frac{f(b) - f(a)}{b - a}
\end{equation}
Adding $\frac{f(b) - f(a)}{b - a}$ to each side, we end up with
\begin{equation}
f'(x)
= \frac{f(b) - f(a)}{b - a}
\end{equation}
This is exactly what was required to be shown. $\square$
