Continuous and differentiable function $f(x)$ in $(x_1, x_2)$. Consider the continuous and differentiable function $f(x)$ in $[x_1, x_2]$.
Let $f'(x)$ be its derivative and
$f'(x_2) = 0$
Show that there exist a number $c >0$ and a $x_3 \in (x_1, x_2)$ such that $f'(x_3) = c(f(x_3)-f(x_1))$.
Apologies but I haven't touched calculus for more than 20 years. One of the few things I remember is that when the 1st derivative is zero, we have a local minimum or maximum.
(I am not a student or anything).
 A: Edited to improve the middle step:
We can assume without loss of generality that $f(x_1) = 0$ (otherwise simply consider $f(x) -f(x_1)$).
Then, if $f(x)$ is identically zero, $x_3 \in (x_1,x_2)$ and $c>0$ may be chosen freely.
Otherwise, the function $\frac{1}{2} f(x)^2$ must be non-zero at some point $\xi $ in the interval $[x_1, x_2]$;  but $\xi > x_1$ because $f(x_1)=0$ and $f(\xi)^2$ is positive because $f(x)^2 \geqslant 0$.  Apply the mean value theorem to obtain $x_3 \in (x_1, \xi)$ satisfying,
\begin{align}
\frac{\frac{1}{2}f(\xi)^2 - \frac{1}{2}f(x_1)^2}
{\xi-x_1} = f'(x_3)f(x_3)
\end{align}
The left side is strictly positive, so we have $x_3 \in (x_1,x_2)$ with
\begin{align}
f'(x_3)f(x_3) > 0 \tag{1}\label{e1}.
\end{align}
It follows neither $f(x_3)$ nor $f'(x_3)$ is zero and both have the same sign.  Accordingly we can define $c = f'(x_3)  / f(x_3) > 0 $ to obtain the required conclusion,
$$f'(x_3) = c f(x_3).$$
I note that this version does not require $f'(x_2)$ to be zero, so I wonder if the OP question ha been mis-transcribed.
