Let f be a differentiable function that is continuous on $[a,b]$. Show that if there is a number $c$ where $a < c \leq b$ such that $f'(c) = 0$, then there is a number j where $a < j < b$ such that $f'(j) = \frac{f(j)-f(a)}{b-a}$
Is this something that involve the converse of mean value theorem? How should one go about this question?
Things I have speculated upon:
c = b
f(j) = f(b)
f is a line
Thanks in advance.
We may assume $f(c)>f(a)$. Let $d\in[a,c]$ satisfy $f(d)=\max\{f(x)\mid x\in[a,c]\}$. Then there is some $e\in(a,d)$ such that $$ f'(e) = \frac{f(d)-f(a)}{d-a} \ge \frac{f(e)-f(a)}{b-a}. $$ On the other hand, $$ f'(c) = 0 < \frac{f(c)-f(a)}{b-a}. $$ It follows that there is some $j\in[e,c)$ such that $f'(j)=\frac{f(j)-f(a)}{b-a}$.
I left the details as an exercise. The last line needs a little trick with two well-known theorems.