Continuous on $[a,b]$ and $f'(x)>0$ on $(a,b)$ except for a $c$ in $(a,b)$ then $f$ is strictly increasing Prove that if $f(x)$ is continuous on $[a, b]$ and $f'(x)>0$ on $(a, b)$ except at a point
$c \in (a, b)$ where $f'(c)=0$, then $f$ is strictly increasing on $[a, b]$.
I can prove the case without the $c$ using MVT, but not sure how to approach this particular case.

Let $a \leq x \lt c$.  Then the Mean Value Theorem tells you $f(c)-f(x)=f'(t)(c-x)$ for some $t \in (x, c)$ and since $f'(t) \gt 0$ we therefore know that $f(c) - f(x) \gt 0 \Rightarrow f(c) \gt f(x)$.  Similarly, for any $y \in (c, b], f(c) \lt f(y)$.  You already know how to prove the function is strictly increasing in the intervals $(a, c)$ and $(c, b)$, so you're done.
