Here's what my textbook asks me to prove:
Suppose that $f$ is continuous on $(a,b)$ and that $f'$ exists and is positive for all but, finite number of points in $(a,b)$. Prove that $f$ is strictly increasing on $(a,b)$.
My attempt:
Let $x_1 , x_2 , \ldots x_n$ be points in $(a,b)$ where $f'(x_i)\le 0$. We eliminate the possibility that $f'(x_i)< 0$ for any $i$. Since $f'(x_i) < 0$, there exist $\delta > 0$ , such that $\frac{f(x)-f(x_i)}{x-x_i} < f'(x_i)/2 < 0$ for all $0<|x-x_i| < \delta $ (This follows immediately from the definition of the limit). That is, $f(x) < 0$ for all $0<|x-x_i| < \delta $ which contradicts our assumption that f is not positive at finitely many points. Thus it must be that $f'(x_i) = 0$ for all $i$.
Without any loss of generality, assume that $x_1 < x_2 < \ldots < x_n $. Since f is continuous on $[x_i , x_{i+1}]$ and is differentiable on $(x_i , x_{i+1})$ and $f'(x) > 0$ on $(x_i , x_{i+1})$, we have that $f$ is strictly increasing on $[x_i , x_{i+1}]$. Thus, $f$ is strictly increasing on $[x_1 , x_n ]$.
I'm stuck at the part $(a, x_1]$ and $[x_n , b)$. I cannot apply the MVT here because $f$ is not continuous at $a$ and $b$. Also, is my rest of the proof correct?