When a function $f$ is strictly increasing

Question

Say $f$ is a real valued function that is differentiable (and continuous) whose domain is $[a,b]$. If \begin{equation} f'(x) > 0, \forall x\in(a,b] \text{ and } f'(a) = 0 \end{equation} Can I say that $f$ is strictly increasing in $[a,b]$ or the only thing that I can derive from the above is that it's strictly increasing in $(a,b]$?

Example: \begin{equation} g(x) = \frac{2 + x^5}{8} \end{equation} and \begin{equation} g'(x) = \frac{5x^4}{8} \end{equation} now it's obvious that $ g'(x) >0, \forall x \in (0,1]$, therefore $g$ is strictly increasing in $(0,1]$. Is there anyway (if it's true) I can say that $g$ is strictly increasing in $[0,1]$?

My effort: Let $x\in(a,b]$, then $f$ is: \begin{equation} \left.\begin{array}{c} \text{continuous in $[a,x]$} \\ \text{differentiable in $(a,x)$} \end{array}\right\} \overset{\text{M.V.T}}{\implies} \text{There exists a $\xi \in (a,x)$ such that} \end{equation} \begin{equation} f'(\xi) = \frac{f(x) - f(a)}{x - a} \end{equation} But we know that $f'(y) > 0, \forall y\in(a,b] \supset (a,b)$ and $x-a > 0$ since $x\in (a,b]$. Therefore it must be that: \begin{equation} f(x) - f(a) >0, \text{ where $x\in(a,b]$} \end{equation} Thus, $f$ is strictly increasing in $[a,b]$. Is my proof correct?

Answer 1

Yes you absolutely can, you just need to have $f \in C^0([a,b])$ and differentiable on $(a,b)$ with $f'>0$. I'm writing an answer but I suggest you try on your own first

Obviously it is strictly increasing on $(a,b)$, so we just need to ensure that $a < x \implies f(a) < f(x)$. If you suppose that $f(a) \geqslant f(x)$, you'll necessarily have $\forall y \in (a,x), \, f(y) < f(a)$, and in particular $$\forall z \in \left(a, a + \frac{x-a}{2}\right],\quad f(z) \leqslant f\left(a + \frac{x-a}{2}\right) < f(a) $$ Then you take the limit for $z \rightarrow a$, and you get $f(a) < f(a)$ which is absurd.

Note that with my proof, I don't even need $f$ to be differentiable, only strictly increasing on $(a,b)$.