Let $f$ be defined on $[a,b]$, Prove that if f has a local maximum at a point $x \in (a,b)$, and if $f'(x)$ exists, then $f'(x)=0$

Question

Is this proof correct:

Let's choose a $\delta$ to that

$a < x - \delta < x < x + \delta < b$

If $ x - \delta < t < x$

then $\frac {f(t) - f(x)} {t-x} \geq 0$

Letting $t \rightarrow x$, we see that $f'(x) \geq 0$

If $ x < t < x + \delta$,

then $\frac {f(t) - f(x)} {t-x} \leq 0$

Which shows that $f'(x) \leq 0$. Hence $f'(x) = 0$

Answer 1

Yes, it is. Maybe you want to say $t\to x^+$ and $t\to x^-$ explicitly.