I think it's very clever to show that if $f$ has a local maximum (or minimum) at a point $x\in(a,b)$ and $f'(x)$ exists, then $f'(x)=0$, for $$a<x-\epsilon <x<x+\epsilon <b,$$ and so the quotient $\frac{f(t)-f(x)}{t-x}\geq0 $, if $x-\epsilon <t<x$
and $\frac{f(t)-f(x)}{t-x}\leq0 $, if $x <t<x+\epsilon$, hence $\frac{f(t)-f(x)}{t-x}=0$.
But what can be said about the existence of local maxima or minima?
If $f$ is a real-valued function defined on $[a,b]$, would the (generalised) mean value theorem be enough to imply that there exists a point $x \in [a,b]$ that is either a local maximum or local minimum, given that $f$ is not constant?
Thanks.
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$\begingroup$I've decided to write an answer to close this question.
From the fairly long discussion in the comment section, for a function $f$ that is continuous on a closed interval, the Extreme Value Theorem states that global maxima and minima exist in this interval, hence local maxima and minima also exist in the interval.
From this other question link, I will refer to the fact that
- If global extrema is not in boundary, i.e. point $x \in (a,b)$, then it is also local extrema in that (open) interval.
And since the proof that, 'if $f$ has local extrema and $f'(x)$ exists, then $f'(x)=0$' is concerning a point $x$ in the open interval $(a,b)$, we don't necessarily have to settle the debate on what happens if global extrema exists on the boundary of the interval.
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