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$f:[0,1] \rightarrow \mathbb{R}$ is continuous.

I know that the "extreme value theorem" holds and that $f$ has a maximum and a minimum.
Is it correct to say that there must be a local minimum/maximum?

(Because, even if the maximum or minimum are in the edges, they are also considered local since the domain of $f$ is $[0, 1]$)

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  • $\begingroup$ A global extremum is also a local extremum $\endgroup$ – charlestoncrabb Jan 26 '16 at 16:35
  • $\begingroup$ Any possible confusion may arise from the fact that (for differentiable functions) an extremum at the boundary may have $f'(x)\ne0$. $\endgroup$ – Hagen von Eitzen Jan 26 '16 at 16:37
  • $\begingroup$ @charlestoncrabb: No it's not ! If $f(x)=x$ on $[0,1]$, then 1 is a global extremum, but not a local extremum. $\endgroup$ – Surb Jan 26 '16 at 16:38
  • $\begingroup$ Sure it is. Why isn't is a local max? $\endgroup$ – charlestoncrabb Jan 26 '16 at 16:41
  • $\begingroup$ @Surb , but the domain of the function above is $[0, 1]$ ... Was I right or wrong in saying that it must have a local extremum? $\endgroup$ – Rebecca Jan 26 '16 at 16:41
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Yes indeed: a global extremum is a local extremum.

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