# Does this function have local minimum/maximum?

$f:[0,1] \rightarrow \mathbb{R}$ 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 $[0, 1]$)

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