Let $f:(0,1)\longrightarrow \mathbb R$ be a continuous function. $f$ does not "have" (Don't know if it's the proper term, but couldn't find another way to translate it) global extrema and also $f({1\over 2})=f({3\over 4})$. Show that $f$ "has" a local minima in the interval $(0,1)$.

I thought of using the extreme value theorem in the close interval $[{1\over 2},{3\over 4}]$, but the question seems to be too easy to prove by using that theorem, and also the choice of such interval seems to be useless since using the theorem would make the information about the equality $f({1\over 2})=f({3\over 4})$ become useless.
However, I can't find an explanation why using the theorem isn't legal.
My question is whether the use of this theorem in the given question is legal, and if not, why?