Let $f:[a,b]\to\mathbb{R}$ be continuous in $[a,b]$, and let $m,M$ be the minimum and maximum of $f$ in $[a,b]$, respectively.
Suppose $m<f(a)<M$, and show that
a) there exists a point $c\in (a,b)$ such that $f(c)=f(a)$
b) show that continuity is essential.
(b) is pretty simple, for example if we take $f:[1,5]\to\mathbb{R}$ $$ f = \begin{cases} x & x\neq 3\\ -8 & x = 3\end{cases} $$ then $5>f(1) = 1 > -8$, but no other $x$ satisfies $f(x)=1$.
For (a), I am having a little trouble. Since $m<f(a)<M$, then $f$ is necessarily not a constant function.
Let $c\in(a,b)$. If $f(c) = f(a)$, we are done. Otherwise, we know that either $f(c)>f(a)$ or $f(c)<f(a)$. WLOG, suppose $f(c)<f(a)$. Then look at $f$ in $[c,b]$. Then $f$ is continuous there, and $m\leq f(c)<f(a)<M$, and therefore, from the IVT, $f$ gets the value $f(a)$ inside $[c,b]$.
Is this semi proof ok? how do I show that the needed point doesn't have to be $b$?
Supose $\;f(x_m)=m\;,\;\;f(x_M)=M\;,\;\;x_m,\,x_M\in (a,b]\;$ , and assume $\;x_m<x_M\;$ , the other case being completely similar. .
Since $\;f(x_m)=m<f(a)<M=f(x_M)\;$, there exists $\; \epsilon>0\;$ such that
$\;m+\epsilon<f(a)<M-\epsilon\;$, so if $\;f(\alpha)=m+\epsilon\,,\,\,f(\beta)=M-\epsilon\;$ there exists then
$\;\;c\in (\alpha,\,\beta)\subsetneq (a,b)\;$ such that $\;f(c)=f(a)\;$ and we've finished.