proof of least and greatest value of function

$f$ is continuous, so it will transform compact sets into compact sets.Thus since $[a,b]$ is compact, $f([a,b])$ is also compact.$f$ will thus attain on the interval $[a,b]$ a maximum and a minimum value because real compact sets are closed and bounded.

Consider the maximum and later use the same argument for $-f$ to consider the minimum.

By a known theorem (http://planetmath.org/FermatsTheoremStationaryPoints) if the maximum is attained in the interior of the domain, $c\in ]a,b[$ then $f(c)\text{is a maximum}⟹f^{\prime }(c)=0$, since $f$ is differentiable.

If the maximum isn’t attained in $]a,b[$ and since it must be attained in $[a,b]$ either $f(a)$ or $f(b)$ is a maximum.

For the minimum consider $-f$ and note that $-f$ will verify all conditions of the theorem and that a maximum of $-f$ corresponds to a minimum of $f$ and that$-f^{\prime }(c)=0\iff f^{\prime }(c)=0$.