proof of Darboux’s theorem

Without loss of generality we migth and shall assume $f_{+}^{\prime }(a)>t>f_{-}^{\prime }(b)$. Let $g(x):=f(x)-tx$.Then $g^{\prime }(x)=f^{\prime }(x)-t$, $g_{+}^{\prime }(a)>0>g_{-}^{\prime }(b)$, and we wish to find a zero of $g^{\prime }$.

Since $g$ is a continuous function on $[a,b]$, it attains a maximum on $[a,b]$.Since $g_{+}^{\prime }(a)>0$ and Fermat’s theorem (http://planetmath.org/FermatsTheoremStationaryPoints) states thatneither $a$ nor $b$ can be points where $f$ has a local maximum.So a maximum is attained at some $c\in (a,b)$. But then $g^{\prime }(c)=0$ again by Fermat’s theorem (http://planetmath.org/FermatsTheoremStationaryPoints).