# 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 $g^{\prime}_{+}(b)<0$ Fermat’s theorem (http://planetmath.org/FermatsTheoremStationaryPoints) states that neither $a$ 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).

Title proof of Darboux’s theorem ProofOfDarbouxsTheorem 2013-03-22 12:45:04 2013-03-22 12:45:04 paolini (1187) paolini (1187) 7 paolini (1187) Proof msc 26A06