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 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
Canonical name	ProofOfDarbouxsTheorem
Date of creation	2013-03-22 12:45:04
Last modified on	2013-03-22 12:45:04
Owner	paolini (1187)
Last modified by	paolini (1187)
Numerical id	7
Author	paolini (1187)
Entry type	Proof
Classification	msc 26A06