# proof of extended mean-value theorem

Let $f:[a,b]\to\mathbb{R}$ and $g:[a,b]\to\mathbb{R}$ be continuous on $[a,b]$ and differentiable on $(a,b)$. Define the function

 $h(x)=f(x)\left(g(b)-g(a)\right)-g(x)\left(f(b)-f(a)\right)-f(a)g(b)+f(b)g(a).$

Because $f$ and $g$ are continuous on $[a,b]$ and differentiable on $(a,b)$, so is $h$. Furthermore, $h(a)=h(b)=0$, so by Rolle’s theorem there exists a $\xi\in(a,b)$ such that $h^{\prime}(\xi)=0$. This implies that

 $f^{\prime}(\xi)\left(g(b)-g(a)\right)-g^{\prime}(\xi)\left(f(b)-f(a)\right)=0$

and, if $g(b)\neq g(a)$,

 $\frac{f^{\prime}(\xi)}{g^{\prime}(\xi)}=\frac{f(b)-f(a)}{g(b)-g(a)}.$
Title proof of extended mean-value theorem ProofOfExtendedMeanvalueTheorem 2013-03-22 13:09:00 2013-03-22 13:09:00 pbruin (1001) pbruin (1001) 6 pbruin (1001) Proof msc 26A06 proof of Cauchy’s mean-value theorem MeanValueTheorem