Differentiable function/Mean value theorem/Quotient version/Fact/Proof

Proof

The statement

${}g(a)\neq g(b)\,$

follows from fact. We consider the auxiliary function

${}h(x):=f(x)-{\frac {f(b)-f(a)}{g(b)-g(a)}}g(x)\,.$

We have

${}{\begin{aligned}h(a)&=f(a)-{\frac {f(b)-f(a)}{g(b)-g(a)}}g(a)\\&={\frac {f(a)g(b)-f(a)g(a)-f(b)g(a)+f(a)g(a)}{g(b)-g(a)}}\\&={\frac {f(a)g(b)-f(b)g(a)}{g(b)-g(a)}}\\&={\frac {f(b)g(b)-f(b)g(a)-f(b)g(b)+f(a)g(b)}{g(b)-g(a)}}\\&=f(b)-{\frac {f(b)-f(a)}{g(b)-g(a)}}g(b)\\&=h(b).\end{aligned}}$

Therefore, ${}h(a)=h(b)$, and fact yields the existence of some ${}c\in {]a,b[}$ with

${}h'(c)=0\,.$

Rearranging proves the claim.

To fact