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I am trying to finish the proof started in the first comment of this post. I have that $$h'(x)=g'(x)[f(b)-f(a)]-f'(x)[g(b)-g(a)]$$ then $$h'(c)=g'(c)[f(b)-f(a)]-f'(c)[g(b)-g(a)]=0$$ by Rolle's theorem. Then $$g'(x)[f(b)-f(a)]=f'(x)[g(b)-g(a)]$$ But then I am stuck... Do I just simply show by algebra that $$\frac{f(b)-f(a)}{g(b)-g(a)}=\frac{f'(c)}{g'(c)}$$ and I am done?

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MathIsHard
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1 Answers1

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From $g'(c)[f(b)-f(a)]-f'(c)[g(b)-g(a)]=0$ we get

$g'(c)[f(b)-f(a)]=f'(c)[g(b)-g(a)]$

It follows that

$\frac{f(b)-f(a)}{g(b)-g(a)}=\frac{f'(c)}{g'(c)}$

(if you have the equation $xy=uv$ then $x/u=v/y$ provided that $u \ne 0 \ne y$

Fred
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