By MVT, $$\frac{f(b)-f(a)}{b-a}=f'(c)$$ $$\frac{g(b)-g(a)}{b-a}=g'(c)$$ so, $$\frac{f(b)-f(a)}{g(b)-g(a)} = \frac{\frac{f(b)-f(a)}{b-a} }{\frac{g(b)-g(a)}{b-a}}=\frac{f'(c)}{g'(c)}$$

I think it is not a valid proof but I am having a discussion with a classmate and I am not sure anymore...

  • 6
    $\begingroup$ It is not valid. What you can say is $\frac{f(b)-f(a)}{b-a}=f'(c)$ and $\frac{g(b)-g(a)}{b-a}=g'(d).$ But you have to prove that $c=d.$ $\endgroup$
    – mfl
    Dec 13 '16 at 21:42
  • $\begingroup$ That is what I was thinking as well. Thank you! $\endgroup$
    – MathIsHard
    Dec 13 '16 at 21:45
  • 2
    $\begingroup$ To convince your classmate consider $f(x)=x^2$ and $g(x)=x^3$ on $[0,1].$ Note that $c=1/2\ne d=1/\sqrt{3}.$ $\endgroup$
    – mfl
    Dec 13 '16 at 21:47
  • 1
    $\begingroup$ Just linking this so it is easeir to find, took me some time .math.stackexchange.com/questions/114694/… $\endgroup$
    – Milan
    May 23 '19 at 16:16

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