If $f$ and $g$ are differentiable in $(0,1)$ satisfying $f(0)=2=g(1), g(0)=0, f…

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If $f$ and $g$ are differentiable in $(0,1)$ satisfying $f(0)=2=g(1), g(0)=0, f(1)=6$, then for some $c\in(0,1)$

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Apply Mean Value Theorem to $f-g$: $\dfrac{(f-g)(1)-(f-g)(0)}{1-0}=\dfrac{(6-2)-(2-0)}{1}=2$ So, $f'(c)-g'(c)=2$ $\Rightarrow f'(c)=2g'(c)$

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