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(B) Let $h(x) = f(x) - 2g(x)$.Since $f$ and $g$ are differentiable on $[0, 1]$,$h(x)$ is also differentiable on $[0, 1]$.Calculate the values of $h(x)

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$f'(c) = 2g'(c)$

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(B) Let $h(x) = f(x) - 2g(x)$.Since $f$ and $g$ are differentiable on $[0, 1]$,$h(x)$ is also differentiable on $[0, 1]$.Calculate the values of $h(x)$ at the endpoints:$h(0) = f(0) - 2g(0) = 2 - 2(0) = 2$.$h(1) = f(1) - 2g(1) = 6 - 2(2) = 6 - 4 = 2$.Since $h(0) = h(1) = 2$,by Rolle's Theorem,there exists at least one $c \in (0, 1)$ such that $h'(c) = 0$.$h'(x) = f'(x) - 2g'(x)$.Setting $h'(c) = 0$ gives $f'(c) - 2g'(c) = 0$,which implies $f'(c) = 2g'(c)$.

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

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