Let f(x) be continuous on [0,4], differentiab | vedclass

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(D) Given that $f(x)$ is continuous on $[0,4]$ and differentiable on $(0,4)$.Since $g(x) = \frac{f(x)}{x+2}$, $g(x)$ is also continuous on $[0,4]$ and

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$-\frac{7}{12}$

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(D) Given that $f(x)$ is continuous on $[0,4]$ and differentiable on $(0,4)$.Since $g(x) = \frac{f(x)}{x+2}$, $g(x)$ is also continuous on $[0,4]$ and differentiable on $(0,4)$ because $x+2 
eq 0$ for $x \in [0,4]$.Calculate the values of $g(x)$ at the endpoints:$g(0) = \frac{f(0)}{0+2} = \frac{4}{2} = 2$$g(4) = \frac{f(4)}{4+2} = \frac{-2}{6} = -\frac{1}{3}$By Lagrange's Mean Value Theorem, there exists at least one $c \in (0,4)$ such that $g^{\prime}(c) = \frac{g(4)-g(0)}{4-0}$.Substituting the values:$g^{\prime}(c) = \frac{-\frac{1}{3} - 2}{4} = \frac{-\frac{7}{3}}{4} = -\frac{7}{12}$.

Let $f(x)$ be continuous on $[0,4]$, differentiable on $(0,4)$, $f(0)=4$ and $f(4)=-2$. If $g(x)=\frac{f(x)}{x+2}$, then the value of $g^{\prime}(c)$ for some Lagrange's constant $c \in (0,4)$ is

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