Let f(x) be continuous on [0,6] and different | vedclass

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(D) Given $g(x) = \frac{f(x)}{x+1}$.Since $f(x)$ is continuous on $[0,6]$ and differentiable on $(0,6)$, $g(x)$ is also continuous on $[0,6]$ and diff

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$-\frac{44}{21}$

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(D) Given $g(x) = \frac{f(x)}{x+1}$.Since $f(x)$ is continuous on $[0,6]$ and differentiable on $(0,6)$, $g(x)$ is also continuous on $[0,6]$ and differentiable on $(0,6)$ because $x+1 
eq 0$ for $x \in [0,6]$.Calculate the values of $g(x)$ at the endpoints:$g(0) = \frac{f(0)}{0+1} = \frac{12}{1} = 12$$g(6) = \frac{f(6)}{6+1} = \frac{-4}{7}$By Lagrange's Mean Value Theorem, there exists at least one $c \in (0,6)$ such that $g^{\prime}(c) = \frac{g(6)-g(0)}{6-0}$.Substituting the values:$g^{\prime}(c) = \frac{-\frac{4}{7} - 12}{6} = \frac{-\frac{4}{7} - \frac{84}{7}}{6} = \frac{-\frac{88}{7}}{6} = -\frac{88}{42} = -\frac{44}{21}$.

Let $f(x)$ be continuous on $[0,6]$ and differentiable on $(0,6)$. Let $f(0)=12$ and $f(6)=-4$. If $g(x)=\frac{f(x)}{x+1}$, then for some Lagrange's constant $c \in(0,6)$, $g^{\prime}(c)=$

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