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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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Verified Answer
The correct answer is:
$\frac{5}{12}$
Let $f(x)$ be continuous on $[0,4]$, differentiable on $(0,4)$
$$
F(0)=4 \text { and } F(4)=-2 \quad g(x)=\frac{f(x)}{x+2}
$$
At $x=0, g(0)=\frac{f(0)}{0+2}=\frac{4}{2}=2$
At $x=4, g(4)=\frac{f(4)}{4+2}=\frac{-2}{6}=\frac{-1}{3}$
Now,
$$
g^{\prime}(c)=\frac{g(4)-g(0)}{4-0}=\frac{\frac{-1}{3}-2}{4}=\frac{-7}{3 \times 4} \quad g^{\prime}(c)=\frac{-7}{12}
$$
$$
F(0)=4 \text { and } F(4)=-2 \quad g(x)=\frac{f(x)}{x+2}
$$
At $x=0, g(0)=\frac{f(0)}{0+2}=\frac{4}{2}=2$
At $x=4, g(4)=\frac{f(4)}{4+2}=\frac{-2}{6}=\frac{-1}{3}$
Now,
$$
g^{\prime}(c)=\frac{g(4)-g(0)}{4-0}=\frac{\frac{-1}{3}-2}{4}=\frac{-7}{3 \times 4} \quad g^{\prime}(c)=\frac{-7}{12}
$$
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