Question

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 $c \in(0,6), g^{\prime}(c)=$

• A. $-\frac{44}{3}$
• B. $-\frac{22}{21}$
• C. $\frac{32}{21}$
• D. $-\frac{44}{21}$

Answer 1

Answer: (D)

According to Lagrange's Mean value theorem
$g^{\prime}(c) =\frac{g(6)-g(0)}{6-0}=\frac{\frac{f(6)}{6+1}-\frac{f(0)}{0+1}}{6} $
$=\frac{-\frac{4}{7}-12}{6}=-\frac{88}{42}=-\frac{44}{21}$