Question

Let $g\left(x\right)=\frac{f\left(x\right)}{x+1}$ where $f\left(x\right)$ is differentiable on $[0,5]$ such that $f\left(0\right)=4,f\left(5\right)=-1$. There exists $c\in (0,5)$ such that $g^{{}^{\prime }}\left(c\right)$ is ?

• A. $-\frac{1}{6}$
• B. $\frac{1}{6}$
• C. $-\frac{5}{6}$
• D. $-1$

Answer 1

The correct option is C $-\frac{5}{6}$

$g\left(x\right)=\frac{f\left(x\right)}{x+1}$

$\Rightarrow g\left(0\right)=\frac{4}{1}=4;g\left(5\right)=\frac{-1}{6}$

By Lagrange's mean value theorem,

$g^{\prime }\left(c\right)=\frac{g\left(5\right)-g\left(0\right)}{5-0}=\frac{\frac{-1}{6}-4}{5}$

$=\frac{-25}{30}=\frac{-5}{6}$