Question

A function $f$ is differentiable in the interval $0 \leq x \leq 5$ such that $f(0)=4$ and $f(5)=-1 .$ If $g(x)=\frac{f(x)}{x+1},$ then there exists some $c \in(0,5)$ such that $g'(c)=$

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

Answer 1

Answer: (B)

Clearly $g(x)$ satisfies condition of LMVT
$\therefore \frac{g(5)-g(0)}{5-0}=g'( c ), c \in(0,5)$
$\Rightarrow \frac{\frac{f(5)}{6}-\frac{f(0)}{1}}{5}=g'( c )$
$\Rightarrow -\frac{5}{6}=g'(c)$