Question

If $f\left(x\right)=x^3-5x^2-3x$, verify conditions of the mean value theorem satisfied for $a=1,b=3$. Find $cϵ(1,3)$ such that $f^{\prime }\left(c\right)=\frac{f\left(3\right)-f\left(1\right)}{3-1}$

• A. $2$
• B. $\frac{5}{4}$
• C. $3$
• D. $\frac{7}{3}$

Answer 1

The correct option is D $\frac{7}{3}$

$f\left(x\right)$ is continuous in $(1,3)$
$f^{\prime }\left(x\right)$ also exist in $(1,3)$
$f^{\prime }\left(c\right)=\frac{-27+7}{2}=-10$
$3c^2-10c-3=-10$
$3c^2-10c+7=0$
$3c^2-3c-7c+7=0$
$3c(c-1)-7(c-1)$
$c=\frac{7}{3}$ , $c=1$

$1$ does not belong to $(1,3)$

$\therefore$ It will not be an answer.

Hence, $c=\frac{7}{3}$.