Question

The value of $c$ of mean value theorem when $f\left(x\right)=x^3-3x-2$ in $[-2,3]$ is

• A. $\sqrt{\frac{7}{3}}$
• B. $\sqrt{\frac{3}{7}}$
• C. $\frac{\sqrt{7}}{3}$
• D. $\frac{\sqrt{3}}{7}$

Answer 1

Answer: (A)

$\sqrt{\frac{7}{3}}$

The definition of the mean value theorem says $f^{\prime }\left(c\right)=\frac{f\left(b\right)-f\left(a\right)}{b-a}$ where $a<c<b$

Here, $f\left(x\right)=x^3-3x-2$ in $[-2,3]$

$\Rightarrow f^{\prime }\left(c\right)=\frac{f\left(3\right)-f(-2)}{3-(-2)}$

$\Rightarrow 3c^2-3=\frac{(27-9-2)-(-8+6-2)}{5}$

$\Rightarrow 3c^2-3=\frac{16+4}{5}=4$

$\Rightarrow 3c^2=7$

$\Rightarrow c=\sqrt{\frac{7}{3}}$