Question

Find the value of c, of mean value theorem, when

$f\left(x\right)=\sqrt{x^2-4}$ in the interval [2, 4].

• A. $\sqrt{6}$
• B. $3$
• C. $\sqrt{8}$
• D. $\sqrt{12}$

Answer 1

The correct option is A $\sqrt{6}$

We have $f\left(x\right)=\sqrt{x^2-4}$
$f^{\prime }\left(x\right)=\frac{1}{2}\frac{1}{\sqrt{x^2-4}}\left(2x\right)=\frac{x}{\sqrt{x^2-4}}$
$\therefore f^{\prime }\left(c\right)=\frac{c}{\sqrt{c^2-4}}$
For $a=2,b=4f\left(2\right)=f\left(a\right)=\sqrt{2^2}-4=0$
$f\left(4\right)=f\left(b\right)=\sqrt{4^2-4}=\sqrt{12}.$
By
mean value theorem, we have $f\left(b\right)=f\left(a\right)+(b-a)f^{\prime }\left(c\right)$
$\sqrt{12}=0+\left(2\right)\frac{c}{\sqrt{c^2-4}}\Rightarrow c=\pm \sqrt{6}$
But $c\in (2,4)\therefore c=\sqrt{6}$