Question

The value of $c$ in the Lagrange's mean value theorem for $f\left(x\right)=\sqrt{x-2}$ in the interval $[2,6]$ is

• A. $\frac{5}{2}$
• B. $3$
• C. $4$
• D. $\frac{9}{2}$

Answer 1

The correct option is B $3$

Given:

$f\left(x\right)=\sqrt{x-2}$

in interval [2,6]

We know that:

Lagrange's mean value theorem,

$f^{\prime }\left(c\right)=\frac{f\left(b\right)-f\left(a\right)}{b-a}$

Solution:

$f\left(x\right)=\sqrt{x-2}$

$f^{\prime }\left(x\right)=\frac{1}{2\sqrt{x-2}}$

Using LMVT,

$f^{\prime }\left(c\right)=\frac{f\left(b\right)-f\left(a\right)}{b-a}$

$\Rightarrow \frac{1}{2\sqrt{c-2}}=\frac{f\left(6\right)-f\left(2\right)}{6-2}$

$\Rightarrow \frac{1}{2\sqrt{c-2}}=\frac{\sqrt{6-2}-\sqrt{2-2}}{6-2}$

$\Rightarrow \frac{1}{2\sqrt{c-2}}=\frac{2-0}{4}$

$\Rightarrow \frac{1}{2\sqrt{c-2}}=\frac{1}{2}$

$\Rightarrow \sqrt{c-2}=1$

$\Rightarrow c-2=1$

$\Rightarrow c=3$