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

Answer 1

The correct option is C

$3$

Explanation for the correct option:

Find the value of $c$.

A function $f\left(x\right)=\sqrt{x-2}$ is given.

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

LaGrange's mean value theorem states that if a function $g\left(x\right)$ is continuous and differentiable in $[a,b]$ then there exists a value $c\in [a,b]$ for which $f\text{'}\left(c\right)=\frac{f\left(a\right)-f\left(b\right)}{a-b}$.

Here $[a,b]=[2,6]$

So,

$$\begin{gathered} \frac{1}{2\sqrt{c-2}}=\frac{\sqrt{2-2}-\sqrt{6-2}}{2-6} \\ \Rightarrow \frac{1}{2\sqrt{c-2}}=\frac{-\sqrt{4}}{-4} \\ \Rightarrow \frac{1}{2\sqrt{c-2}}=\frac{1}{2} \\ \Rightarrow \frac{1}{\sqrt{c-2}}=1 \\ \Rightarrow \sqrt{c-2}=1 \\ \Rightarrow c-2=1 \\ \Rightarrow c=3 \end{gathered}$$

Therefore, the required value of $c$ is $3$.

Hence, option $C$ is the correct answer.