Question

The value of '$c$' of Lagranges Mean value theorem for $f\left(x\right)=\sqrt{x}$, when $a=1$ and $b=4$ is:

• A. $\frac{1}{2}$
• B. $\frac{9}{4}$
• C. $\frac{1}{4}$
• D. $\frac{3}{2}$

Answer 1

The correct option is B $\frac{9}{4}$

$f\left(x\right)=\sqrt{x}$, where $a=1,b=4$

Checking conditions for Mean value theorem.

Condition $1:$

$f\left(x\right)=\sqrt{x}$ is continuous at $(1,4)$

Since $f\left(x\right)$ is polynomial.

It is continuous in $(1,4)$

Condition $2:$

If $f\left(x\right)$ is differentiable

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

$f\left(x\right)$ is a polynomial and every polynomial function is differentiable

$\Rightarrow$ $f\left(x\right)$ is differentiable at $x\in [1,4]$

Condition $3:$

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

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

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

$f\left(a\right)=f\left(1\right)$

$=\sqrt{1}=1$

$f\left(b\right)=f\left(4\right)$

$=\sqrt{4}$

$=2$

By Lagranges mean value theorem,

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

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

$\frac{1}{2\sqrt{c}}=\frac{1}{3}$

$\sqrt{c}=\frac{3}{2}$

$\therefore$ $c=\frac{9}{4}$