Question

In the mean value theorem, $f\left(b\right)-f\left(a\right)=(b-a)f’\left(c\right)$if $a=4,b=9$ and $f\left(x\right)=\sqrt{x}$then the value of $c$ is

• A. $8.00$
• B. $5.25$
• C. $4.00$
• D. $6.25$

Answer 1

The correct option is D

$6.25$

Explanation for the correct option:

Finding the value of $c$:

Given $f\left(b\right)-f\left(a\right)=(b-a)f’\left(c\right)$if $a=4,b=9$ and $f\left(x\right)=\sqrt{x}$

$\because f\left(x\right)=\sqrt{x}$ then we get, $f\text{'}\left(x\right)=\frac{1}{2\sqrt{x}}$ after differentiating w.r.t $x$

$$\begin{gathered} \Rightarrow f\left(b\right)-f\left(a\right)=(b-a)f\text{'}\left(c\right) \\ \Rightarrow f\text{'}\left(c\right)=\frac{f\left(b\right)-f\left(a\right)}{(b-a)} \\ \Rightarrow f\text{'}\left(c\right)=\frac{\sqrt{b}-\sqrt{a}}{(b-a)}\left[\because f\left(x\right)=\sqrt{x}\right] \\ \Rightarrow \frac{1}{2\sqrt{c}}=\frac{\sqrt{9}-\sqrt{4}}{(9-4)}\left[\because f\text{'}\left(x\right)=\frac{1}{2\sqrt{x}}\&a=4,b=9\right] \\ \Rightarrow \sqrt{c}=\frac{5}{2} \\ \Rightarrow c=6.25 \end{gathered}$$

Hence, option (D) is the correct answer.