Question

A function $f\left(x\right)=1+\left(\frac{1}{x}\right)$ is defined on the closed interval $[1,3]$. A point in the interval, where the function satisfies the mean value theorem, is

• A. $x=\frac{1}{3}$
• B. $x=\frac{1}{\sqrt{3}}$
• C. $x=\sqrt{3}$
• D. None of these

Answer 1

The correct option is C

$x=\sqrt{3}$

Explanation for the correct answer:

Step 1: Finding the derivative

The function is $f\left(x\right)=1+\left(\frac{1}{x}\right)$, and the interval is $\left[a,b\right]=[1,3]$

Take the first derivative with respect to $x$.

$$\begin{gathered} \Rightarrow f\text{'}\left(x\right)=0-\frac{1}{x^2} \\ =-\frac{1}{x^2} \end{gathered}$$

Now put $x=c,f\text{'}\left(x\right)$ becomes

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

Step 2: Find $f\left(a\right),f\left(b\right)$

Put $a=1$in $f\left(x\right)$,

$$\begin{gathered} \Rightarrow f\left(a\right)=1+\frac{1}{1} \\ =2 \end{gathered}$$

Put $b=3$ in $f\left(x\right)$,

$$\begin{gathered} \Rightarrow f\left(b\right)=1+\frac{1}{3} \\ =\frac{4}{3} \end{gathered}$$

Step 3: Apply mean value theorem

The formula for mean value theorem is

$\mathit{f}\mathbf{\text{'}}\mathbf{\left(}\mathit{c}\mathbf{\right)}\mathbf{=}\frac{\mathbf{f}\mathbf{\left(}\mathbf{b}\mathbf{\right)}\mathbf{-}\mathbf{f}\mathbf{\left(}\mathbf{a}\mathbf{\right)}}{\mathbf{b}\mathbf{-}\mathbf{a}}$

On substituting the values of $f\text{'}\left(c\right),f\left(a\right),f\left(b\right)$

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

Hence, a point in the interval where the function satisfies the mean value theorem is option (c) $\sqrt{\mathbf{3}}$ is the correct answer.