Question

If $f\left(x\right)=x+\frac{1}{x},x>0$ then its greatest value is

• A. $-2$
• B. $0$
• C. $3$
• D. None of these

Answer 1

The correct option is D

None of these

Explanation for the correct option:

Step 1: Finding the $f\text{'}\left(x\right)$ and the critical point:

Given that,

$f\left(x\right)=x+\frac{1}{x}$

Differentiate the given function with respect to $x$.

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

For critical point $f\text{'}\left(x\right)=0$,

$\begin{array}{rcl}1-\frac{1}{x^2}& =& 0\\ & \Rightarrow & x=-1or1\end{array}$

Since, $x>0$, we can take $x=1$.

Step 2: Find the $f\text{'}\text{'}\left(x\right)$:

Differentiate the $f\text{'}\left(x\right)$ with respect to $x$,

$\begin{array}{rcl}f\text{'}\text{'}\left(x\right)& =& 0-\left(\frac{-2}{x^3}\right)\\ & =& \frac{2}{x^3}\end{array}$

Step 3: Finding the greatest value of the function:

For the greatest value, $f\text{'}\text{'}\left(x\right)<0$

Now, substitute $x=1$in the $f\text{'}\text{'}\left(x\right)$, we get

$\begin{array}{rcl}f\text{'}\text{'}\left(1\right)& =& \frac{2}{1^3}\\ & =& 2\end{array}$

Here, $f\text{'}\text{'}\left(x\right)>0,$ so there is no greatest value of the function.

Hence, the correct option is D.