91Ó°ÊÓ

We have given the following function :-

$f\left(x\right)=3x\left(x+\frac{1}{x}\right)$

We have to find the points for which $f\text{'}\left(x\right)=0$.

Also we have to find the points for which $f\text{'}\left(x\right)$does not exist.

Firstly we will find the derivative, then we will find the required points.

We have given the following function :-

$f\left(x\right)=3x\left(x+\frac{1}{x}\right)$

We can simplify this function as following :-

$$\begin{gathered} f\left(x\right)=3x\left(x\right)+3x\left(\frac{1}{x}\right) \\ ⇒f\left(x\right)=3x^2+3 \end{gathered}$$

Use power rule to differentiate this function, then we have :-

$$\begin{gathered} f\text{'}\left(x\right)=\frac{d}{dx}3x^2+\frac{d}{dx}3 \\ ⇒f\text{'}\left(x\right)=3\left(2x\right)+0 \\ ⇒f\text{'}\left(x\right)=6x \end{gathered}$$

We find that :-

$f\text{'}\left(x\right)=6x$

Now put $f\text{'}\left(x\right)=0$, then we have :-

$$\begin{gathered} 6x=0 \\ ⇒x=0 \end{gathered}$$

We find that :-

$f\text{'}\left(x\right)=6x$.

We know that a function does not exist where the denominator is equals to zero. But here denominator is constant that is $1$. So it cannot be equals to zero.

So we can conclude that there is no point for which $f\text{'}\left(x\right)$does not exist.