Question

Finding critical points: For each of the following functions $f$, find all of the $x$-values for which $f\text{'}\left(x\right)=0$and all of the $x$-values for which $f\text{'}\left(x\right)$does not exist.

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

Short answer

The point for which $f\text{'}\left(x\right)=0$is:-

$x=0$.

There is no points for which $f\text{'}\left(x\right)$does not exist.

Step-by-step solution

Step 1. Given Information

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.

Step 2. Find the derivative of the given function

We have given the following function :-

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

We can simplify this function as following :-

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

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

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

Step 3. Put f'(x)=0

We find that :-

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

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

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

Step 4. The values for which f'(x) does not exist    

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.