Answer
$$
f(x) =x-\frac{1}{x}
$$
Since $f(x) $ is not defined at $x=0 $. therefore, there are
no critical numbers and no relative extrema.
Work Step by Step
$$
f(x) =x-\frac{1}{x}
$$
To find the intervals where the function is increasing and decreasing.
First, we find $f^{\prime}(x) $,
$$
\begin{aligned}
f^{\prime}(x) &=1+\frac{1}{x^{2}}\\
\end{aligned}
$$
To find the critical numbers, we first find any values of $x$ that make $f^{\prime} =0 $. But, the equation $f^{\prime} $ is never equal zero.
Next, we find any values of $x$ where $f^{\prime}(x) $ fails to exist. This occurs whenever the denominator of $f^{\prime}(x) $ is $0$, here we observe that the denominator $x^{2}$ equal to 0, when $x=0$ So,
The only critical number : $x=0$ .
Since $f(x) =x-\frac{1}{x}$ is not defined at $x=0 $. therefore, there are
no critical numbers and no relative extrema.
