Answer
$$
f(x)=2 x+\ln x
$$
$f(x)$ is always increasing on its domain. So $f$ has no relative extrema.
Work Step by Step
$$
f(x)=2 x+\ln x
$$
To find the intervals where the function is increasing and decreasing.
First, we find $f^{\prime}(x) $,
$$
\begin{aligned}
f^{\prime}(x) &=2+\frac{1}{x} \\
&=\frac{2 x+1}{x}
\end{aligned}
$$
To find the critical numbers, we first find any values of $x$ that make $f^{\prime} =0 $. Now, solve the equation $f^{\prime} =0 $ to get
$$
\begin{aligned}
f^{\prime}(x) =\frac{2 x+1}{x}&=0\\
\Rightarrow\quad\quad\quad\quad\quad\\
2 x+1 &=0\\
\Rightarrow\quad\quad\quad\quad\quad\\
2 x& =-1 \\
x=\frac{-1}{2}\\
\end{aligned}
$$
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 equal to 0, when $x=0$. But , $f(x) $ is not defined also at $x=0 $. Since, the domain of $f(x)$ is $(0, \infty) .$
Therefore $f^{\prime}(x)$ is never zero in the domain of $f(x)$.
Next,
$f^{\prime}(1)=3>0. $
Since $f(x)$ is always increasing, so $f$ has no relative extrema.
