Answer
$x= 1$ is a local minimum
Work Step by Step
Given $$ g(x)=x-\ln x$$
Since
\begin{align*}
g^{\prime}(x)&= 1-\frac{1}{x}
\end{align*}
Then $g(x)$ has critical points when
\begin{align*}
g'(x)&=0\\
\frac{x-1}{x}&=0
\end{align*}
Then $x=1$ is a critical point.
Now, we use the second derivative to check $x=1 $
\begin{align*}
g''(x) &=\frac{1}{x^2}
\end{align*}
Hence $$g''(1)=1>0 $$
Since the second derivative at the critical point is positive, we have a minimum.
Then $x= 1$ is a local minimum.
