Calculus (3rd Edition)

$x= 1$ is a local minimum
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.