## Calculus (3rd Edition)

$$x=\pm \sqrt 2.$$
The zeros of $f(x)=2x^2-4$ can be obtained by putting $$2x^2-4=0\Longrightarrow x=\pm \sqrt 2.$$ Since $f(-x)=f(x)$, then the function is symmetric about the y-axis and increasing on the interval $(0,\infty)$ as in the figure below.