# Question #ceea8

Nov 25, 2014

Since you can plug in any real number for $x$, the domain of $f$ is

${D}_{f} = \left(- \infty , \infty\right)$.

Since ${x}^{2} \ge 0$, we have

$f \left(x\right) = {x}^{2} - 2 \ge - 2$,

which means that the range of $f$ is

${R}_{f} = \left[- 2 , \infty\right)$

I hope that this was helpful.