## Precalculus: Mathematics for Calculus, 7th Edition

$-\sqrt{2} < x < \sqrt{2}$
Factor out $e^{x}:$ $e^{x}(x^{2}-2) < 0$ The function $f(x)=e^{x}$ is always positive, never zero, so LHS $< 0$ only if $x^{2}-2 < 0$ $x^{2}<2$ This is valid only if $|x| < 2$, that is, if $-\sqrt{2} < x < \sqrt {2}$