## Precalculus: Mathematics for Calculus, 7th Edition

$(-\infty, -1) ∪ (0, \infty)$
$$x^4>x^2$$ Subtract $x^2$ $$x^4-x^2>0$$ Factor out $$x^2(x^2-1)>0$$ $$x^2(x-1)(x+1)>0$$ So, we have key points: $x+1=0$ ; $x=-1$ $x=0$ $x-1=0$ ; $x=1$ Which gives us following intervals: $(-\infty, -1)$ - Positive $(-1, 0)$ - Negative $(0, 1)$ - Negative $(1, \infty)$ - Positive We need open interval of all positive set of values, so the answer is: $(-\infty, -1) ∪ (0, \infty)$