Absolute-value inequality with five zeros

Text

Solve the inequality $x|x^3-x|-3x|x^2-1|>0.$ The graph of $y=x|x^3-x|-3x|x^2-1|$ touches the $x$-axis at $5$ points: use the factorisation to find them.

Solution

Since $x^3-x=x(x^2-1)$, we have $|x^3-x|=|x|\,|x^2-1|$. Factor out: $x|x|\,|x^2-1|-3x|x^2-1|=|x^2-1|\cdot x(|x|-3).$ The factor $|x^2-1|\ge0$ vanishes at $x=\pm1$; the factor $x(|x|-3)$ vanishes at $x=0,\pm3$: these are the 5 zeros $-3,-1,0,1,3$. Since $|x^2-1|\ge0$, the sign (for $x\neq\pm1$) equals that of $x(|x|-3)$:

• for $x>0$: $x(x-3)>0 \iff x>3$;
• for $x<0$: $x(-x-3)>0 \iff -3<x<0$.

Hence the (strict) inequality holds for $\boxed{-3<x<0\ (x\neq-1)\ \ \vee\ \ x>3}$