## Precalculus: Mathematics for Calculus, 7th Edition

The solution is $(-4,0)\cup(4,\infty)$ The graph is:
$16x\le x^{3}$ Take $16x$ to the right side: $0\le x^{3}-16x$ $x^{3}-16x\ge0$ Factor the left side: $x(x^{2}-16)\ge0$ $x(x-4)(x+4)\ge0$ Find the intervals. The factors are $x$, $x-4$ and $x+4$. Set them equal to $0$ and solve for $x$: $x=0$ $x-4=0$ $x=4$ $x+4=0$ $x=-4$ The factors are zero when $x=0,4,-4$. These three numbers divide the real line into the following intervals: $(-\infty,-4)$ $,$ $(-4,0)$ $,$ $(0,4)$ $,$ $(4,\infty)$ Elaborate a diagram, using test points to determine the sign of each factor in each interval: (refer to the attached image below) It can be seen from the diagram that the intervals $(-4,0)$ and $(4,\infty)$ satisfy the inequality. Also, the inequality involves $\ge$ so the endpoints satisfy the inequality too. The solution is $(-4,0)\cup(4,\infty)$