Answer
The solution is $(-4,0)\cup(4,\infty)$
The graph is:
Work Step by Step
$16x\le x^{3}$
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 $(-4,0)\cup(4,\infty)$
