College Algebra, Chapter 1, 1.6, Section 1.6, Problem 56

Solve the nonlinear inequality $16x \leq x^3$. Express the solution using interval notation and graph the solution set.

$
\begin{equation}
\begin{aligned}
16x &\leq x^3\\
\\
0 &\leq x^3 - 16 x && \text{Subtract } 16x\\
\\
0 &\leq x (x^2 - 16) && \text{Factor } x\\
\\
0 &\leq (x+4)(x-4) && \text{Difference of square}
\end{aligned}
\end{equation}
$


The factors on the left hand side are $x$, $x+4$ and $x-4$. These factors are zero when $x$ is 0, -4 and 4 respectively. These numbers divide the real line into intervals
$(-\infty,-4],(-4,0),(0,4),[4,\infty)$




From the diagram, the solution of the inequality $x^2 (x+4)(x-4) \geq 0$ are
$(-\infty, -4] \bigcup [4, \infty)$

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