Question

Solve the nonlinear inequality. Express the solution using interval notation and graph the solution set.

$16x\le x^3$

Short answer

The solution set is $\left[-\text{4,0}\right]\cup \left[\text{4,}\infty \right]$.

The required graph is shown below:

Step-by-step solution

Step 1. Set the inequality equal to zero and factor out.

The inequality can be written as: $16x-x^{\text{3}}\le \text{0}$

Set the inequality equal to zero and factor out.

$\begin{array}{l}16x-x^{\text{3}}=\text{0}\\ x\left(\text{16}-x^{\text{2}}\right)=\text{0}\\ x\left(4-x\right)\left(4+x\right)=\text{0}\\ x=-4,\text{0},4\end{array}$

Step 2. Determine the sign.

The zeros divide the number line into 4 equal parts, i.e. $\left(-\infty ,-\text{4}\right),\left(-4,0\right),\left(0,4\right),\left(4,\infty \right)$.

Check the signs:

For x belongs to $\left(-\infty ,-\text{4}\right)$we get:

$16\left(-\text{5}\right)-{\left(-\text{5}\right)}^{\text{3}}=-\text{80}+\text{125}$(Positive number)

For x belongs to $\left(-4,0\right)$we get:

$16\left(-\text{2}\right)-{\left(-\text{2}\right)}^{\text{3}}=-\text{32}+\text{8}$(Negative number)

For x belongs to $\left(0,4\right)$we get:

$16\left(\text{1}\right)-{\left(\text{1}\right)}^{\text{3}}$(Positive number)

For x belongs to $\left(4,\infty \right)$we get:

$16\left(\text{5}\right)-{\left(\text{5}\right)}^{\text{3}}=\text{80}-\text{125}$(Negative number)

Step 3. Determine the solution set.

Since the value of inequality should be less than or equal to zero, therefore the required solution interval will be $\left[-4,0\right]\cup \left[4,\infty \right)$.

The required graph is shown below: