Chapter 2.8, Problem 48E

Solution

To solve: The given inequality using sign chart.

The solution set is equal to $\left(-\infty ,-4^{\frac{1}{3}}\right]\cup \left(0,\infty \right)$ .

Given information:

Consider the given inequality.

  $x^2+\frac{4}{x}\ge 0$

Calculation:

Rewrite the given inequality:

  $\frac{x^3+4}{x}\ge 0$

To set numerator and the denominator equal to zero. These are the values that make the quotient zero or undefined and solve for $x$ :

  $\begin{array}{c}{x}^3+4=0\\ x^3=-4\\ x={\left(-4\right)}^{\frac{1}{3}}\\ x=-4^{\frac{1}{3}}\end{array}$

And

  $x=0$

So, the boundary points are $-4^{\frac{1}{3}},0$ .

Now, locate these boundary points on the number line (or sign chart) and dividing the number line into intervals and check the function $f\left(x\right)$ is positive or negative at each interval below:

  

The boundary points divide the number line into three test intervals. Including the boundary points (because of the given less than or equal to sign), the intervals are below:

  $\left(-\infty ,-4^{\frac{1}{3}}\right],\left[-4^{\frac{1}{3}},0\right),\left(0,\infty \right)$

Now, take one representative number within each test interval and substitute that number into the inequality below:

  $\begin{array}{l}\left(-\infty ,-4^{\frac{1}{3}}\right],\left[-4^{\frac{1}{3}},0\right),\left(0,\infty \right)\\ {\left(-2\right)}^2+\frac{4}{-2}\ge 0{\left(-1\right)}^2+\frac{4}{-1}\ge 0{\left(1\right)}^2+\frac{4}{1}\ge 0\\ 2\ge 0\text{True }-3\ge 0\text{False}5\ge 0\text{True}\end{array}$

Therefore, the intervals $\left(-\infty ,-4^{\frac{1}{3}}\right]$ and $\left(0,\infty \right)$ belongs to the solution set and the intervals $\left[-4^{\frac{1}{3}},0\right)$ does not belong to the solution set.

Hence, the solution set is $\left(-\infty ,-4^{\frac{1}{3}}\right]\cup \left(0,\infty \right)$ .