Chapter 2.8, Problem 47E

Solution

To solve: The given inequality using sign chart.

The solution set are equal to $\left(-\infty ,0\right)\cup \left(2^{\frac{1}{3}},\infty \right)$ .

Given information:

Consider the given inequality.

  $x^2-\frac{2}{x}>0$

Calculation:

Rewrite the given inequality:

  $\frac{x^3-2}{x}>0$

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

  $\begin{array}{c}{x}^3-2=0\\ x^3=2\\ x=2^{\frac{1}{3}}\end{array}$

And

  $x=0$

The boundary points are $0,2^{\frac{1}{3}}$ .

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 below:

  $\left(-\infty ,0\right),\left(0,2^{\frac{1}{3}}\right),\left(2^{\frac{1}{3}},\infty \right)$

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

  $\begin{array}{l}\left(-\infty ,0\right),\left(0,2^{\frac{1}{3}}\right),\left(2^{\frac{1}{3}},\infty \right)\\ {\left(-4\right)}^2-\frac{2}{-4}>0{\left(1\right)}^2-\frac{2}{1}>0{\left(2\right)}^2-\frac{2}{2}>0\\ \frac{33}{2}>0\text{True }-1>0\text{False}3>0\text{True}\end{array}$

Therefore, the intervals $\left(-\infty ,0\right)$ and $\left(2^{\frac{1}{3}},\infty \right)$ belongs to the solution set and the interval $\left(0,2^{\frac{1}{3}}\right)$ does not belong to the solution set.

Hence, the solution set are $\left(-\infty ,0\right)\cup \left(2^{\frac{1}{3}},\infty \right)$ .