Chapter 2.8, Problem 45E

Solution

To solve: The given inequality using sign chart.

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

Given information:

Consider the given inequality.

  $\frac{x^3\left(x-2\right)}{{\left(x+3\right)}^2}<0$

Calculation:

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

  $\begin{array}{c}{x}^3\left(x-2\right)=0\\ x=0,2\end{array}$

And

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

The boundary points are $-3,0,2$ .

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 four test intervals below:

  $\left(-\infty ,-3\right),\left(-3,0\right),\left(0,2\right),\left(2,\infty \right)$

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

  $\begin{array}{l}\left(-\infty ,-3\right),\left(-3,0\right),\left(0,2\right),\left(2,\infty \right)\\ \frac{{\left(-4\right)}^3\left(-4-2\right)}{{\left(-4+3\right)}^2}<0\frac{{\left(-1\right)}^3\left(-1-2\right)}{{\left(-1+3\right)}^2}<0\frac{{\left(1\right)}^3\left(1-2\right)}{{\left(1+3\right)}^2}<0\frac{{\left(3\right)}^3\left(3-2\right)}{{\left(3+3\right)}^2}<0\\ -384<0\text{True }\frac{3}{4}<0\text{False}-\frac{1}{16}<0\text{True}\frac{3}{4}<0\text{False}\end{array}$

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

Hence, the solution set are $\left(-\infty ,-3\right)\cup \left(0,2\right)$ .