Chapter 2.8, Problem 41E

Solution

To solve: The given inequality using sign chart.

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

Given information:

Consider the given inequality.

  $x\left|x-2\right|>0$

Calculation:

To set given expression equal to zero. These are the values that make the expression zero and solve for $x$ :

  $\begin{array}{c}x\left|x-2\right|=0\\ x=0,2\end{array}$

So, the boundary points are $0,2$ .

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 three test intervals. The intervals are below:

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

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

  $\begin{array}{l}\left(-\infty ,0\right),\left(0,2\right),\left(2,\infty \right)\\ -1\left|-1-2\right|>01\left|1-2\right|>03\left|3-2\right|>0\\ -3>0\text{False }1>0\text{True}3>0\text{True}\end{array}$

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

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