Chapter 4, Problem 44RE

To determine

To solve:The inequality $\frac{(x-2)(x-1)}{x-3}\ge 0$ . Also, graph the solution set.

Answer to Problem 44RE

The interval is $(-\infty ,1]\cup [2,\infty )$ .

Explanation of Solution

Given information:

The inequality $\frac{(x-2)(x-1)}{x-3}\ge 0$ .

Formula used:

The zeros of the function $f(x)=0$ .

Calculation:

Consider ,

  $\frac{(x-2)(x-1)}{x-3}\ge 0$

Step 1.

Re-arranging the inequality

  $\begin{array}{l}=(x-2)(x-1)\ge 0\\ =x(x-1)-2(x-1)\ge 0\\ =x^2-x-2x+2\ge 0\\ =x^2-3x+2\ge 0\end{array}$ .

Step 2.

The zeros of $f(x)=x^2-3x+2$ .

  $\begin{array}{l}=x^2-3x+2=0\\ =x^2-2x-x+2=0\\ =x(x-2)-1(x-2)=0\\ =(x-1)(x-2)=0\end{array}$

  $\begin{array}{l}=x-1=0\\ =x=1\end{array}$ or $\begin{array}{l}=x-2=0\\ =x=2\end{array}$

Step 3.

Using, the zeros to separate the real number line into one interval:

  $(-\infty ,1);(1,2);(2,\infty )$

Step 4.

Evaluate the $f(x)=x^2-3x+2$ at each number.

Interval	$\left(-\infty ,1\right)$	$\left(1,2\right)$	$\left(2,\infty \right)$
Number chosen	$0$	$1.5$	$4$
Value of $f$	$f(x)=0-3(0)+2=2$	$\begin{array}{l}f(x)={(}^{1.5}-3(1.5)+2\\ =2.25-4.5+2=-0.25\end{array}$	$\begin{array}{l}f(x)={(}^4-3(4)+2\\ =16-12+2=6\end{array}$
Conclusion	Positive	Negative	Positive

Now, using the above values we can draw a number line as follows:

  

Figure 1.

Step 5.

Using the information gathered from Step 1 − Step 4 we concluded that $f(x)\ge 0$ for all numbers x for $x\le 1$ and $x\ge 2$ . Hence , the solution set of the inequality $\frac{(x-2)(x-1)}{x-3}\ge 0$ is $(-\infty ,1]\cup [2,\infty )$ .