Chapter 4, Problem 42RE

To determine

To solve:The inequality $\frac{6}{x+3}\ge 1$ . Also, graph the solution set.

Answer to Problem 42RE

The interval is $(-\infty ,3)\cup (3,\infty )$ .

Explanation of Solution

Given information:

The inequality $\frac{6}{x+3}\ge 1$ .

Formula used:

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

Calculation:

Consider ,

  $\frac{6}{x+3}\ge 1$

Step 1.

Re-arranging the inequality

  $\begin{array}{l}=6\ge x+3\\ =6-x-3\ge x+3-x-3\\ =3-x\ge 0\end{array}$ .

Step 2.

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

  $\begin{array}{l}=3-x=0\\ =x=3\end{array}$

Step 3.

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

  $(-\infty ,3);(3,\infty )$

Step 4.

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

Interval	$\left(-\infty ,3\right)$	$\left(3,\infty \right)$
Number chosen	$0$	$4$
Value of $f$	$f(x)=3-0=3$	$f(x)=3-4=-1$
Conclusion	Positive	Negative

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 $x<3$ . Hence , the solution set of the inequality $\frac{6}{x+3}\ge 1$ is $(-\infty ,3)\cup (3,\infty )$ .