Chapter 2, Problem 26CR

To determine

To solve and graph: the given inequality

Answer to Problem 26CR

The required result is $f\ge 3\text{ and }f\le 9$

  

Explanation of Solution

Given:

  $4|f-6|\le 12$

Calculation:

In solving absolute value inequalities, remember that $|ax+b|<c$ provided that $c>0$ will either be:

  $ax+b>-c\text{ AND }ax+b<c$

Also, $|ax+b|>c$ provided that $c>0$ will either be:

  $ax+b<-c\text{ OR }ax+b>c$

In graphing the inequalities, remember that inequalities joined by "and" is the intersection of the inequalities while those joined by "or" is the union of the inequalities.

  $4|f-6|\le 12$

  $\frac{4|f-6|}{4}\le \frac{12}{4}$

Divide both sides by 4

  $f-6\ge -3$ and $f-6\le 3$

Write a compound inequality

  $f-6+6\ge -3+6$ and $f-6+6\le 3+6$

Add both sides by 3

  $f\ge 3\text{ and }f\le 9$

Simplify

Since the inequalities are joined by "and", take the intersection of the inequalities as its solution set.

Therefore, its graph should have a closed point at 3 and a shaded line going to the right until the closed point at 9, as in:

  

Conclusion:

The required result is $f\ge 3\text{ and }f\le 9$