Chapter 1.6, Problem 70PPSE

Solution

To write an absolute value inequality to represent each situation.

  $\left|c-11\right|\le 1$

Given:

The workers at a hardware store take their morning break no earlier than 10 A.M. and no later than noon.

Concept Used:

• If there is the strict inequality symbol $<\text{ or}>$, then use a parenthesis $\left(\right)$ on the interval
• If there is the inequality symbol $\le \text{ or}\ge$, then use a bracket $\left[\right]$ on the interval
• To graph the endpoint with strict inequality $<\text{ or}>$, use the open parenthesis or a hollow circle at that endpoint.
• To graph the endpoint with inequality symbol $\le \text{ or}\ge$, use the bracket or a solid circle at that endpoint.
• Multiplying or dividing an inequality by a negative number reverse the inequality sign.
• The absolute value property: $\text{if }\left|x\right|>a,\text{then}x>a\text{ or }x<-a$
• The absolute value property: $\text{if }\left|x\right|<a,\text{then}-a<x<a$

Calculation:

Let c represent the time the workers take their break.

Note that the workers take their morning break no earlier than 10 A.M. and no later than noon, that is, at 12 P.M, so the midpoint of 10 and 12 is 11 and the difference of each endpoint from the midpoint is 1 unit.

So, the absolute value inequality that represents the situation is given by $\left|c-11\right|\le 1$ .

Thus, the required absolute value inequality that represent the given situation is $\left|c-11\right|\le 1$ .