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Chapter 1: Problem 84

Write an inequality of the form \(|x-a| < k\) or of the form \(|x-a| > k\) so that the inequality has the given solution set. HINT: \(|x-a| < k\) means that \(x\) is less than \(k\) units from \(a\) and \(|x-a|>k\) means that \(x\) is more than \(k\) units from \(a\) on the number line. $$(-\infty,-1) \cup(5, \infty)$$

Short Answer

Expert verified
The inequality is \(|x+1| > 6\) or \(|x-5| > 6\).

Step by step solution

01

Understand the Solution Set on the Number Line

The given solution set is \((-\infty, -1) \cup (5, \infty)\). This means that x must be less than -1 or greater than 5.
02

Identify the Center (Midpoint) of the Intervals

The midpoint between -1 and 5 is \[ \frac{-1+5}{2} = 2 \], but in this case, we should consider the points -1 and 5 separately due to the characteristics of the solution set.
03

Determine the Distance from x to Points -1 and 5

The inequality \(|x+1| > k\) ensures that x is more than k units away from -1. Similarly, the inequality \(|x-5| > k\) ensures that x is more than k units away from 5.
04

Formulate the Inequality

Recognizing that the union of two intervals means x is outside the range defined by these points, write the inequality reflecting the absolute distance. Here, \(|x + 1| > 6\) or \(|x - 5| > 6 \) ensures x is outside the interval between -1 and 5.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

number line
When dealing with absolute value inequalities, it's helpful to visualize the problem on a number line.
A number line allows us to see where the values of x are in relation to certain points. For example, in this problem, we are given the solution set (\(-\forall, -1) \bigcup (5, \forall\)).
This means that x must be either less than -1 or greater than 5. We use the number line to place these critical points, -1 and 5.
Everything to the left of -1 and everything to the right of 5 is part of the solution.
Visualizing the problem on a number line helps us see the ranges and better understand the absolute distances involved.
solution set
The solution set of an inequality tells us all of the possible values that x can take to make the inequality true.
For the given problem, the solution set is (\(-\forall, -1) \bigcup (5, \forall\)), which includes all values less than -1 and greater than 5. To express this mathematically, we use set notation and interval notation.

This means x can be any number in two separate ranges:
  • \(x < -1\)
  • \(x > 5\)

The union symbol (\(\bigcup\)) means we combine these intervals to represent all the values that satisfy our inequality.
inequality formulation
Formulating an inequality involving absolute values is about capturing the distance between x and a reference point.

In this problem, the reference points are -1 and 5. To express that we want values of x that are more than a certain distance away from these points, we write two separate absolute value inequalities:
(|x + 1| > k) and (|x - 5| > k)
where k represents the critical distances from these points.

Knowing that x must be either less than -1 or greater than 5, we have to determine k, so that these conditions are met. These inequalities tell us that the distance of x from both -1 and 5 must be greater than some k.
absolute distance
Absolute value measures how far a number is from zero, regardless of direction.
In the context of this problem, we're interested in how far x is from -1 and 5.

For example, the inequality (|x + 1| > k) means the distance between x and -1 is greater than k units. Similarly, (|x - 5| > k) means the distance between x and 5 is greater than k units.
We use these absolute distances to express that x must stay either to the left of -1 or to the right of 5 on the number line.

In the final step, replacing k with 6 gives us the inequalities (|x + 1|> 6) and (|x - 5|> 6), ensuring x falls within the intended ranges.

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Most popular questions from this chapter

Find an exact solution to each problem. If the solution is irrational, then find an approximate solution also. Initial Velocity of a Basketball Player Vince Carter is famous for his high leaps and "hang time," especially while slam-dunking. If Carter leaped for a dunk and reached a peak height at the basket of \(1.07 \mathrm{m},\) what was his upward velocity in meters per second at the moment his feet left the floor? The formula \(v_{1}^{2}=v_{0}^{2}+2 g S\) gives the relationship between final velocity \(v_{1}\), initial velocity \(v_{0}\), acceleration of gravity \(g,\) and his height \(S\). Use \(g=-9.8 \mathrm{m} / \mathrm{sec}^{2}\) and the fact that his final velocity was zero at his peak height. Use \(S=\frac{1}{2} g t^{2}+v_{0} t\) to find the amount of time he was in the air. (figure cannot copy)

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