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Chapter 1: Problem 49

The set of real numbers satisfying the given inequality is one or more intervals on the number line. Show the interval(s) on a number line. $$|x|<4$$

Short Answer

Expert verified
The solution interval is \((-4, 4)\) on the number line.

Step by step solution

01

Interpret the Absolute Value Inequality

The inequality \(|x| < 4\) means that the distance of \(x\) from 0 on the number line is less than 4.
02

Remove the Absolute Value

The expression \(|x| < 4\) can be rewritten as two separate inequalities: \(x > -4\) and \(x < 4\). This is because \(|x|\) represents how far away \(x\) is from 0 regardless of direction.
03

Identify the Solution Interval

Combine the inequalities from Step 2 to form the compound inequality \(-4 < x < 4\). This represents the set of all \(x\) values that satisfy both \(x > -4\) and \(x < 4\).
04

Express the Solution Interval Using Interval Notation

The solution to the inequality is expressed in interval notation as \((-4, 4)\). This interval includes all real numbers between -4 and 4, but not inclusive of -4 and 4 themselves.
05

Visualize on a Number Line

On a number line, sketch the interval \((-4, 4)\). Typically, open circles are drawn at -4 and 4 to indicate that these endpoints are not included in the interval. A line is drawn between these circles to show all numbers between them are included.

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Key Concepts

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

Number Line Representation
Representing inequalities on a number line can make understanding them much simpler. A number line visually shows where the set of all solutions falls along the real number continuum. For the inequality \(|x| < 4\), you want to represent all the values of \(x\) that have an absolute distance from 0 that is less than 4. To do this:
  • Place open circles on -4 and 4 to represent that these values are not included in the solution (since the inequality uses \(<\), not \(\leq\)).
  • Draw a line connecting these open circles to show that every value between -4 and 4 is part of the solution set.
  • This line visually confirms that any number you pick inside this highlighted segment will satisfy \(|x| < 4\).
This visualization is sometimes more intuitive than reading the symbolic notation because you can directly "see" the range of solutions.
Interval Notation
Interval notation is a shorthand way to write the solution set of an inequality, capturing the beginning and end of an interval. For \(|x| < 4\), the interval notation is \((-4, 4)\). This notation presents:
  • A parenthesis \(()\) on both sides, indicating that -4 and 4 are not included in the solutions since it鈥檚 an open interval due to the \(<\) operator.
  • The values inside the parentheses, -4 and 4, are the boundaries of the solution set.
When interpreting interval notation, remember:
  • Brackets \([\cdot, \cdot]\) would be used if 4 and -4 were included, which would occur with \(\leq\) or \(\geq\) inequalities.
  • Always consider the context in question to determine if you need open or closed intervals.
This concise format is very helpful when communicating solutions in a standardized mathematical language.
Compound Inequalities
Compound inequalities involve combining two separate inequalities into one. The expression \(|x| < 4\) creates two conditions for the variable \(x\):
  • \(x > -4\)
  • \(x < 4\)
These are simple inequalities that together describe the solutions where both conditions hold true simultaneously, creating the compound inequality \(-4 < x < 4\). This compound inequality portrays:
  • "And" logic, meaning \(x\) must meet both conditions at the same time.
  • Provides a clear picture of the set of all \(x\) values that satisfy the inequality.
When solving such problems, isolating the variable within this combined framework gives clarity about the range of values meeting the original condition, which in this case, describes all numbers strictly between -4 and 4.

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Most popular questions from this chapter

(As background for this exercise, you might want to work Exercise \(23 .)\) Prove that $$ \max (a, b)=\frac{a+b+|a-b|}{2} $$ Hint: Consider three separate cases: \(a=b ; a>b ;\) and \(b>a\).

The set of real numbers satisfying the given inequality is one or more intervals on the number line. Show the interval(s) on a number line. $$|x|<2$$

Show that the slope of the line passing through the two points \(\left(a, a^{3}\right)\) and \(\left(x, x^{3}\right)\) is \(x^{2}+a x+a^{2} .\) Hint: You'll need to use difference of cubes factoring from intermediate algebra. If you need a review, see Appendix B.4.

Use a graphing utility to graph the equations and to approximate the \(x\) -intercepts. In approximating the \(x\) -intercepts, use a "solve" key or a sufficiently magnified view to ensure that the values you give are correct in the first three decimal places. Remark: None of the \(x\) -intercepts for these four equations can be obtained using factoring techniques.) $$y=x^{3}-3 x+1$$

Graph the equation after determining the \(x\) - and \(y\) -intercepts and whether the graph possesses any of the three types of symmetry described on page 58 $$y=|x|-2$$
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