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Graph \(|x|<2\). [Hint: Remember that \(|x|<2\) is equivalent to \(-2

Short Answer

Expert verified
Graph an open interval between -2 and 2 on the number line.

Step by step solution

01

Understand the Expression

The expression \(|x| < 2\) means we want all the values of \(x\) where the absolute value of \(x\) is less than 2. The absolute value of \(x\) represents the distance of \(x\) from zero, regardless of direction. Hence, it is always non-negative.
02

Convert Absolute Inequality to Compound Inequality

The inequality \(|x| < 2\) can be rewritten as a compound inequality. According to the property of absolute value, \(|x| < c\) is equivalent to \(-c < x < c\). Therefore, \(|x| < 2\) becomes \(-2 < x < 2\). This shows that \(x\) can take any value between \(-2\) and \(2\), but not including \(-2\) and \(2\) themselves.
03

Determine the Interval

The inequality \(-2 < x < 2\) represents the interval where the values of \(x\) lie. Specifically, \(x\) belongs to the open interval \((-2, 2)\), which means \(x\) can be any real number between \(-2\) and \(2\).
04

Graph the Solution on the Number Line

To graph \(-2 < x < 2\) on a number line, you draw an open circle at \(-2\) (since \(x\) does not equal \(-2\)) and another open circle at \(2\) (since \(x\) does not equal \(2\)), and shade the line between these points. The shaded region represents all the values of \(x\) satisfying \(|x| < 2\).

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Key Concepts

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

Understanding Absolute Value Inequalities
Absolute value inequalities involve expressions where an absolute value is compared to a number. Understanding the absolute value is essential. It describes how far a number is from zero on a number line, without considering the direction. In simple terms, \(|x|\) represents the non-negative value of \(|x|\).

When we say \(|x| < 2\), we are interested in finding all possible values of \(x\) that are within a distance of 2 from zero on the number line. This does not just mean numbers like 1 or -1, which are clearly within 2 units from zero, but all real numbers that fit between -2 and 2.

To simplify solving these inequalities, we can use properties of absolute values to convert them into compound inequalities.
Fundamentals of Compound Inequalities
Compound inequalities are formed when two simple inequalities are combined into one. They express a range of values and are often encountered when dealing with absolute value inequalities.

For instance, the expression \(|x| < 2\) can be rewritten as a compound inequality: \(-2 < x < 2\). This implies \(x\) can have any value between -2 and 2, but not including those endpoints.

Such expressions are crucial because they provide a direct way to interpret the solution set of an absolute value inequality graphically on a number line. Compound inequalities help us see the relationships between values and provide clarity on the span of possible solutions.
  • Example: If \(|x| \leq 5\), this converts to \(-5 \leq x \leq 5\), covering all values from -5 to 5, including -5 and 5 themselves.
  • Solution Steps: First, remove the absolute value by writing two inequalities. Second, solve these inequalities separately if needed. Third, represent the solution set clearly, both algebraically and graphically.
Graphing on a Number Line
When it comes to graphing inequalities on a number line, it allows us to visualize which numbers satisfy the inequality. Let's focus on how to graph the inequality \(-2 < x < 2\).

To properly graph this, follow these steps:
  • Place open circles on -2 and 2. These open circles indicate that these values are not included in the solution set.
  • Shade the line between these two points. This shaded line represents all solutions that make the inequality true, namely all numbers greater than -2 and less than 2.
Graphing in this manner shows the range of solutions visually. You quickly grasp which numbers are included, and this clarity aids understanding, especially for visual learners.

A clear number line graph is an essential tool in seeing how inequalities form spans or intervals and provides a quick method to check solutions against inequality constraints.

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Most popular questions from this chapter

(a) Graph \(y=2 x-3, y=2 x+3, y=2 x-6\), and \(y=2 x+5\) on the same set of axes. (b) Graph \(y=-3 x+1, y=-3 x+4, y=-3 x-2\), and \(y=-3 x-5\) on the same set of axes. (c) Graph \(y=\frac{1}{2} x+3, y=\frac{1}{2} x-4, y=\frac{1}{2} x+5\), and \(y=\frac{1}{2} x-2\) on the same set of axes. (d) What relationship exists among all lines of the form \(y=3 x+b\), where \(b\) is any real number?

The cost (c) of playing an online computer game for a time \((t)\) in hours is given by the equation \(c=3 t+5\). Label the horizontal axis \(t\) and the vertical axis \(c\), and graph the equation for nonnegative values of \(t\).

What is the graph of \(x=0\) ? What is the graph of \(y=0\) ? Explain your answers.

The problem of finding the perpendicular bisector of a line segment presents itself often in the study of analytic geometry. As with any problem of writing the equation of a line, you must determine the slope of the line and a point that the line passes through. A perpendicular bisector passes through the midpoint of the line segment and has a slope that is the negative reciprocal of the slope of the line segment. The problem can be solved as follows: Find the perpendicular bisector of the line segment between the points for the following. Write the equation in standard form. (a) \((-1,2)\) and \((3,0)\) Find the perpendicular bisector of the line segment between the points \((1,-2)\) and \((7,8)\). The midpoint of the line segment is \(\left(\frac{1+7}{2}, \frac{-2+8}{2}\right)\) \(=(4,3)\). The slope of the line segment is \(m=\frac{8-(-2)}{7-1}\) \(=\frac{10}{6}=\frac{5}{3}\). Hence the perpendicular bisector will pass through the point \((4,3)\) and have a slope of \(m=-\frac{3}{5}\). $$ \begin{aligned} y-3 &=-\frac{3}{5}(x-4) \\ 5(y-3) &=-3(x-4) \\ 5 y-15 &=-3 x+12 \\ 3 x+5 y &=27 \end{aligned} $$ Thus the equation of the perpendicular bisector of the line segment between the points \((1,-2)\) and \((7,8)\) is \(3 x+5 y=27\). (b) \((6,-10)\) and \((-4,2)\) (c) \((-7,-3)\) and \((5,9)\) (d) \((0,4)\) and \((12,-4)\)

Hal purchased a 500 -minute calling card for \(\$ 17.50\). After he used all the minutes on that card, he purchased another card from the same company at a price of \(\$ 26.25\) for 750 minutes. Let \(y\) represent the cost of the card in dollars and \(x\) represent the number of minutes.

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