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Mathematics LibreTexts

1.9.2: Absolute Value

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  • Page ID 59846

  • David Arnold
  • College of the Redwoods

Now that we have the fundamentals of piecewise-defined functions in place, we are ready to introduce the absolute value function. First, let’s state a trivial reminder of what it means to take the absolute value of a real number.

In a sense, the absolute value of a number is a measure of its magnitude, sans (without) its sign. Thus,

\[|7|=7 \qquad \text { and } \qquad |-7| =7 \nonumber \]

Here is the formal definition of the absolute value of a real number.

Definition: Absolute Value

To find the absolute value of any real number, first locate the number on the real line.

WeChatfe21ab7cdaad1f1de591d514b27d3961.png

The absolute value of the number is defined as its distance from the origin.

For example, to find the absolute value of 7, locate 7 on the real line and then find its distance from the origin.

WeChata761aa5433a8c3d11266fd5abcd2913b.png

To find the absolute value of −7, locate −7 on the real line and then find its distance from the origin.

WeChatc39901c112b7c3e585128057661b2373.png

Some like to say that taking the absolute value “produces a number that is always positive.” However, this ignores an important exception, that is,

\[|0|=0 \nonumber \]

Thus, the correct statement is “the absolute value of any real number is either positive or it is zero,” i.e., the absolute value of a real number is “not negative.”2 Instead of using the phrase “not negative,” mathematicians prefer the word “nonnegative.” When we take the absolute value of a number, the result is always nonnegative; that is, the result is either positive or zero. In symbols,

\[|x| \geq 0 \text { for all real numbers } x \nonumber \]

This makes perfect sense in light of Definition 2. Distance is always nonnegative.

However, the discussion above is not of sufficient depth to handle more sophisticated problems involving absolute value.

A Piecewise Definition of Absolute Value

Because absolute value is intimately connected with distance, mathematicians and scientists find it an invaluable tool for measurement and error analysis. However, we will need a formulaic definition of the absolute value if we want to use this tool in a meaningful way. We need to develop a piecewise definition of the absolute value function, one that will define the absolute value for any arbitrary real number x.

We begin with a few observations. Remember, the absolute value of a number is always nonnegative (positive or zero).

  • If a number is negative, negating that number will make it positive. | − 5| = −(−5) = 5, and similarly, | − 12| = −(−12) = 12. Thus, if x < 0 (if x is negative), then |x| = −x.
  • If x = 0, then |x| = 0.
  • If a number is positive, taking the absolute value of that number will not change a thing.

\[|5|=5, \text { and similarly, }|12|=12 \nonumber \]

Thus, if \(x>0\) (if \(x\) is positive), then \(|x|=x\).

We can summarize these three cases with a piecewise definition.

\[|x|=\left\{\begin{array}{ll}{-x,} & {\text { if } x<0} \\ {0,} & {\text { if } x=0} \\ {x,} & {\text { if } x>0}\end{array}\right. \nonumber \]

It is the first line in our piecewise definition (4) that usually leaves students scratching their heads. They might say “I thought absolute value makes a number positive (or zero), yet you have \(|x| = −x\); that is, you have the absolute value of x equal to a negative x.” Try as they might, this seems contradictory. Does it seem so to you?

However, there is no contradiction. If x < 0, that is, if x is a negative number, then −x is a positive number, and our intuitive notion of absolute value is not dissimilar to that of our piecewise definition (4). For example, if x = −8, then −x = 8, and even though we say “negative x,” in this case −x is a positive number.

If this still has you running confused, consider the simple fact that x and −x must have “opposite signs.” If one is positive, the other is negative, and vice versa. Consequently,

  • if x is positive, then −x is negative, but
  • if x is negative, then −x is positive.

Let’s summarize what we’ve learned thus far.

Summarizing the Definition on a Number Line

We like to use a number line to help summarize the definition of the absolute value of x.

WeChatd6ba895737ff52a6a145279fafa8da67.png

Some remarks are in order for this summary on the number line.

  • We first draw the real line then mark the “critical value” for the expression inside the absolute value bars on the number line. The number zero is a critical value for the expression x, because x changes sign as you move from one side of zero to the other.
  • To the left of zero, x is a negative number. We indicate this with the minus sign below the number line. To the right of zero, x is a positive number, indicated with a plus sign below the number line.
  • Above the number line, we simplify the expression |x|. To the left of zero, x is a negative number (look below the line), so |x| = −x. Note how the result −x is placed above the line to the left of zero. Similarly, to the right of zero, x is a positive number (look below the line), so |x| = x. Note how the result x is placed above the line to the right of zero.

In the piecewise definition of |x| in (4), note that we have three distinct pieces, one for each case discussed above. However, because |0| = 0, we can include this case with the piece |x| = x, if we adjust the condition to include zero.

\[|x|=\left\{\begin{array}{ll}{-x,} & {\text { if } x<0} \\ {x,} & {\text { if } x \geq 0}\end{array}\right. \nonumber \]

Note that this piecewise definition agrees with our discussion to date.

  • In the first line of equation (6), if x is a negative number (i.e., if \(x < 0\)), then the absolute value must change x to a positive number by negating. That is, |x| = −x.
  • In the second line of equation (6), if x is positive or zero (i.e., if \(x \geq 0\)), then there’s nothing to do except remove the absolute value bars. That is, |x| = x.

Because |0| = −0, we could just as well include the case for zero on the left, defining the absolute value with

\[|x|=\left\{\begin{array}{ll}{-x,} & {\text { if } x \leq 0} \\ {x,} & {\text { if } x>0}\end{array}\right. \nonumber \]

However, in this text we will always include the critical value on the right, as shown in Definition 5.

Constructing Piecewise Definitions

Let’s see if we can determine piecewise definitions for other expressions involving absolute value.

Example \(\PageIndex{1}\)

Determine a piecewise definition for |x − 2|.

First, set the expression inside the absolute value bars equal to zero and solve for x.

\[\begin{aligned} x-2 &=0 \\ x &=2 \end{aligned} \nonumber \]

Note that x − 2 = 0 at x = 2. This is the “critical value” for this expression. Draw a real line and mark this critical value of x on the line. Place the expression x − 2 below the line at its left end.

WeChatcde0f7a6577ffaa8ef9bef52276b8462.png

Next, determine the sign of x − 2 for values of x on each side of 2. This is easily done by “testing” a point on each side of 2 in the expression x − 2.

  • Take x = 1, which lies to the left of the critical value 2 on our number line. Substitute this value of x in the expression x − 2, obtaining

\[x-2=1-2=-1 \nonumber \]

which is negative. Indeed, regardless of which x-value you pick to the left of 2, when inserted into the expression x − 2, you will get a negative result (you should check this for other values of x to the left of 2). We indicate that the expression x − 2 is negative for values of x to the left of 2 by placing a minus (−) sign below the number line to the left of 2.

WeChata4b5e37ed22bde976a06d30e0de4c72e.png

  • Next, pick x = 3, which lies to the right of the critical value 2 on the number line. Substitute this value of x into the expression x − 2, obtaining

\[x-2=3-2=1 \nonumber \]

which is positive. Indeed, regardless of which x-value you pick to the right of 2, when inserted into the expression x − 2, you will get a positive result (you should check this for other values of x to the right of 2). We indicate that the expression x − 2 is positive for values of x to the right of 2 by placing a plus (+) sign below the number line to the right of 2 (see the number line above).

The next step is to remove the absolute value bars from the expression |x−2|, depending on the sign of x − 2.

  • To the left of 2, the expression x − 2 is negative (note the minus sign (−) below the number line), so |x − 2| = −(x − 2). That is, we have to negate x − 2 to make it positive. This is indicated by placing −(x − 2) above the line to the left of 2.

WeChat5d2a2507db05ef85ac6bd6f581abc31d.png

  • To the right of 2, the expression x − 2 is positive (note the plus sign (+) below the line), so |x − 2| = x − 2. That is, we simply remove the absolute value bars because the quantity inside is already positive. This is indicated by placing x − 2 above the line to the right of 2 (see the number line above).

We can use this last number line summary to construct a piecewise definition of the expression |x − 2|.

\[|x-2|=\left\{\begin{array}{ll}{-(x-2),} & {\text { if } x<2,} \\ {x-2,} & {\text { if } x \geq 2}\end{array}=\left\{\begin{array}{ll}{-x+2,} & {\text { if } x<2} \\ {x-2,} & {\text { if } x \geq 2}\end{array}\right.\right. \nonumber \]

Our number line and piecewise definition agree: |x − 2| = −(x − 2) to the left of 2 and |x − 2| = x − 2 to the right of 2. Further, note how we’ve included the critical value of 2 “on the right” in our piecewise definition.

Let’s summarize the method we followed to construct the piecewise function above.

Constructing a Piecewise Definition for Absolute Value

When presented with the absolute value of an algebraic expression, perform the following steps to remove the absolute value bars and construct an equivalent piecewise definition.

  • Take the expression that is inside the absolute value bars, and set that expression equal to zero. Then solve for x. This value of x is called a “critical value.” (Note: The expression inside the absolute value bars could have more than one critical value. We will not encounter such problems in this text.)
  • Place your critical value on a number line.
  • Place the expression inside the absolute value bars below the number line at the left end.
  • Test the sign of the expression inside the absolute value bars by inserting a value of x from each side of the critical value and marking the result with a plus (+) or minus (−) sign below the number line.
  • Place the original expression, the one including the absolute value bars, above the number line at the left end.
  • Use the sign of the expression inside the absolute value bars (indicated by the plus and minus signs below the number line) to remove the absolute value bars, placing the results above the number line on each side of the critical value.
  • Construct a piecewise definition that mimics the results on the number line.

Let’s apply this technique to another example.

Example \(\PageIndex{2}\)

Determine a piecewise definition for |3 − 2x|.

Step 1 : First set the expression inside the absolute value bars equal to zero and solve for x.

\[\begin{aligned} 3-2 x &=0 \\ x &=3 / 2 \end{aligned} \nonumber \]

Note that 3 − 2x = 0 at x = 3/2. This is the “critical value” for this expression.

Steps 2 and 3 : Draw a number line and mark this critical value on the line. The next step requires that we place the expression inside the absolute value bars, namely 3 − 2x, underneath the line at its left end.

WeChatf8b53e9cd145037bb92c2487876b4dcc.png

Step 4 : Next, determine the sign of 3 − 2x for values of x on each side of 3/2. This is easily done by “testing” a point on each side of 3/2 in the expression 3 − 2x.

  • Take x = 1, which lies to the left of 3/2. Substitute this value of x into the expression 3 − 2x, obtaining \[3-2 x=3-2(1)=1 \nonumber \] which is positive. Indicate this result by placing a plus sign (+) below the number line to the left of 3/2.

WeChat8da60f46fba3967ac0b8402201967d12.png

  • Next, pick x = 2, which lies to the right of 3/2. Substitute this value of x into the expression 3 − 2x, obtaining \[3-2 x=3-2(2)=-1 \nonumber \] which is negative. Indicate this result by placing a negative sign (−) below the line to the right of 3/2 (see the number line above).

Steps 5 and 6 : Place the original expression, namely |3 − 2x|, above the number line at the left end. The next step is to remove the absolute value bars from the expression |3 − 2x|.

  • To the left of 3/2, the expression 3 − 2x is positive (note the plus sign (+) below the number line), so |3−2x| = 3−2x. Indicate this result by placing the expression 3 − 2x above the number line to the left of 3/2.

WeChate88fa939bcad9116505b9db0e36610c4.png

  • To the right of 3/2, the expression 3−2x is negative (note the minus sign (−) below the numberline), so |3−2x| = −(3−2x). That is, we have to negate 3−2x to make it positive. This is indicated by placing the expression −(3 − 2x) above the line to the right of 3/2 (see the number line above).

Step 7 : We can use this last number line summary to write a piecewise definition for the expression |3 − 2x|.

\[|3-2 x|=\left\{\begin{array}{ll}{3-2 x,} & {\text { if } x<3 / 2 .} \\ {-(3-2 x),} & {\text { if } x \geq 3 / 2}\end{array}=\left\{\begin{array}{ll}{3-2 x,} & {\text { if } x<3 / 2} \\ {-3+2 x,} & {\text { if } x \geq 3 / 2}\end{array}\right.\right. \nonumber \]

Again, note how we’ve included the critical value of 3/2 “on the right.”

Drawing the Graph of an Absolute Value Function

Now that we know how to construct a piecewise definition for an expression containing absolute value bars, we can use what we learned in the previous section to draw the graph.

Example \(\PageIndex{3}\)

Sketch the graph of the function f(x) = |3 − 2x|.

In Example \(\PageIndex{2}\), we constructed the following piecewise definition.

\[f(x)=|3-2 x|=\left\{\begin{array}{ll}{3-2 x,} & {\text { if } x<3 / 2} \\ {-3+2 x,} & {\text { if } x \geq 3 / 2}\end{array}\right. \nonumber \]

We now sketch each piece of this function.

  • If x < 3/2, then f(x) = 3 − 2x (see equation (10)). This is a ray, starting at x = 3/2 and extending to the left. At x = 3/2,

\[f(3 / 2)=3-2(3 / 2)=3-3=0 \nonumber \]

Thus, the endpoint of the ray is located at (3/2, 0).

Next, pick a value of x that lies to the left of 3/2. At x = 0,

\[f(0)=3-2(0)=3-0=3 \nonumber \]

Thus, a second point on the ray is (0, 3).

A table containing the two evaluated points and a sketch of the accompanying ray are shown in Figure \(\PageIndex{1}\). Because f(x) = 3 − 2x only if x is strictly less than 3/2, the point at (3/2, 0) is unfilled.

WeChat2c8e9cc2c837b034a14b7ab0d6cba1d9.png

  • If x ≥ 3/2, then f(x) = −3 + 2x (see equation (10)). This is a ray, starting at x = 3/2 and extending to the right. At x = 3/2, \[f(3 / 2)=-3+2(3 / 2)=-3+3=0 \nonumber \]

Next, pick a value of x that lies to the right of 3/2. At x = 3, \[f(3)=-3+2(3)=-3+6=3 \nonumber \]

Thus, a second point on the ray is (3, 3). A table containing the two evaluated points and a sketch of the accompanying ray are shown in Figure \(\PageIndex{2}\). Because f(x) = −3 + 2x for all values of x that are greater than or equal to 3/2, the point at (3/2, 0) is filled in this plot.

WeChat3613b3f7c1afba3a162e113ba0c40789.png

  • To sketch the graph of f(x) = |3 − 2x|, we need only combine the two pieces from Figures \(\PageIndex{1}\) and \(\PageIndex{2}\). The result is shown in Figure \(\PageIndex{3}\).

WeChatb2a984d19c2b20f7acf88440c04a68d2.png

Note the “V-shape” of the graph. We will refer to the point at the tip of the “V” as the vertex of the absolute value function.

In Figure \(\PageIndex{3}\), the equation of the left-hand branch of the “V” is y = 3 − 2x. An alternate approach to drawing this branch is to note that its graph is contained in the graph of the full line y = 3 − 2x, which has slope −2 and y-intercept at (0, 3). Thus, one could draw the full line using the slope and y-intercept, then erase that part of the line that lies to the right of x = 3/2. A similar strategy would work for the right-hand branch of y = |3 − 2x|.

Using Transformations

Consider again the basic definition of the absolute value of x.

\[f(x)=|x|=\left\{\begin{array}{ll}{-x,} & {\text { if } x<0} \\ {x,} & {\text { if } x \geq 0}\end{array}\right. \nonumber \]

Some basic observations are:

  • If x < 0, then f(x) = −x. This ray starts at the origin and extends to the left with slope −1. Its graph is pictured in Figure \(\PageIndex{4}\)(a).
  • If \(x \geq 0\), then f(x) = x. This ray starts at the origin and extends to the right with slope 1. Its graph is pictured in Figure \(\PageIndex{4}\)(b).
  • We combine the graphs in Figures \(\PageIndex{4}\)(a) and \(\PageIndex{4}\)(b) to produce the graph of f(x) = |x| in Figure \(\PageIndex{4}\)(c).

WeChatb35cead7dee4a378f294d349de9abfd4.png

You should commit the graph of f(x) = |x| to memory. Things to note:

  • The graph of f(x) = |x| is “V-shaped.”
  • The vertex of the graph is at the point (0, 0).
  • The left-hand branch has equation y = −x and slope −1.
  • The right-hand branch has equation y = x and slope 1.
  • Each branch of the graph of f(x) = |x| forms a perfect 45◦ angle with the x-axis.

Now that we know how to draw the graph of f(x) = |x|, we can use the transformations we learned in Chapter 2 (sections 5 and 6) to sketch a number of simple graphs involving absolute value.

Example \(\PageIndex{4}\)

Sketch the graph of f(x) = |x − 3|.

First, sketch the graph of y = f(x) = |x|, as shown in Figure \(\PageIndex{5}\)(a). Note that if f(x) = |x|, then

\[y=f(x-3)=|x-3| \nonumber \]

To sketch the graph of y = f(x − 3) = |x − 3|, shift the graph of y = f(x) = |x| three units to the right, producing the result shown in Figure \(\PageIndex{5}\)(b).

WeChat524d23968d3e5cdb28aaa257acc37620.png

We can check this result using the graphing calculator. Load the function f(x) = |x − 3| into Y1 in the Y= menu on your graphing calculator as shown in Figure \(\PageIndex{6}\)(a). Push the MATH button, right-arrow to the NUM menu, then select 1:abs( (see Figure \(\PageIndex{6}\)(b)) to enter the absolute value in Y1. Push the ZOOM button, then select 6:ZStandard to produce the image shown in Figure \(\PageIndex{6}\)(c).

WeChat16d0696921470053ece851cd33a69cf9.png

Let’s look at another simple example.

Example \(\PageIndex{5}\)

Sketch the graph of f(x) = |x| − 4.

First, sketch the graph of y = f(x) = |x|, as shown in Figure \(\PageIndex{7}\)(a). Note that if f(x) = |x|, then \[y=f(x)-4=|x|-4 \nonumber \]

To sketch the graph of y = f(x) − 4 = |x| − 4, shift the graph of y = f(x) = |x| downward 4 units, producing the result shown in Figure \(\PageIndex{5}\)(b).

WeChat58998e76a8afc038b41ce420e0686609.png

Let’s look at one final example.

Example \(\PageIndex{6}\)

Sketch the graph of f(x) = −|x| + 5. State the domain and range of this function.

  • First, sketch the graph of y = f(x) = |x|, as shown in Figure \(\PageIndex{8}\)(a).
  • Next, sketch the graph of y = −f(x) = −|x|, which is a reflection of the graph of y = f(x) = |x| across the x-axis and is pictured in Figure \(\PageIndex{8}\)(b).
  • Finally, we will want to sketch the graph of y = −f(x) + 5 = −|x| + 5. To do this, we shift the graph of y = −f(x) = −|x| in Figure \(\PageIndex{8}\)(b) upward 5 units to produce the result in Figure \(\PageIndex{8}\)(c).

To find the domain of f(x) = −|x| + 5, project all points on the graph onto the x-axis, as shown in Figure \(\PageIndex{9}\)(a). Thus, the domain of f is \((-\infty, \infty)\). To find the range, project all points on the graph onto the y-axis, as shown in Figure \(\PageIndex{9}\)(b). Thus, the range is \((-\infty, 5]\).

WeChat1ba1933705a816e6e2028e4ee7afd8f4.png

For each of the functions in Exercises 1 - 8 , as in Examples 7 and 8 in the narrative, mark the “critical value” on a number line, then mark the sign of the expression inside of the absolute value bars below the number line. Above the number line, remove the absolute value bars according to the sign of the expression you marked below the number line. Once your number line summary is finished, create a piecewise definition for the given absolute value function.

Exercise \(\PageIndex{1}\)

f(x) = |x+1|

Screen Shot 2019-09-09 at 3.12.13 PM.png

\[f(x)=\left\{\begin{array}{ll}{-x-1,} & {\text { if } x<-1} \\ {x+1,} & {\text { if } x \ge -1} \nonumber \end{array}\right. \nonumber \]

Exercise \(\PageIndex{2}\)

f(x) = |x−4|

Exercise \(\PageIndex{3}\)

g(x) = |4−5x|

Screen Shot 2019-09-09 at 3.14.38 PM.png

\[g(x)=\left\{\begin{array}{ll}{4-5x,} & {\text { if } x<\frac{4}{5}} \\ {-4+5x,} & {\text { if } x \ge \frac{4}{5}} \nonumber \end{array}\right. \nonumber \]

Exercise \(\PageIndex{4}\)

g(x) = |3−2x|

Exercise \(\PageIndex{5}\)

h(x) = |−x−5|

Screen Shot 2019-09-09 at 3.18.17 PM.png

\[h(x)=\left\{\begin{array}{ll}{-x-5,} & {\text { if } x<-5} \\ {x+5,} & {\text { if } x \ge -5} \nonumber \end{array}\right. \nonumber \]

Exercise \(\PageIndex{6}\)

h(x) = |−x−3|

Exercise \(\PageIndex{7}\)

f(x) = x+|x|

Screen Shot 2019-09-09 at 3.20.24 PM.png

\[f(x)=\left\{\begin{array}{ll}{0,} & {\text { if } x<0} \\ {2x,} & {\text { if } x \ge 0} \nonumber \end{array}\right. \nonumber \]

Exercise \(\PageIndex{8}\)

\(f(x) = \frac{|x|}{x}\)

For each of the functions in Exercises 9 - 16 , perform each of the following tasks.

  • Create a piecewise definition for the given function, using the technique in Exercises 1 - 8 and Examples 7 and 8 in the narrative.
  • Following the lead in Example 9 in the narrative, use your piecewise definition to sketch the graph of the given function on a sheet of graph paper. Please place each exercise on its own coordinate system.

Exercise \(\PageIndex{9}\)

f(x) = |x−1|

\[f(x)=\left\{\begin{array}{ll}{-x+1,} & {\text { if } x<1} \\ {x-1,} & {\text { if } x \ge 1} \nonumber \end{array}\right. \nonumber \]

Screen Shot 2019-09-09 at 3.22.13 PM.png

Exercise \(\PageIndex{10}\)

f(x) = |x+2|

Exercise \(\PageIndex{11}\)

g(x) = |2x−1|

\[g(x)=\left\{\begin{array}{ll}{-2x+1,} & {\text { if } x<\frac{1}{2}} \\ {2x-1,} & {\text { if } x \ge \frac{1}{2}} \nonumber \end{array}\right. \nonumber \]

Screen Shot 2019-09-09 at 3.23.54 PM.png

Exercise \(\PageIndex{12}\)

g(x) = |5−2x|

Exercise \(\PageIndex{13}\)

h(x) = |1−3x|

\[h(x)=\left\{\begin{array}{ll}{1-3x,} & {\text { if } x<\frac{1}{3}} \\ {-1+3x,} & {\text { if } x \ge \frac{1}{3}} \nonumber \end{array}\right. \nonumber \]

Screen Shot 2019-09-09 at 3.25.35 PM.png

Exercise \(\PageIndex{14}\)

h(x) = |2x+1|

Exercise \(\PageIndex{15}\)

f(x) = x−|x|

\[f(x)=\left\{\begin{array}{ll}{2x,} & {\text { if } x<0} \\ {0,} & {\text { if } x \ge 0} \nonumber \end{array}\right. \nonumber \]

Screen Shot 2019-09-09 at 3.26.40 PM.png

Exercise \(\PageIndex{16}\)

f(x) = x+|x−1|

Exercise \(\PageIndex{17}\)

Use a graphing calculator to draw the graphs of y = |x|, y = 2|x|, y = 3|x|, and y = 4|x| on the same viewing window. In your own words, explain what you learned in this exercise.

Multiplying by a factor of a > 1, as in y = a|x|, stretches the graph of y = |x| vertically by a factor of a. The higher the value of a, the more it stretches vertically.

Exercise \(\PageIndex{18}\)

Use a graphing calculator to draw the graphs of y = |x|, y = (1/2)|x|, y = (1/3)|x|, and y = (1/4)|x| on the same viewing window. In your own words, explain what you learned in this exercise.

Exercise \(\PageIndex{19}\)

Use a graphing calculator to draw the graphs of y = |x|, y = |x−2|, y =|x−4|, and y = |x−6| on the same viewing window. In your own words, explain what you learned in this exercise.

Subtracting a positive value of a, as in y = |x−a|, shifts the graph a units to the right.

Exercise \(\PageIndex{20}\)

Use a graphing calculator to draw the graphs of y = |x|, y = |x+2|, y = |x+4|, and y = |x+6| on the same view- ing window. In your own words, explain what you learned in this exercise.

In Exercises 21 - 36 , perform each of the following tasks. Feel free to check your work with your graphing calculator, but you should be able to do all of the work by hand.

  • Set up a coordinate system on a sheet of graph paper. Label and scale each axis. Create an accurate plot of the function y = |x| on your coordinate system and label this graph with its equation.
  • Use the technique of Examples 12,13, and 14 in the narrative to help select the appropriate geometric transformations to transform the equation y = |x| into the form of the function given in the exercise. On the same coordinate system, use a different colored pencil or pen to draw the graph of the function resulting from your applied transformation. Label the resulting graph with its equation.
  • Use interval notation to describe the domain and range of the given function.

Exercise \(\PageIndex{21}\)

f(x) = |−x|

The graphs of y = |x| and y = |−x| coincide. The domain is \((−\infty, \infty)\) and the range is \([0, \infty)\).

Screen Shot 2019-09-09 at 3.30.12 PM.png

Exercise \(\PageIndex{22}\)

f(x) = −|x|

Exercise \(\PageIndex{23}\)

\(f(x) = \frac{1}{2}|x|\)

The domain is \((−\infty, \infty)\) and the range is \([0, \infty)\).

Screen Shot 2019-09-09 at 3.33.19 PM.png

Exercise \(\PageIndex{24}\)

f(x) = −2|x|

Exercise \(\PageIndex{25}\)

f(x) = |x+4|

Screen Shot 2019-09-09 at 3.34.19 PM.png

Exercise \(\PageIndex{26}\)

f(x) = |x−2|

Exercise \(\PageIndex{27}\)

f(x) = |x|+2

The domain is \((−\infty, \infty)\) and the range is \([2, \infty)\).

Screen Shot 2019-09-09 at 3.35.07 PM.png

Exercise \(\PageIndex{28}\)

f(x) = |x|−3

Exercise \(\PageIndex{29}\)

f(x) = |x+3|+2

Screen Shot 2019-09-09 at 3.36.48 PM.png

Exercise \(\PageIndex{30}\)

f(x) = |x−3|−4

Exercise \(\PageIndex{31}\)

f(x) = −|x−2|

The domain is \((−\infty, \infty)\) and the range is \((−\infty, 0]\).

Screen Shot 2019-09-09 at 3.37.32 PM.png

Exercise \(\PageIndex{32}\)

f(x) = −|x|−2

Exercise \(\PageIndex{33}\)

f(x) = −|x|+4

The domain is \((−\infty, \infty)\) and the range is \((−\infty, 4]\).

Screen Shot 2019-09-09 at 3.38.48 PM.png

Exercise \(\PageIndex{34}\)

f(x) = −|x+4| ​​​​​​

Exercise \(\PageIndex{35}\)

f(x) = −|x−1|+5

The domain is \((−\infty, \infty)\) and the range is \((−\infty, 5]\).

Screen Shot 2019-09-09 at 3.39.57 PM.png

Exercise \(\PageIndex{36}\)

f(x) = −|x+5|+2

Absolute Value

Absolute value means ....

... only how far a number is from zero:

"6" is 6 away from zero, and "βˆ’6" is also 6 away from zero.

So the absolute value of 6 is 6 , and the absolute value of βˆ’6 is also 6

More Examples:

  • The absolute value of βˆ’9 is 9
  • The absolute value of 3 is 3
  • The absolute value of 0 is 0
  • The absolute value of βˆ’156 is 156

No Negatives!

So in practice "absolute value" means to remove any negative sign in front of a number, and to think of all numbers as positive (or zero).

Absolute Value Symbol

To show that we want the absolute value of something, we put "|" marks either side (they are called "bars" and are found on the right side of a keyboard), like these examples:

Sometimes absolute value is also written as "abs()", so abs(βˆ’1) = 1 is the same as |βˆ’1| = 1

Try It Yourself

Subtract either way around.

And it doesn't matter which way around we do a subtraction, the absolute value will always be the same:

|8βˆ’3| = 5       (8βˆ’3 = 5)

|3βˆ’8| = 5       (3βˆ’8 = βˆ’5, and |βˆ’5| = 5)

More Examples

Here are some more examples of how to handle absolute values:

|βˆ’3Γ—6| = 18 Because βˆ’3 Γ— 6 = βˆ’18, and |βˆ’18| = 18

βˆ’|5βˆ’2| = βˆ’3 Because 5βˆ’2 = 3 and then the first minus gets us βˆ’3

βˆ’|2βˆ’5| = βˆ’3 Because 2βˆ’5 = βˆ’3 , |βˆ’3| = 3 , and then the first minus gets us βˆ’3

βˆ’|βˆ’12| = βˆ’12 Because |βˆ’12| = 12 and then the first minus gets us βˆ’12

Learn more at Absolute Value in Algebra

The Relationship Between Absolute Value and Order

Related Topics: Lesson Plans and Worksheets for Grade 6 Lesson Plans and Worksheets for all Grades More Lessons for Grade 6 Common Core For Grade 6

Videos to help Grade 6 students understand the relationship between absolute value and order and the statements of order in the real world.

New York State Common Core Math Module 3, Grade 6, Lesson 12

Worksheets for Grade 6

Lesson 12 Student Outcomes Students understand that the order of positive numbers is the same as the order of their absolute values. Students understand that the order of negative numbers is the opposite order of their absolute values. Students understand that negative numbers are always less than positive numbers. Lesson 12 Opening Exercise Record your integer values in order from least to greatest in the space below. 7, -12, 0, -5, -2, 5, -1, -9, 8, 2 Example 1: Comparing Order of Integers to the Order of their Absolute Values

Write an inequality statement relating the ordered integers from the Opening Exercise. Below each integer write its absolute value.

Are the absolute values of your integers in order? Explain. Circle the absolute values that are in increasing numerical order and their corresponding integers, and then describe the circled values.

Rewrite the integers that are not circled in the space below. How do these integers differ from the ones you circled? Rewrite the negative integers in ascending order and their absolute values in ascending order below them. Describe how the order of the absolute values compares to the order of the negative integers. Example 2: The Order of Negative Integers and their Absolute Values

Draw arrows starting at the dashed line (zero) to represent each of the integers shown on the number line below. The arrows that correspond with and have been modeled for you.

As you approach zero from the left on the number line, the integers increase, but the absolute values of those integers decrease. This means that the order of negative integers is opposite the order of their absolute values.

Complete the steps below to order the numbers.

a. Separate the set of numbers into positive and negative values and zero in the top cells below. b. Write the absolute values of the rational numbers (order does not matter) in the bottom cells below. c. Order each subset of absolute values. d. Order each subset of rational numbers. e. Order the whole given set of rational numbers. Lesson 12 Summary

The absolute values of positive numbers will always have the same order as the positive numbers themselves. Negative numbers, however, have exactly the opposite order as their absolute values. The absolute values of numbers on the number line increase as you move away from zero in either direction.

Lesson 12 Problem Set

  • Micah and Joel each have a set of five rational numbers. Although their sets are not the same, their sets of numbers have absolute values that are the same. Show an the sets in order and the absolute values in order.
  • For each pair of rational numbers below, place each number in the Venn diagram based on how it compares to the other

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1-2 Order of Operations and Evaluating Expressions

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Algebra 1 Curriculum

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  • 2.7 Linear Inequalities and Absolute Value Inequalities
  • Introduction to Prerequisites
  • 1.1 Real Numbers: Algebra Essentials
  • 1.2 Exponents and Scientific Notation
  • 1.3 Radicals and Rational Exponents
  • 1.4 Polynomials
  • 1.5 Factoring Polynomials
  • 1.6 Rational Expressions
  • Key Equations
  • Key Concepts
  • Review Exercises
  • Practice Test
  • Introduction to Equations and Inequalities
  • 2.1 The Rectangular Coordinate Systems and Graphs
  • 2.2 Linear Equations in One Variable
  • 2.3 Models and Applications
  • 2.4 Complex Numbers
  • 2.5 Quadratic Equations
  • 2.6 Other Types of Equations
  • Introduction to Functions
  • 3.1 Functions and Function Notation
  • 3.2 Domain and Range
  • 3.3 Rates of Change and Behavior of Graphs
  • 3.4 Composition of Functions
  • 3.5 Transformation of Functions
  • 3.6 Absolute Value Functions
  • 3.7 Inverse Functions
  • Introduction to Linear Functions
  • 4.1 Linear Functions
  • 4.2 Modeling with Linear Functions
  • 4.3 Fitting Linear Models to Data
  • Introduction to Polynomial and Rational Functions
  • 5.1 Quadratic Functions
  • 5.2 Power Functions and Polynomial Functions
  • 5.3 Graphs of Polynomial Functions
  • 5.4 Dividing Polynomials
  • 5.5 Zeros of Polynomial Functions
  • 5.6 Rational Functions
  • 5.7 Inverses and Radical Functions
  • 5.8 Modeling Using Variation
  • Introduction to Exponential and Logarithmic Functions
  • 6.1 Exponential Functions
  • 6.2 Graphs of Exponential Functions
  • 6.3 Logarithmic Functions
  • 6.4 Graphs of Logarithmic Functions
  • 6.5 Logarithmic Properties
  • 6.6 Exponential and Logarithmic Equations
  • 6.7 Exponential and Logarithmic Models
  • 6.8 Fitting Exponential Models to Data
  • Introduction to Systems of Equations and Inequalities
  • 7.1 Systems of Linear Equations: Two Variables
  • 7.2 Systems of Linear Equations: Three Variables
  • 7.3 Systems of Nonlinear Equations and Inequalities: Two Variables
  • 7.4 Partial Fractions
  • 7.5 Matrices and Matrix Operations
  • 7.6 Solving Systems with Gaussian Elimination
  • 7.7 Solving Systems with Inverses
  • 7.8 Solving Systems with Cramer's Rule
  • Introduction to Analytic Geometry
  • 8.1 The Ellipse
  • 8.2 The Hyperbola
  • 8.3 The Parabola
  • 8.4 Rotation of Axes
  • 8.5 Conic Sections in Polar Coordinates
  • Introduction to Sequences, Probability and Counting Theory
  • 9.1 Sequences and Their Notations
  • 9.2 Arithmetic Sequences
  • 9.3 Geometric Sequences
  • 9.4 Series and Their Notations
  • 9.5 Counting Principles
  • 9.6 Binomial Theorem
  • 9.7 Probability

Learning Objectives

In this section, you will:

  • Use interval notation
  • Use properties of inequalities.
  • Solve inequalities in one variable algebraically.
  • Solve absolute value inequalities.

It is not easy to make the honor roll at most top universities. Suppose students were required to carry a course load of at least 12 credit hours and maintain a grade point average of 3.5 or above. How could these honor roll requirements be expressed mathematically? In this section, we will explore various ways to express different sets of numbers, inequalities, and absolute value inequalities.

Using Interval Notation

Indicating the solution to an inequality such as x β‰₯ 4 x β‰₯ 4 can be achieved in several ways.

We can use a number line as shown in Figure 2 . The blue ray begins at x = 4 x = 4 and, as indicated by the arrowhead, continues to infinity, which illustrates that the solution set includes all real numbers greater than or equal to 4.

We can use set-builder notation : { x | x β‰₯ 4 } , { x | x β‰₯ 4 } , which translates to β€œall real numbers x such that x is greater than or equal to 4.” Notice that braces are used to indicate a set.

The third method is interval notation , in which solution sets are indicated with parentheses or brackets. The solutions to x β‰₯ 4 x β‰₯ 4 are represented as [ 4 , ∞ ) . [ 4 , ∞ ) . This is perhaps the most useful method, as it applies to concepts studied later in this course and to other higher-level math courses.

The main concept to remember is that parentheses represent solutions greater or less than the number, and brackets represent solutions that are greater than or equal to or less than or equal to the number. Use parentheses to represent infinity or negative infinity, since positive and negative infinity are not numbers in the usual sense of the word and, therefore, cannot be β€œequaled.” A few examples of an interval , or a set of numbers in which a solution falls, are [ βˆ’2 , 6 ) , [ βˆ’2 , 6 ) , or all numbers between βˆ’2 βˆ’2 and 6 , 6 , including βˆ’2 , βˆ’2 , but not including 6 ; 6 ; ( βˆ’ 1 , 0 ) , ( βˆ’ 1 , 0 ) , all real numbers between, but not including βˆ’1 βˆ’1 and 0 ; 0 ; and ( βˆ’ ∞ , 1 ] , ( βˆ’ ∞ , 1 ] , all real numbers less than and including 1. 1. Table 1 outlines the possibilities.

Using Interval Notation to Express All Real Numbers Greater Than or Equal to a

Use interval notation to indicate all real numbers greater than or equal to βˆ’2. βˆ’2.

Use a bracket on the left of βˆ’2 βˆ’2 and parentheses after infinity: [ βˆ’2 , ∞ ) . [ βˆ’2 , ∞ ) . The bracket indicates that βˆ’2 βˆ’2 is included in the set with all real numbers greater than βˆ’2 βˆ’2 to infinity.

Use interval notation to indicate all real numbers between and including βˆ’3 βˆ’3 and 5. 5.

Using Interval Notation to Express All Real Numbers Less Than or Equal to a or Greater Than or Equal to b

Write the interval expressing all real numbers less than or equal to βˆ’1 βˆ’1 or greater than or equal to 1. 1.

We have to write two intervals for this example. The first interval must indicate all real numbers less than or equal to 1. So, this interval begins at βˆ’ ∞ βˆ’ ∞ and ends at βˆ’1 , βˆ’1 , which is written as ( βˆ’ ∞ , βˆ’1 ] . ( βˆ’ ∞ , βˆ’1 ] .

The second interval must show all real numbers greater than or equal to 1 , 1 , which is written as [ 1 , ∞ ) . [ 1 , ∞ ) . However, we want to combine these two sets. We accomplish this by inserting the union symbol, βˆͺ , βˆͺ , between the two intervals.

Express all real numbers less than βˆ’2 βˆ’2 or greater than or equal to 3 in interval notation.

Using the Properties of Inequalities

When we work with inequalities, we can usually treat them similarly to but not exactly as we treat equalities. We can use the addition property and the multiplication property to help us solve them. The one exception is when we multiply or divide by a negative number; doing so reverses the inequality symbol.

Properties of Inequalities

These properties also apply to a ≀ b , a ≀ b , a > b , a > b , and a β‰₯ b . a β‰₯ b .

Demonstrating the Addition Property

Illustrate the addition property for inequalities by solving each of the following:

  • ⓐ x βˆ’ 15 < 4 x βˆ’ 15 < 4
  • β“‘ 6 β‰₯ x βˆ’ 1 6 β‰₯ x βˆ’ 1
  • β“’ x + 7 > 9 x + 7 > 9

The addition property for inequalities states that if an inequality exists, adding or subtracting the same number on both sides does not change the inequality.

  • ⓐ x βˆ’ 15 < 4 x βˆ’ 15 + 15 < 4 + 15 Add 15 to both sides . x < 19 x βˆ’ 15 < 4 x βˆ’ 15 + 15 < 4 + 15 Add 15 to both sides . x < 19
  • β“‘ 6 β‰₯ x βˆ’ 1 6 + 1 β‰₯ x βˆ’ 1 + 1 Add 1 to both sides . 7 β‰₯ x 6 β‰₯ x βˆ’ 1 6 + 1 β‰₯ x βˆ’ 1 + 1 Add 1 to both sides . 7 β‰₯ x
  • β“’ x + 7 > 9 x + 7 βˆ’ 7 > 9 βˆ’ 7 Subtract 7 from both sides . x > 2 x + 7 > 9 x + 7 βˆ’ 7 > 9 βˆ’ 7 Subtract 7 from both sides . x > 2

Solve: 3 x βˆ’2 < 1. 3 x βˆ’2 < 1.

Demonstrating the Multiplication Property

Illustrate the multiplication property for inequalities by solving each of the following:

  • ⓐ 3 x < 6 3 x < 6
  • β“‘ βˆ’2 x βˆ’ 1 β‰₯ 5 βˆ’2 x βˆ’ 1 β‰₯ 5
  • β“’ 5 βˆ’ x > 10 5 βˆ’ x > 10
  • ⓐ 3 x < 6 1 3 ( 3 x ) < ( 6 ) 1 3 x < 2 3 x < 6 1 3 ( 3 x ) < ( 6 ) 1 3 x < 2
  • β“‘ βˆ’ 2 x βˆ’ 1 β‰₯ 5 βˆ’ 2 x β‰₯ 6 ( βˆ’ 1 2 ) ( βˆ’ 2 x ) β‰₯ ( 6 ) ( βˆ’ 1 2 ) Multiply by βˆ’ 1 2 . x ≀ βˆ’ 3 Reverse the inequality . βˆ’ 2 x βˆ’ 1 β‰₯ 5 βˆ’ 2 x β‰₯ 6 ( βˆ’ 1 2 ) ( βˆ’ 2 x ) β‰₯ ( 6 ) ( βˆ’ 1 2 ) Multiply by βˆ’ 1 2 . x ≀ βˆ’ 3 Reverse the inequality .
  • β“’ 5 βˆ’ x > 10 βˆ’ x > 5 ( βˆ’ 1 ) ( βˆ’ x ) > ( 5 ) ( βˆ’ 1 ) Multiply by βˆ’ 1. x < βˆ’ 5 Reverse the inequality . 5 βˆ’ x > 10 βˆ’ x > 5 ( βˆ’ 1 ) ( βˆ’ x ) > ( 5 ) ( βˆ’ 1 ) Multiply by βˆ’ 1. x < βˆ’ 5 Reverse the inequality .

Solve: 4 x + 7 β‰₯ 2 x βˆ’ 3. 4 x + 7 β‰₯ 2 x βˆ’ 3.

Solving Inequalities in One Variable Algebraically

As the examples have shown, we can perform the same operations on both sides of an inequality, just as we do with equations; we combine like terms and perform operations. To solve, we isolate the variable.

Solving an Inequality Algebraically

Solve the inequality: 13 βˆ’ 7 x β‰₯ 10 x βˆ’ 4. 13 βˆ’ 7 x β‰₯ 10 x βˆ’ 4.

Solving this inequality is similar to solving an equation up until the last step.

The solution set is given by the interval ( βˆ’ ∞ , 1 ] , ( βˆ’ ∞ , 1 ] , or all real numbers less than and including 1.

Solve the inequality and write the answer using interval notation: βˆ’ x + 4 < 1 2 x + 1. βˆ’ x + 4 < 1 2 x + 1.

Solving an Inequality with Fractions

Solve the following inequality and write the answer in interval notation: βˆ’ 3 4 x β‰₯ βˆ’ 5 8 + 2 3 x . βˆ’ 3 4 x β‰₯ βˆ’ 5 8 + 2 3 x .

We begin solving in the same way we do when solving an equation.

The solution set is the interval ( βˆ’ ∞ , 15 34 ] . ( βˆ’ ∞ , 15 34 ] .

Solve the inequality and write the answer in interval notation: βˆ’ 5 6 x ≀ 3 4 + 8 3 x . βˆ’ 5 6 x ≀ 3 4 + 8 3 x .

Understanding Compound Inequalities

A compound inequality includes two inequalities in one statement. A statement such as 4 < x ≀ 6 4 < x ≀ 6 means 4 < x 4 < x and x ≀ 6. x ≀ 6. There are two ways to solve compound inequalities: separating them into two separate inequalities or leaving the compound inequality intact and performing operations on all three parts at the same time. We will illustrate both methods.

Solving a Compound Inequality

Solve the compound inequality: 3 ≀ 2 x + 2 < 6. 3 ≀ 2 x + 2 < 6.

The first method is to write two separate inequalities: 3 ≀ 2 x + 2 3 ≀ 2 x + 2 and 2 x + 2 < 6. 2 x + 2 < 6. We solve them independently.

Then, we can rewrite the solution as a compound inequality, the same way the problem began.

In interval notation, the solution is written as [ 1 2 , 2 ) . [ 1 2 , 2 ) .

The second method is to leave the compound inequality intact, and perform solving procedures on the three parts at the same time.

We get the same solution: [ 1 2 , 2 ) . [ 1 2 , 2 ) .

Solve the compound inequality: 4 < 2 x βˆ’ 8 ≀ 10. 4 < 2 x βˆ’ 8 ≀ 10.

Solving a Compound Inequality with the Variable in All Three Parts

Solve the compound inequality with variables in all three parts: 3 + x > 7 x βˆ’ 2 > 5 x βˆ’ 10. 3 + x > 7 x βˆ’ 2 > 5 x βˆ’ 10.

Let's try the first method. Write two inequalities :

The solution set is βˆ’4 < x < 5 6 βˆ’4 < x < 5 6 or in interval notation ( βˆ’4 , 5 6 ) . ( βˆ’4 , 5 6 ) . Notice that when we write the solution in interval notation, the smaller number comes first. We read intervals from left to right, as they appear on a number line. See Figure 3 .

Solve the compound inequality: 3 y < 4 βˆ’ 5 y < 5 + 3 y . 3 y < 4 βˆ’ 5 y < 5 + 3 y .

Solving Absolute Value Inequalities

As we know, the absolute value of a quantity is a positive number or zero. From the origin, a point located at ( βˆ’ x , 0 ) ( βˆ’ x , 0 ) has an absolute value of x , x , as it is x units away. Consider absolute value as the distance from one point to another point. Regardless of direction, positive or negative, the distance between the two points is represented as a positive number or zero.

An absolute value inequality is an equation of the form

Where A , and sometimes B , represents an algebraic expression dependent on a variable x. Solving the inequality means finding the set of all x x - values that satisfy the problem. Usually this set will be an interval or the union of two intervals and will include a range of values.

There are two basic approaches to solving absolute value inequalities: graphical and algebraic. The advantage of the graphical approach is we can read the solution by interpreting the graphs of two equations. The advantage of the algebraic approach is that solutions are exact, as precise solutions are sometimes difficult to read from a graph.

Suppose we want to know all possible returns on an investment if we could earn some amount of money within $200 of $600. We can solve algebraically for the set of x- values such that the distance between x x and 600 is less than or equal to 200. We represent the distance between x x and 600 as | x βˆ’ 600 | , | x βˆ’ 600 | , and therefore, | x βˆ’ 600 | ≀ 200 | x βˆ’ 600 | ≀ 200 or

This means our returns would be between $400 and $800.

To solve absolute value inequalities, just as with absolute value equations, we write two inequalities and then solve them independently.

Absolute Value Inequalities

For an algebraic expression X, and k > 0 , k > 0 , an absolute value inequality is an inequality of the form

These statements also apply to | X | ≀ k | X | ≀ k and | X | β‰₯ k . | X | β‰₯ k .

Determining a Number within a Prescribed Distance

Describe all values x x within a distance of 4 from the number 5.

We want the distance between x x and 5 to be less than or equal to 4. We can draw a number line, such as in Figure 4 , to represent the condition to be satisfied.

The distance from x x to 5 can be represented using an absolute value symbol, | x βˆ’ 5 | . | x βˆ’ 5 | . Write the values of x x that satisfy the condition as an absolute value inequality.

We need to write two inequalities as there are always two solutions to an absolute value equation.

If the solution set is x ≀ 9 x ≀ 9 and x β‰₯ 1 , x β‰₯ 1 , then the solution set is an interval including all real numbers between and including 1 and 9.

So | x βˆ’ 5 | ≀ 4 | x βˆ’ 5 | ≀ 4 is equivalent to [ 1 , 9 ] [ 1 , 9 ] in interval notation.

Describe all x- values within a distance of 3 from the number 2.

Solving an Absolute Value Inequality

Solve | x βˆ’ 1 | ≀ 3 | x βˆ’ 1 | ≀ 3 .

Using a Graphical Approach to Solve Absolute Value Inequalities

Given the equation y = βˆ’ 1 2 | 4 x βˆ’ 5 | + 3 , y = βˆ’ 1 2 | 4 x βˆ’ 5 | + 3 , determine the x -values for which the y -values are negative.

We are trying to determine where y < 0 , y < 0 , which is when βˆ’ 1 2 | 4 x βˆ’ 5 | + 3 < 0. βˆ’ 1 2 | 4 x βˆ’ 5 | + 3 < 0. We begin by isolating the absolute value.

Next, we solve for the equality | 4 x βˆ’ 5 | = 6. | 4 x βˆ’ 5 | = 6.

Now, we can examine the graph to observe where the y- values are negative. We observe where the branches are below the x- axis. Notice that it is not important exactly what the graph looks like, as long as we know that it crosses the horizontal axis at x = βˆ’ 1 4 x = βˆ’ 1 4 and x = 11 4 , x = 11 4 , and that the graph opens downward. See Figure 5 .

Solve βˆ’ 2 | k βˆ’ 4 | ≀ βˆ’ 6. βˆ’ 2 | k βˆ’ 4 | ≀ βˆ’ 6.

Access these online resources for additional instruction and practice with linear inequalities and absolute value inequalities.

  • Interval notation
  • How to solve linear inequalities
  • How to solve an inequality
  • Absolute value equations
  • Compound inequalities
  • Absolute value inequalities

2.7 Section Exercises

When solving an inequality, explain what happened from Step 1 to Step 2:

Step 1 βˆ’ 2 x > 6 Step 2 x < βˆ’ 3 Step 1 βˆ’ 2 x > 6 Step 2 x < βˆ’ 3

When solving an inequality, we arrive at:

x + 2 < x + 3 2 < 3 x + 2 < x + 3 2 < 3

Explain what our solution set is.

When writing our solution in interval notation, how do we represent all the real numbers?

x + 2 > x + 3 2 > 3 x + 2 > x + 3 2 > 3

Describe how to graph y = | x βˆ’ 3 | y = | x βˆ’ 3 |

For the following exercises, solve the inequality. Write your final answer in interval notation.

4 x βˆ’ 7 ≀ 9 4 x βˆ’ 7 ≀ 9

3 x + 2 β‰₯ 7 x βˆ’ 1 3 x + 2 β‰₯ 7 x βˆ’ 1

βˆ’2 x + 3 > x βˆ’ 5 βˆ’2 x + 3 > x βˆ’ 5

4 ( x + 3 ) β‰₯ 2 x βˆ’ 1 4 ( x + 3 ) β‰₯ 2 x βˆ’ 1

βˆ’ 1 2 x ≀ βˆ’ 5 4 + 2 5 x βˆ’ 1 2 x ≀ βˆ’ 5 4 + 2 5 x

βˆ’5 ( x βˆ’ 1 ) + 3 > 3 x βˆ’ 4 βˆ’ 4 x βˆ’5 ( x βˆ’ 1 ) + 3 > 3 x βˆ’ 4 βˆ’ 4 x

βˆ’3 ( 2 x + 1 ) > βˆ’2 ( x + 4 ) βˆ’3 ( 2 x + 1 ) > βˆ’2 ( x + 4 )

x + 3 8 βˆ’ x + 5 5 β‰₯ 3 10 x + 3 8 βˆ’ x + 5 5 β‰₯ 3 10

x βˆ’ 1 3 + x + 2 5 ≀ 3 5 x βˆ’ 1 3 + x + 2 5 ≀ 3 5

For the following exercises, solve the inequality involving absolute value. Write your final answer in interval notation.

| x + 9 | β‰₯ βˆ’6 | x + 9 | β‰₯ βˆ’6

| 2 x + 3 | < 7 | 2 x + 3 | < 7

| 3 x βˆ’ 1 | > 11 | 3 x βˆ’ 1 | > 11

| 2 x + 1 | + 1 ≀ 6 | 2 x + 1 | + 1 ≀ 6

| x βˆ’ 2 | + 4 β‰₯ 10 | x βˆ’ 2 | + 4 β‰₯ 10

| βˆ’2 x + 7 | ≀ 13 | βˆ’2 x + 7 | ≀ 13

| x βˆ’ 7 | < βˆ’4 | x βˆ’ 7 | < βˆ’4

| x βˆ’ 20 | > βˆ’1 | x βˆ’ 20 | > βˆ’1

| x βˆ’ 3 4 | < 2 | x βˆ’ 3 4 | < 2

For the following exercises, describe all the x -values within or including a distance of the given values.

Distance of 5 units from the number 7

Distance of 3 units from the number 9

Distance of 10 units from the number 4

Distance of 11 units from the number 1

For the following exercises, solve the compound inequality. Express your answer using inequality signs, and then write your answer using interval notation.

βˆ’4 < 3 x + 2 ≀ 18 βˆ’4 < 3 x + 2 ≀ 18

3 x + 1 > 2 x βˆ’ 5 > x βˆ’ 7 3 x + 1 > 2 x βˆ’ 5 > x βˆ’ 7

3 y < 5 βˆ’ 2 y < 7 + y 3 y < 5 βˆ’ 2 y < 7 + y

2 x βˆ’ 5 < βˆ’11 or 5 x + 1 β‰₯ 6 2 x βˆ’ 5 < βˆ’11 or 5 x + 1 β‰₯ 6

x + 7 < x + 2 x + 7 < x + 2

For the following exercises, graph the function. Observe the points of intersection and shade the x -axis representing the solution set to the inequality. Show your graph and write your final answer in interval notation.

| x βˆ’ 1 | > 2 | x βˆ’ 1 | > 2

| x + 3 | β‰₯ 5 | x + 3 | β‰₯ 5

| x + 7 | ≀ 4 | x + 7 | ≀ 4

| x βˆ’ 2 | < 7 | x βˆ’ 2 | < 7

| x βˆ’ 2 | < 0 | x βˆ’ 2 | < 0

For the following exercises, graph both straight lines (left-hand side being y1 and right-hand side being y2) on the same axes. Find the point of intersection and solve the inequality by observing where it is true comparing the y -values of the lines.

x + 3 < 3 x βˆ’ 4 x + 3 < 3 x βˆ’ 4

x βˆ’ 2 > 2 x + 1 x βˆ’ 2 > 2 x + 1

x + 1 > x + 4 x + 1 > x + 4

1 2 x + 1 > 1 2 x βˆ’ 5 1 2 x + 1 > 1 2 x βˆ’ 5

4 x + 1 < 1 2 x + 3 4 x + 1 < 1 2 x + 3

For the following exercises, write the set in interval notation.

{ x | βˆ’1 < x < 3 } { x | βˆ’1 < x < 3 }

{ x | x β‰₯ 7 } { x | x β‰₯ 7 }

{ x | x < 4 } { x | x < 4 }

{ x | x is all real numbers } { x | x is all real numbers }

For the following exercises, write the interval in set-builder notation.

( βˆ’ ∞ , 6 ) ( βˆ’ ∞ , 6 )

( 4 , ∞ ) ( 4 , ∞ )

[ βˆ’3 , 5 ) [ βˆ’3 , 5 )

[ βˆ’4 , 1 ] βˆͺ [ 9 , ∞ ) [ βˆ’4 , 1 ] βˆͺ [ 9 , ∞ )

For the following exercises, write the set of numbers represented on the number line in interval notation.

For the following exercises, input the left-hand side of the inequality as a Y1 graph in your graphing utility. Enter y2 = the right-hand side. Entering the absolute value of an expression is found in the MATH menu, Num, 1:abs(. Find the points of intersection, recall (2 nd CALC 5:intersection, 1 st curve, enter, 2 nd curve, enter, guess, enter). Copy a sketch of the graph and shade the x -axis for your solution set to the inequality. Write final answers in interval notation.

| x + 2 | βˆ’ 5 < 2 | x + 2 | βˆ’ 5 < 2

βˆ’ 1 2 | x + 2 | < 4 βˆ’ 1 2 | x + 2 | < 4

| 4 x + 1 | βˆ’ 3 > 2 | 4 x + 1 | βˆ’ 3 > 2

| x βˆ’ 4 | < 3 | x βˆ’ 4 | < 3

| x + 2 | β‰₯ 5 | x + 2 | β‰₯ 5

Solve | 3 x + 1 | = | 2 x + 3 | | 3 x + 1 | = | 2 x + 3 |

Solve x 2 βˆ’ x > 12 x 2 βˆ’ x > 12

x βˆ’ 5 x + 7 ≀ 0 , x βˆ’ 5 x + 7 ≀ 0 , x β‰  βˆ’7 x β‰  βˆ’7

p = βˆ’ x 2 + 130 x βˆ’ 3000 p = βˆ’ x 2 + 130 x βˆ’ 3000 is a profit formula for a small business. Find the set of x -values that will keep this profit positive.

Real-World Applications

In chemistry the volume for a certain gas is given by V = 20 T , V = 20 T , where V is measured in cc and T is temperature in ΒΊC. If the temperature varies between 80ΒΊC and 120ΒΊC, find the set of volume values.

A basic cellular package costs $20/mo. for 60 min of calling, with an additional charge of $.30/min beyond that time.. The cost formula would be C = 20 + .30 ( x βˆ’ 60 ) . C = 20 + .30 ( x βˆ’ 60 ) . If you have to keep your bill no greater than $50, what is the maximum calling minutes you can use?

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Access for free at https://openstax.org/books/college-algebra-2e/pages/1-introduction-to-prerequisites
  • Authors: Jay Abramson
  • Publisher/website: OpenStax
  • Book title: College Algebra 2e
  • Publication date: Dec 21, 2021
  • Location: Houston, Texas
  • Book URL: https://openstax.org/books/college-algebra-2e/pages/1-introduction-to-prerequisites
  • Section URL: https://openstax.org/books/college-algebra-2e/pages/2-7-linear-inequalities-and-absolute-value-inequalities

Β© Jan 9, 2024 OpenStax. Textbook content produced by OpenStax is licensed under a Creative Commons Attribution License . The OpenStax name, OpenStax logo, OpenStax book covers, OpenStax CNX name, and OpenStax CNX logo are not subject to the Creative Commons license and may not be reproduced without the prior and express written consent of Rice University.

Module 1: Algebraic Expressions, Linear Equations and Mathematical Models

1.2 – absolute value, learning objectives, (1.2.1) – evaluating expressions with absolute value signs.

  • Β [latex]|3| = 3[/latex]
  • Β [latex]|-5| = 5[/latex]
  • Β [latex]|0| = 0[/latex]

Recall that the absolute value of a quantity is always positive or 0.

FindΒ [latex]|7- 10|[/latex].

FindΒ [latex]-|3-2|[/latex].

When you see an absolute value expression included within a larger expression, treat the absolute value like a grouping symbol and evaluate the expression within the absolute value sign first. Then take the absolute value of that expression. The example below shows how this is done.

Simplify [latex]\Large\frac{3+\left|2-6\right|}{2\left|3\cdot1.5\right|-\left(-3\right)}[/latex].

Grouping symbols, including absolute value, are handled first. Simplify the numerator, then the denominator.

Evaluate [latex]\left|2–6\right|[/latex].

[latex]\Large\begin{array}{c}\frac{3+\left|2-6\right|}{2\left|3\cdot1.5\right|-\left(-3\right)}\\\\\frac{3+\left|-4\right|}{2\left|3\cdot1.5\right|-\left(-3\right)}\end{array}[/latex]

Take the absolute value of [latex]\left|βˆ’4\right|[/latex].

[latex]\Large\begin{array}{c}\frac{3+\left|-4\right|}{2\left|3\cdot1.5\right|-\left(-3\right)}\\\\\frac{3+4}{2\left|3\cdot1.5\right|-\left(-3\right)}\end{array}[/latex]

Add the numbers in the numerator.

[latex]\Large\begin{array}{c}\frac{3+4}{2\left|3\cdot1.5\right|-\left(-3\right)}\\\\\frac{7}{2\left| 3\cdot 1.5 \right|-(-3)}\end{array}[/latex]

Now that the numerator is simplified, turn to the denominator.

Evaluate the absolute value expression first. [latex]3 \cdot 1.5 = 4.5[/latex], giving

Β [latex]\Large\begin{array}{c}\frac{7}{2\left|{3\cdot{1.5}}\right|-(-3)}\\\\\frac{7}{2\left|{ 4.5}\right|-(-3)}\end{array}[/latex]

The expression β€œ[latex]2\left|4.5\right|[/latex]” reads β€œ2 times the absolute value of 4.5.” Multiply 2 times 4.5.

[latex]\Large\begin{array}{c}\frac{7}{2\left|4.5\right|-\left(-3\right)}\\\\\frac{7}{9-\left(-3\right)}\end{array}[/latex]

[latex]\Large\begin{array}{c}\frac{7}{9-\left(-3\right)}\\\\\frac{7}{12}\end{array}[/latex]

[latex]\Large\frac{3+\left|2-6\right|}{2\left|3\cdot1.5\right|-3\left(-3\right)}=\frac{7}{12}[/latex]

The following video uses the order of operations to simplify an expression in fraction form that contains absolute value terms. Note how the absolute values are treated like parentheses and brackets when using the order of operations.

  • Math Article

Absolute Values

The absolute value of a number refers to the distance of a number from the origin of a number line.Β It is represented as |a|, which defines the magnitude of any integer ‘a’. The absolute value of any integer, whether positive or negative, will be the real numbers , regardless of which sign it has.Β Β It is represented by two vertical lines |a|, which is known as the modulus of a.

For example: 5 is the absolute value for both 5 and -5.

|-5| = +5 and |+ 5| = +5

Absolute value

In this article, we will learn what is the absolute value of a number, symbol, examples, absolute value in number line, absolute value of real numbers, absolute value of complex numbers in detail.

What is the Absolute Value of a Number?

The absolute value of a number or integer is the actual distance of the integer from zero, in a number line. Therefore, the absolute value is always a positive value and not a negative number.

We can define the absolute values like the following:

{ a if a β‰₯ 0 }

|a| = { -a if a < 0 }

Note: There is no absolute value for 0 because the absolute value changes the sign of the numbers into positive and zero has no sign.

If the number is positive then it will result in a positive number only. And if the number is negative, then the modulus of this number will also be a positive number. It is denoted as |n|, where n is an integer.

Absolute Value Symbol

The symbol of absolute value is represented by the modulus symbol, β€˜| |’, with the numbers between it. For example, the absolute value of 9 is denoted as |9|.

The distance of any number from the origin on the number line is the absolute value of that number. It also shows the polarity of the number whether it is positive or negative. It can be negative ever a s it shows the distance and the distance can’t be negative. So, it is always positive.

Absolute value of a Number Examples

Let us see some examples of the absolute value of a number.

  • |7-2| = |5| = 5
  • |2+3| = |5| = 5
  • |-3×5| = |-15| = 15

Related Articles:

Absolute value function.

The absolute value function is given by f(x) = |x| such that: Β 

  • |x| = +x for x > 0Β 
  • |x| = -x for x < 0

Absolute Value Properties

If x and y are real numbers and then the absolute values are satisfying the following properties:

Absolute Value of a Real Number

If a real number x, the absolute value will satisfy the following conditions.

| x | = x, if x β‰₯ 0

| x | = – x, if x < 0

Let’s look at the absolute value of 2 in the number line given below. Here, |2| is the distance of 2 from 0(zero). So, both +2 and -2 is the distance of 2 from the origin. But it would be taken as 2 because distance is never measured in negative.

assignment 1 number order and absolute value

Absolute Value of Complex Number

Complex numbers consist of real numbers and imaginary numbers. Hence, unlike integers, it is difficult to find the absolute value for them. Suppose, x+iy is the given complex number.

The absolute value of z will be;

|z| = √(x 2 +y 2 )

Where x and y are the real numbers.

Absolute Value in Number Line

The graph of absolute values is called the absolute value graph. As we know the absolute value of any real number is positive, so the absolute value of any number or function graph will lie on the positive side only.

Example: Graph the absolute value of the number -9.

Solution:Β  Absolute value of |-9| is +9.

So the graph for the absolute value of -9 will look like following

Absolute value example

Problems and Solutions

Let’s understand this topic better with the help of examples.

Question 1:

Arrange the given numbers in ascending order.

-|-14|, |12|, |7|, |-91|, |-5|, |-8|, |-65|, |6|

Initially, find the absolute values of the given numbers

-14, 12, 7, 91, 5, 8, 65, 6

Now, arrange the numbers in ascending order (smallest to the largest number)

-14, 5, 6, 7, 8, 12, 65, 91

Question 2:

-|-24|, |21|, 17|, |-109|, |-15|, |-19|, |-75|, |16|

Find the absolute values of the given numbers

-24, 21, 17, 109, 15, 19, 75, 16

-24, 15, 16, 17,19, 21, 75, 109.

Find the absolute value of a number -12/5

To find: |-12/5|

|-12/5| = 12/5

Hence, the absolute value of a number -12/5 is 12/5

Find the absolute value for the following numbers:

The absolute value of the numbers are:

(a) |-1/2| = 1/2

(b) |72| = 72

(c) |3/4| = 3/4

Practice Questions

1. What is the absolute value of -13?

2. Find the absolute value of 100.

3. Arrange the following in ascending order.

Β  Β  Β  Β  Β  Β |3|, |-9|, |1|, |-2|, |-5|

4. Plot the absolute value of -7 on the number line.

Frequently Asked Questions on Absolute Values

What is an absolute value of a number, what is the absolute value of 4, what is the absolute value of -12, what is the symbol for absolute value, what is the absolute value of -7, is absolute value negative.

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  1. Absolute Value Formula

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  2. Absolute Value of Numbers

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  3. What Is Absolute Value? Definition and Examples

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  4. 7th 1.1 Lesson Ordering Absolute Value

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  5. How To Teach Absolute Value

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  6. Solving Absolute Value Equations: Complete Guide

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  1. 4. Solving Absolute Value Pt 2

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  3. Absolute Value

  4. 1.4 Putting the V in Absolute Value (Activity 4.2)

  5. Number line and absolute value #math #prealgebra #sat #shorts

  6. FINDING ABSOLUTE VALUE OF A NUMBER

COMMENTS

  1. Number Order and Absolute Value Flashcards

    The Rational numbers include which of the following? Positive Integers, Negative Integers, and Fractions. Select and place the symbol that will make the statement true. 6____8. 6<8. Select and place the symbol that will make the statement true. 3_____4_____5. 3<4<5.

  2. Number order and absolute value: algebra 2 Flashcards

    Terms in this set (8) Do you rational numbers include which of the following? Positive integers negative integers and fractions. Make the statement true. 3<4<5. Evaluate absolute value of -3. |-3|. 3. Evaluate absolute value of -7+3 times the absolute value of four.

  3. NUMBER ORDER AND ABSOLUTE VALUE Flashcards

    NUMBER ORDER AND ABSOLUTE VALUE Flashcards | Quizlet NUMBER ORDER AND ABSOLUTE VALUE means that the number a is to the left of b on the number line Click the card to flip πŸ‘† a < b Click the card to flip πŸ‘† 1 / 22 Flashcards Learn Test Match Q-Chat Created by km1409 Students also viewed Number Order and Absolute Value 16 terms coryml Preview

  4. Identify and order absolute values (video)

    Use a number line to find absolute value and then order absolute values from least to greatest. Created by Sal Khan and Monterey Institute for Technology and Education. Questions Tips & Thanks Want to join the conversation? Sort by: Top Voted TheFourthDimension 11 years ago An absolute value always makes a number positive.

  5. Intro to absolute value equations and graphs

    To solve absolute value equations, find x values that make the expression inside the absolute value positive or negative the constant. To graph absolute value functions, plot two lines for the positive and negative cases that meet at the expression's zero. The graph is v-shaped. Created by Sal Khan and CK-12 Foundation. Questions Tips & Thanks

  6. 1.1 Real Numbers: Algebra Essentials

    Any rational number can be represented as either: ⓐ a terminating decimal: 15 8 = 1.875, 15 8 = 1.875, or. β“‘ a repeating decimal: 4 11 = 0.36363636 … = 0. 36 Β―. 4 11 = 0.36363636 … = 0. 36 Β―. We use a line drawn over the repeating block of numbers instead of writing the group multiple times.

  7. 3.6 Absolute Value Functions

    The absolute value function is commonly thought of as providing the distance the number is from zero on a number line. Algebraically, for whatever the input value is, the output is the value without regard to sign. Knowing this, we can use absolute value functions to solve some kinds of real-world problems. Absolute Value Function

  8. 1.1: Review of Real Numbers and Absolute Value

    Solution. Think of a real number whose distance to the origin is 6 units. There are two solutions: the distance to the right of the origin and the distance to the left of the origin, namely \ {Β±6\}. The symbol Β± is read " plus or minus " and indicates that there are two answers, one positive and one negative.

  9. Intro to absolute value (article)

    Intro to absolute value. Learn how to think about absolute value as distance from zero, and practice finding absolute values. The absolute value of a number is its distance from 0 . This seems kind of obvious. Of course the distance from 0 to 4 is 4 . Where absolute value gets interesting is with negative numbers.

  10. Order and absolute value

    In this lesson, we will learn to interpret the absolute value of a number and understand how to order negative numbers using inequality notation.

  11. 2.7 Solve Absolute Value Inequalities

    The absolute value of a number is its distance from zero on the number line. The absolute value of a number n is written as | n | and | n | β‰₯ 0 for all numbers. Absolute values are always greater than or equal to zero. We learned that both a number and its opposite are the same distance from zero on the number line.

  12. 1.6: Absolute Value Functions

    Example \(\PageIndex{1}\): Determine a Number within a Prescribed Distance. Describe all values \(x\) within or including a distance of 4 from the number 5. Solution. ... After determining that the absolute value is equal to 4 at \(x=1\) and \(x=9\), we know the graph can change only from being less than 4 to greater than 4 at these values. ...

  13. Algebra

    Assignment Problems; Show/Hide; Show all Solutions/Steps/etc. Hide all Solutions/Steps/etc. ... First Order DE's. 2.1 Linear Equations; 2.2 Separable Equations; 2.3 Exact Equations; ... In the case the quantities inside the absolute value were the same number but opposite signs. However, upon taking the absolute value we got the same number and ...

  14. 1.9.2: Absolute Value

    For example, to find the absolute value of 7, locate 7 on the real line and then find its distance from the origin. To find the absolute value of βˆ’7, locate βˆ’7 on the real line and then find its distance from the origin. Some like to say that taking the absolute value "produces a number that is always positive."

  15. Absolute Value

    So the absolute value of 6 is 6, and the absolute value of βˆ’6 is also 6. More Examples: The absolute value of βˆ’9 is 9; The absolute value of 3 is 3; The absolute value of 0 is 0; The absolute value of βˆ’156 is 156; No Negatives! So in practice "absolute value" means to remove any negative sign in front of a number, and to think of all ...

  16. The Relationship Between Absolute Value and Order

    This means that the order of negative integers is opposite the order of their absolute values. Exercise 1. Complete the steps below to order the numbers. a. Separate the set of numbers into positive and negative values and zero in the top cells below. b. Write the absolute values of the rational numbers (order does not matter) in the bottom ...

  17. 1-2 Order of Operations and Evaluating Expressions

    Order of Operations and Evaluating Expressions - Word Docs & PowerPoints. 1-2 Assignment - Order of Operations and Evaluating Expressions. 1-2 Bell Work - Order of Operations and Evaluating Expressions. 1-2 Exit Quiz - Order of Operations and Evaluating Expressions. 1-2 Guide Notes SE - Order of Operations and Evaluating Expressions.

  18. Math / Number Order and Absolute Value Flashcards

    Graph the following expression on the number line by placing the dot in the proper locations. | x | = 2. points at -2 and 2. Study with Quizlet and memorize flashcards containing terms like The Rational numbers include which of the following? fractions all square roots positive integers negative integers, Evaluate |-3| Β±3 -3 3, Evaluate |-7 ...

  19. 2.7 Linear Inequalities and Absolute Value Inequalities

    Solving Absolute Value Inequalities. As we know, the absolute value of a quantity is a positive number or zero. From the origin, a point located at (βˆ’ x, 0) (βˆ’ x, 0) has an absolute value of x, x, as it is x units away. Consider absolute value as the distance from one point to another point.

  20. 1.2

    The following video uses the order of operations to simplify an expression in fraction form that contains absolute value terms. Note how the absolute values are treated like parentheses and brackets when using the order of operations. Simplify an Expression in Fraction Form with Absolute Values.

  21. Solving Absolute Value Equations

    The absolute value of any number is either positive or zero. But this equation suggests that there is a number whose absolute value is negative. Can you think of any numbers that can make the equation true? Well, there is none. Since there's no value of [latex]x [/latex] that can satisfy the equation, we say that it has no solution.

  22. Absolute Value of a Number

    The absolute value of a number is the distance of the number from zero, in a number line. Visit BYJU'S to learn the definition, symbol, properties, and examples in detail. ... Now, arrange the numbers in ascending order (smallest to the largest number)-24, 15, 16, 17,19, 21, 75, 109. Example 3: Find the absolute value of a number -12/5 ...