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== 2.5#h(0.6em)The Absolute Value Function


    
      
=== Introduction


      The absolute value function is used to model problems involving distance. Recall that the absolute value of a number gives the distance from the origin to that number on the number line.


      #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
#emph[Distance and Absolute Value]


        The distance between two points #math.equation(block: false, alt: "x")[$x$] and #math.equation(block: false, alt: "a")[$a$] is given by #math.equation(block: false, alt: "vertical bar x minus a vertical bar")[$| x − a |$].
]


      For example, the equation #math.equation(block: false, alt: "vertical bar x minus 2 vertical bar equals 6")[$| x − 2 | = 6$] means "the distance between #math.equation(block: false, alt: "x")[$x$] and #math.equation(block: false, alt: "2")[$2$] is #math.equation(block: false, alt: "6")[$6$] units." The number #math.equation(block: false, alt: "x")[$x$] could be to the left or the right of #math.equation(block: false, alt: "2")[$2$] on the number line. Thus, the equation has two solutions, #math.equation(block: false, alt: "8")[$8$] and #math.equation(block: false, alt: "minus 4")[$− 4$], as shown below.


      #figure(figph[number line showing 6 units from 2], alt: "number line showing 6 units from 2", caption: none)


      
        To find the absolute value of a number, we can


        \_\_\_\_\_


        #solutionbox[
To find the absolute value of a number, we can find its distance from 0.
]


      
      
        To find the absolute value of a number, we can


        + change its sign.
+ subtract it from 0.
+ find its distance from 0.
+ square it.


      
      #examplebox("Example 1")[][
Write each statement using absolute value notation. Illustrate the solutions on a number line.


          + #math.equation(block: false, alt: "x")[$x$] is three units from the origin.
+ #math.equation(block: false, alt: "p")[$p$] is two units from #math.equation(block: false, alt: "5")[$5$].
+ #math.equation(block: false, alt: "a")[$a$] is within four units of #math.equation(block: false, alt: "minus 2")[$− 2$].


        
        #solutionbox[
First, restate each statement in terms of distance.


          + The distance between #math.equation(block: false, alt: "x")[$x$] and the origin is three units, or #math.equation(block: false, alt: "vertical bar x vertical bar equals 3")[$| x | = 3$]. Thus, #math.equation(block: false, alt: "x")[$x$] can be #math.equation(block: false, alt: "3")[$3$] or #math.equation(block: false, alt: "minus 3")[$− 3$]. #figure(figph[number line showing absolute value of x is 3], alt: "number line showing absolute value of x is 3", caption: none)
+ The distance between #math.equation(block: false, alt: "p")[$p$] and #math.equation(block: false, alt: "5")[$5$] is two units, or #math.equation(block: false, alt: "vertical bar p minus 5 vertical bar equals 2")[$" " | p − 5 | = 2$]. If we count two units on either side of #math.equation(block: false, alt: "5")[$5$], we see that #math.equation(block: false, alt: "p")[$p$] can be #math.equation(block: false, alt: "3")[$3$] or #math.equation(block: false, alt: "7")[$7$]. #figure(figph[number line showing 2 units from 5], alt: "number line showing 2 units from 5", caption: none)
+ The distance between #math.equation(block: false, alt: "a")[$a$] and #math.equation(block: false, alt: "minus 2")[$− 2$] is less than four units, or #math.equation(block: false, alt: "vertical bar a minus open parenthesis minus 2 close parenthesis vertical bar less than 4")[$" " | a − ( − 2 ) | < 4 " "$], or #math.equation(block: false, alt: "vertical bar a plus 2 vertical bar less than 4")[$" " | a + 2 | < 4$]. Count four units on either side of #math.equation(block: false, alt: "minus 2")[$− 2$], to find #math.equation(block: false, alt: "minus 6")[$− 6$] and #math.equation(block: false, alt: "2")[$2$]. Then #math.equation(block: false, alt: "a")[$a$] is between #math.equation(block: false, alt: "minus 6")[$− 6$] and #math.equation(block: false, alt: "2")[$2$], or #math.equation(block: false, alt: "minus 6 less than a less than 2")[$− 6 < a < 2$].#figure(figph[numner line showing 4 units from -2], alt: "numner line showing 4 units from -2", caption: none)
]
]


      
        The notation #math.equation(block: false, alt: "vertical bar x minus 3 vertical bar equals 5")[$| x − 3 | = 5$] means


        \_\_\_\_\_


        #solutionbox[
The notation #math.equation(block: false, alt: "vertical bar x minus 3 vertical bar equals 5")[$| x − 3 | = 5$] means that the distance between #math.equation(block: false, alt: "x")[$x$] and 3 is 5 units.
]


      
      
        The notation #math.equation(block: false, alt: "vertical bar x minus 3 vertical bar equals 5")[$" " | x − 3 | = 5 " "$] means


        + #math.equation(block: false, alt: "x")[$x$] is 3 units bigger than 5.
+ the distance between #math.equation(block: false, alt: "x")[$x$] and 5 is 3 units.
+ 5 and 3 are #math.equation(block: false, alt: "x")[$x$] units apart.
+ the distance between #math.equation(block: false, alt: "x")[$x$] and 3 is 5 units.


      
      
        Write each statement using absolute value notation; then illustrate the solutions on a number line.


        + #math.equation(block: false, alt: "x")[$x$] is five units away from #math.equation(block: false, alt: "minus 3")[$− 3$]. #linebreak() \_\_\_\_\_#math.equation(block: false, alt: "equals")[$=$]\_\_\_\_\_
+ #math.equation(block: false, alt: "x")[$x$] is at least six units away from #math.equation(block: false, alt: "4")[$4$]. #linebreak() \_\_\_\_\_#math.equation(block: false, alt: "greater than or equal to")[$≥$]\_\_\_\_\_


        Use "abs(x)" to get #math.equation(block: false, alt: "vertical bar x vertical bar")[$| x |$], etc.


        #solutionbox[
+ #math.equation(block: false, alt: "vertical bar x plus 3 vertical bar equals 5")[$| x + 3 | = 5$] #linebreak() A number line is shown below.
+ #math.equation(block: false, alt: "vertical bar x minus 4 vertical bar greater than or equal to 6")[$| x − 4 | ≥ 6$] #linebreak() A number line is shown below.
]


        Number lines:


        + #figure(figph[number line], alt: "number line", caption: none)
+ #figure(figph[number line], alt: "number line", caption: none)


      
      
        Write each statement using absolute value notation; then illustrate the solutions on a number line.


        + #math.equation(block: false, alt: "x")[$x$] is five units away from #math.equation(block: false, alt: "minus 3")[$− 3$].
+ #math.equation(block: false, alt: "x")[$x$] is at least six units away from #math.equation(block: false, alt: "4")[$4$].


        #solutionbox[
+ #math.equation(block: false, alt: "vertical bar x plus 3 vertical bar equals 5")[$| x + 3 | = 5$]#figure(figph[number line showing 5 units from -3], alt: "number line showing 5 units from -3", caption: none)
+ #math.equation(block: false, alt: "vertical bar x minus 4 vertical bar greater than or equal to 6")[$| x − 4 | ≥ 6$]#figure(figph[number line showing 6 units from 4], alt: "number line showing 6 units from 4", caption: none)
]


      
    
    
      
=== Absolute Value Equations


      We can use distances on a number line to solve simple equations such as


      #math.equation(block: true, alt: "vertical bar 3 x minus 6 vertical bar equals 9")[$| 3 x − 6 | = 9$]
      First, we factor out the coefficient of #math.equation(block: false, alt: "x")[$x$], to get #math.equation(block: false, alt: "vertical bar 3 open parenthesis x minus 2 close parenthesis vertical bar equals 9")[$" " | 3 ( x − 2 ) | = 9$]. Because of the multiplicative property of the absolute value, namely that 
    #math.equation(block: false, alt: "vertical bar a b vertical bar equals vertical bar a vertical bar vertical bar b vertical bar")[$| a b | = | a | | b |$], we can write the left side as


      #math.equation(block: true, alt: "vertical bar 3 vertical bar vertical bar x minus 2 vertical bar, equals 9; 3 vertical bar x minus 2 vertical bar, equals 9, Divide both sides by 3.; vertical bar x minus 2 vertical bar, equals 3")[$| 3 | | x − 2 | & = 9 \ 3 | x − 2 | & = 9 &  & "Divide both sides by 3." \ | x − 2 | & = 3$]
      which tells us that the distance between #math.equation(block: false, alt: "x")[$x$] and #math.equation(block: false, alt: "2")[$2$] is #math.equation(block: false, alt: "3")[$3$] units, so the solutions are #math.equation(block: false, alt: "x equals minus 1")[$x = − 1$] and #math.equation(block: false, alt: "x equals 5")[$x = 5$].


      Alternatively, we can use graphs when working with absolute values. For example, we know that the simple equation #math.equation(block: false, alt: "vertical bar x vertical bar equals 5")[$| x | = 5$] has two solutions, #math.equation(block: false, alt: "x equals 5")[$x = 5$] and #math.equation(block: false, alt: "x equals minus 5")[$x = − 5$].


      In fact, we can see from the graph at right that the equation #math.equation(block: false, alt: "vertical bar x vertical bar equals k")[$| x | = k$] has two solutions if #math.equation(block: false, alt: "k greater than 0")[$k > 0$], one solution if #math.equation(block: false, alt: "k equals 0")[$k = 0$], and no solution if #math.equation(block: false, alt: "k less than 0")[$k < 0$].


      #figure(figph[abs value graph and horizontal line], alt: "abs value graph and horizontal line", caption: none)


      
        Which statement is true?


        \_\_\_\_\_


        #solutionbox[
Depending on #math.equation(block: false, alt: "x")[$x$], #math.equation(block: false, alt: "vertical bar 2 x minus 8 vertical bar")[$| 2 x − 8 |$] can equal #math.equation(block: false, alt: "2 x minus 8")[$2 x − 8$] or #math.equation(block: false, alt: "8 minus 2 x")[$8 − 2 x$].
]


      
      
        Which statement is true?


        + The graph of #math.equation(block: false, alt: "y equals vertical bar 2 x minus 8 vertical bar")[$" " y = | 2 x − 8 | " "$] has no negative inputs.
+ The equation #math.equation(block: false, alt: "vertical bar 2 x minus 8 vertical bar equals minus 4")[$" " | 2 x − 8 | = − 4 " "$] has two solutions.
+ Depending on #math.equation(block: false, alt: "x , vertical bar 2 x minus 8 vertical bar")[$" " x , " " " " " " | 2 x − 8 | " "$] can equal #math.equation(block: false, alt: "2 x minus 8")[$2 x − 8$] or #math.equation(block: false, alt: "8 minus 2 x")[$8 − 2 x$].
+ The graph of #math.equation(block: false, alt: "y equals vertical bar 2 x minus 8 vertical bar")[$" " y = | 2 x − 8 | " "$] is a straight line.


      
      #examplebox("Example 2")[][
+ Use a graph of #math.equation(block: false, alt: "y equals vertical bar 3 x minus 6 vertical bar")[$" " y = | 3 x − 6 | " "$] to solve the equation #math.equation(block: false, alt: "vertical bar 3 x minus 6 vertical bar equals 9")[$" " | 3 x − 6 | = 9$].
+ Use a graph of #math.equation(block: false, alt: "y equals vertical bar 3 x minus 6 vertical bar")[$" " y = | 3 x − 6 | " "$] to solve the equation #math.equation(block: false, alt: "vertical bar 3 x minus 6 vertical bar equals minus 2")[$" " | 3 x − 6 | = − 2$].


        
        #solutionbox[
#figure(figph[graph of absolute value of 3x-6], alt: "graph of absolute value of 3x-6", caption: none)


          + The graph shows the graphs of #math.equation(block: false, alt: "y equals vertical bar 3 x minus 6 vertical bar")[$y = | 3 x − 6 |$] and #math.equation(block: false, alt: "y equals 9")[$y = 9$]. We see that there are two points on the graph of #math.equation(block: false, alt: "y equals vertical bar 3 x minus 6 vertical bar")[$" " y = | 3 x − 6 | " "$] that have #math.equation(block: false, alt: "y equals 9")[$y = 9$], and those points have #math.equation(block: false, alt: "x")[$x$]-coordinates #math.equation(block: false, alt: "x equals minus 1")[$x = − 1$] and #math.equation(block: false, alt: "x equals 5")[$x = 5$]. We can verify algebraically that the solutions are #math.equation(block: false, alt: "minus 1")[$− 1$] and #math.equation(block: false, alt: "5")[$5$]. 
        #math.equation(block: true, alt: "x, equals minus 1 : vertical bar 3 open parenthesis minus 1 close parenthesis minus 6 vertical bar equals vertical bar minus 9 vertical bar equals 9; x, equals 5 : vertical bar 3 open parenthesis 5 close parenthesis minus 6 vertical bar equals vertical bar 9 vertical bar equals 9")[$x & = − 1 ": " " " " " | 3 ( − 1 ) − 6 | = | − 9 | = 9 \ x & = 5 ": " " " " " | 3 ( 5 ) − 6 | = | 9 | = 9$]
+ There are no points on the graph of #math.equation(block: false, alt: "y equals vertical bar 3 x minus 6 vertical bar")[$" " y = | 3 x − 6 | " "$] with #math.equation(block: false, alt: "y equals minus 2")[$y = − 2$], so the equation #math.equation(block: false, alt: "vertical bar 3 x minus 6 vertical bar equals minus 2")[$" " | 3 x − 6 | = − 2 " "$] has no solutions.
]
]


      #notebox("Technology", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
#emph[Solving Absolute Value Equations]


        We can use a graphing utility to solve the equations in Example.


        The graph shows the graphs of 
        #math.equation(block: false, alt: "Y sub 1 equals abs open parenthesis 3 X minus 6 close parenthesis")[$Y_(1) = "abs" ( 3 X − 6 )$] and #math.equation(block: false, alt: "Y sub 2 equals 9")[$Y_(2) = 9$] in the window


        #math.equation(block: true, alt: "Xmin, equals minus 2.7, Xmax equals 6.7; Ymin, equals minus 2, Ymax equals 12")[$"Xmin" & = − 2.7 &  & "Xmax" = 6.7 \ "Ymin" & = − 2 &  & "Ymax" = 12$]
        We use the Trace or the #emph[intersect] feature to locate the intersection points at #math.equation(block: false, alt: "open parenthesis minus 1 , 9 close parenthesis")[$( − 1 , 9 )$] and #math.equation(block: false, alt: "open parenthesis 5 , 9 close parenthesis")[$( 5 , 9 )$].


        #figure(figph[graphing calculator graph of absolute value of 3x-6], alt: "graphing calculator graph of absolute value of 3x-6", caption: none)
]


      
        + The graph of #math.equation(block: false, alt: "y equals vertical bar 2 x plus 7 vertical bar")[$y = | 2 x + 7 |$] is #linebreak() \_\_\_\_\_
+ Use your graph to solve the equation #math.equation(block: false, alt: "vertical bar 2 x plus 7 vertical bar equals 11")[$| 2 x + 7 | = 11$]. #linebreak() #math.equation(block: false, alt: "x equals")[$x =$]\_\_\_\_\_


        Separate different values with a comma.


        #solutionbox[
+ A graph is also shown below for part (b).
+ #math.equation(block: false, alt: "x equals minus 9")[$x = − 9$] or #math.equation(block: false, alt: "x equals 2")[$x = 2$]
]


        A graph for part (b):


        #figure(figph[abs(2x+7)], alt: "abs(2x+7)", caption: none)


      
      
        + Graph #math.equation(block: false, alt: "y equals vertical bar 2 x plus 7 vertical bar")[$" " y = | 2 x + 7 | " "$] for #math.equation(block: false, alt: "minus 12 less than or equal to x less than or equal to 8")[$" " − 12 ≤ x ≤ 8$].
+ Use your graph to solve the equation #math.equation(block: false, alt: "vertical bar 2 x plus 7 vertical bar equals 11")[$" " | 2 x + 7 | = 11$].


        #solutionbox[
+ #figure(figph[abs(2x+7)], alt: "abs(2x+7)", caption: none)
+ #math.equation(block: false, alt: "x equals minus 9")[$x = − 9$] or #math.equation(block: false, alt: "x equals 2")[$x = 2$]
]


      
      To solve an absolute value equation algebraically, we use the definition of absolute value.


      #examplebox("Example 3")[][
Solve the equation #math.equation(block: false, alt: "vertical bar 3 x minus 6 vertical bar equals 9")[$" " | 3 x − 6 | = 9 " "$] algebraically.


        
        #solutionbox[
We write the piecewise definition of #math.equation(block: false, alt: "vertical bar 3 x minus 6 vertical bar")[$| 3 x − 6 |$].


          #math.equation(block: true, alt: "vertical bar 3 x minus 6 vertical bar equals open brace 3 x minus 6, if 3 x minus 6 greater than or equal to 0 , or x greater than or equal to 2; minus open parenthesis 3 x minus 6 close parenthesis, if 3 x minus 6 less than 0 , or x less than 2")[$| 3 x − 6 | = \{ 3 x − 6 & "if " 3 x − 6 ≥ 0 ", or " x ≥ 2 \ − ( 3 x − 6 ) & "if " 3 x − 6 < 0 ", or " x < 2$]
          Thus, the absolute value equation #math.equation(block: false, alt: "vertical bar 3 x minus 6 vertical bar equals 9")[$" " | 3 x − 6 | = 9 " "$] is equivalent to two regular equations:


          #math.equation(block: true, alt: "3 x minus 6 equals 9 or minus open parenthesis 3 x minus 6 close parenthesis equals 9")[$3 x − 6 = 9 " " " " " " " or " " " " " " " − ( 3 x − 6 ) = 9$]
          or, by simplifying the second equation,


          #math.equation(block: true, alt: "3 x minus 6 equals 9 or 3 x minus 6 equals minus 9")[$3 x − 6 = 9 " " " " " " " or " " " " " " " 3 x − 6 = − 9$]
          Solving these two equations gives us the same solutions we found in Example, namely #math.equation(block: false, alt: "x equals 5")[$x = 5$] and #math.equation(block: false, alt: "minus 1")[$− 1$].
]
]


      In general, we have the following strategy for solving absolute value equations.


      #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
#emph[Absolute Value Equations]


        The equation


        #math.equation(block: true, alt: "vertical bar a x plus b vertical bar equals c open parenthesis c greater than 0 close parenthesis")[$| a x + b | = c #hide($b r e a k$) ( c > 0 )$]
        is equivalent to


        #math.equation(block: true, alt: "a x plus b equals c or a x plus b equals minus c")[$a x + b = c " " " " " or " " " " " a x + b = − c$]
]


      
        Solve  #math.equation(block: false, alt: "vertical bar 2 x plus 7 vertical bar equals 11")[$| 2 x + 7 | = 11$] algebraically.


        #math.equation(block: false, alt: "x equals")[$x =$]\_\_\_\_\_


        Separate different values with a comma.


        #solutionbox[
#math.equation(block: false, alt: "x equals minus 9")[$x = − 9$] or #math.equation(block: false, alt: "x equals 2")[$x = 2$]
]


      
      
        Solve  #math.equation(block: false, alt: "vertical bar 2 x plus 7 vertical bar equals 11")[$" " | 2 x + 7 | = 11 " "$] algebraically.


        #solutionbox[
The absolute value equation is equivalent to two regular equations:


          #math.equation(block: true, alt: "2 x plus 7 equals 11 or 2 x plus 7 equals minus 11")[$2 x + 7 = 11 " " " " " " " or " " " " " " " 2 x + 7 = − 11$]
          whose solutions are #math.equation(block: false, alt: "x equals minus 9")[$x = − 9$] or #math.equation(block: false, alt: "x equals 2")[$x = 2$]
]


      
      
        Explain how the equation #math.equation(block: false, alt: "vertical bar a x plus b vertical bar equals c")[$| a x + b | = c$] can have two solutions, one solution, or no solutions.


        \_\_\_\_\_


      
      
        Explain how the equation #math.equation(block: false, alt: "vertical bar a x plus b vertical bar equals c")[$" " | a x + b | = c " "$] can have two solutions, one solution, or no solutions.


      
    
    
      
=== Absolute Value Inequalities


      We can also use graphs to solve absolute value inequalities. Look again at the graph of #math.equation(block: false, alt: "y equals vertical bar 3 x minus 6 vertical bar")[$" " y = | 3 x − 6 | " "$] in figure (a) below.


      #figure(figph[abs-3x-6], alt: "abs-3x-6", caption: none)


      Because of the V-shape of the graph, all points with #math.equation(block: false, alt: "y")[$y$]-values less than #math.equation(block: false, alt: "9")[$9$] lie between the two solutions of #math.equation(block: false, alt: "vertical bar 3 x minus 6 vertical bar equals 9")[$" " | 3 x − 6 | = 9 " "$], that is, between #math.equation(block: false, alt: "minus 1")[$− 1$] and #math.equation(block: false, alt: "5")[$5$]. Thus, the solutions of the inequality #math.equation(block: false, alt: "vertical bar 3 x minus 6 vertical bar less than 9")[$" " | 3 x − 6 | < 9 " "$] are #math.equation(block: false, alt: "minus 1 less than x less than 5")[$− 1 < x < 5$]. (In the Homework Problems, you will be asked to show this algebraically.)


      On the other hand, to solve the inequality #math.equation(block: false, alt: "vertical bar 3 x minus 6 vertical bar greater than 9")[$" " | 3 x − 6 | > 9 " "$], we look for points on the graph with #math.equation(block: false, alt: "y")[$y$]-values greater than #math.equation(block: false, alt: "9")[$9$]. In figure (b), we see that these points have #math.equation(block: false, alt: "x")[$x$]-values outside the interval between #math.equation(block: false, alt: "minus 1")[$− 1$] and #math.equation(block: false, alt: "5")[$5$]. In other words, the solutions of the inequality #math.equation(block: false, alt: "vertical bar 3 x minus 6 vertical bar greater than 9")[$" " | 3 x − 6 | > 9 " "$] are #math.equation(block: false, alt: "x less than minus 1")[$x < − 1$] or #math.equation(block: false, alt: "x greater than 5")[$x > 5$].


      Thus, we can solve an absolute value inequality by first solving the related equation.


      #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
#emph[Absolute Value Inequalities]


        Suppose the solutions of the equation #math.equation(block: false, alt: "vertical bar a x plus b vertical bar equals c")[$| a x + b | = c$] are #math.equation(block: false, alt: "r")[$r$] and #math.equation(block: false, alt: "s")[$s$], with #math.equation(block: false, alt: "r less than s")[$r < s$]. Then


        + The solutions of #math.equation(block: false, alt: "vertical bar a x plus b vertical bar less than c")[$" " | a x + b | < c " "$] are 
            #math.equation(block: true, alt: "r less than x less than s")[$r < x < s$]
+ The solutions of #math.equation(block: false, alt: "vertical bar a x plus b vertical bar greater than c")[$" " | a x + b | > c " "$] are
            #math.equation(block: true, alt: "x less than r or x greater than s")[$x < r " " " " " or " " " " " x > s$]
]


      
        If #math.equation(block: false, alt: "vertical bar x minus 3 vertical bar greater than 5")[$| x − 3 | > 5$], then


        \_\_\_\_\_


        #solutionbox[
If #math.equation(block: false, alt: "vertical bar x minus 3 vertical bar greater than 5")[$| x − 3 | > 5$], then either #math.equation(block: false, alt: "x minus 3 greater than 5")[$x − 3 > 5$] or #math.equation(block: false, alt: "x minus 3 less than minus 5")[$x − 3 < − 5$].
]


      
      
        If #math.equation(block: false, alt: "vertical bar x minus 3 vertical bar greater than 5")[$" " | x − 3 | > 5 " "$], then


        + either #math.equation(block: false, alt: "x minus 3 greater than 5")[$" " x − 3 > 5 " "$] or #math.equation(block: false, alt: "x minus 3 greater than minus 5")[$" " x − 3 > − 5$].
+ #math.equation(block: false, alt: "5 greater than x minus 3 greater than minus 5")[$" " 5 > x − 3 > − 5$].
+ either #math.equation(block: false, alt: "x minus 3 greater than 5")[$" " x − 3 > 5 " "$] or #math.equation(block: false, alt: "x minus 3 less than minus 5")[$" " x − 3 < − 5$].
+ #math.equation(block: false, alt: "minus 5 greater than x minus 3 less than 5")[$" " − 5 > x − 3 < 5$].


      
      #examplebox("Example 4")[][
Solve #math.equation(block: false, alt: "vertical bar 4 x minus 15 vertical bar less than 0.01")[$" " | 4 x − 15 | < 0.01$]


        
        #solutionbox[
First, we solve the equation #math.equation(block: false, alt: "vertical bar 4 x minus 15 vertical bar equals 0.01")[$" " | 4 x − 15 | = 0.01$]. There are two cases:


          #math.equation(block: true, alt: "4 x minus 15, equals 0.01, or, 4 x minus 15, equals minus 0.01; 4 x, equals 15.01, 4 x, equals 14.99; x, equals 3.7525, x, equals 3.7475")[$4 x − 15 & = 0.01 & "or" &  & 4 x − 15 & = − 0.01 \ 4 x & = 15.01 &  &  & 4 x & = 14.99 \ x & = 3.7525 &  &  & x & = 3.7475$]
          Because the inequality symbol is #math.equation(block: false, alt: "less than")[$<$], the solutions of the inequality are between these two values: #math.equation(block: false, alt: "3.7475 less than x less than 3.7525")[$3.7475 < x < 3.7525$]. In interval notation, the solutions are #math.equation(block: false, alt: "open parenthesis 3.7475 , 3.7525 close parenthesis")[$( 3.7475 , 3.7525 )$].
]
]


      
        + Solve the inequality #math.equation(block: false, alt: "vertical bar 2 x plus 7 vertical bar less than 11")[$| 2 x + 7 | < 11$] #linebreak() \_\_\_\_\_
+ Solve the inequality #math.equation(block: false, alt: "vertical bar 2 x plus 7 vertical bar greater than 11")[$| 2 x + 7 | > 11$] #linebreak() \_\_\_\_\_


        You may use inequality symbols or enter your answers using interval notation. When using interval notation, use "inf" for #math.equation(block: false, alt: "∞")[$upright(∞)$] and use "U" (an upper case letter u) for the union symbol #math.equation(block: false, alt: "union")[$∪$].


        #solutionbox[
+ #math.equation(block: false, alt: "open parenthesis minus 9 , 2 close parenthesis")[$( − 9 , 2 )$], or #math.equation(block: false, alt: "minus 9 less than x less than 2")[$− 9 < x < 2$]
+ #math.equation(block: false, alt: "open parenthesis minus ∞ , minus 9 close parenthesis union open parenthesis 2 , ∞ close parenthesis")[$( − upright(∞) , − 9 ) ∪ ( 2 , upright(∞) )$]; as inequalities, #math.equation(block: false, alt: "x less than minus 9")[$x < − 9$] or #math.equation(block: false, alt: "x greater than 2")[$x > 2$]
]


      
      
        + Solve the inequality #math.equation(block: false, alt: "vertical bar 2 x plus 7 vertical bar less than 11")[$" " | 2 x + 7 | < 11$]
+ Solve the inequality #math.equation(block: false, alt: "vertical bar 2 x plus 7 vertical bar greater than 11")[$" " | 2 x + 7 | > 11$]


        #solutionbox[
+ #math.equation(block: false, alt: "open parenthesis minus 9 , 2 close parenthesis")[$( − 9 , 2 )$], or #math.equation(block: false, alt: "minus 9 less than x less than 2")[$− 9 < x < 2$]
+ #math.equation(block: false, alt: "open parenthesis minus ∞ , minus 9 close parenthesis union open parenthesis 2 , ∞ close parenthesis")[$( − upright(∞) , − 9 ) ∪ ( 2 , upright(∞) )$]; as inequalities, #math.equation(block: false, alt: "x less than minus 9")[$x < − 9$] or #math.equation(block: false, alt: "x greater than 2")[$x > 2$]
]


      
    
    
      
=== Using the Absolute Value in Modeling


      In the next Example, we use the absolute value function to model a problem about distances.


      #examplebox("Example 5")[][
Marlene is driving to a new outlet mall on Highway 17. There is a gas station at Marlene's on-ramp, where she buys gas and resets her odometer to zero before getting on the highway. The mall is only 15 miles from Marlene’s on-ramp, but she mistakenly drives past the mall and continues down the highway. Marlene's distance from the mall is a function of how far she has driven on Highway 17.


          #figure(figph[highway 17 with gas station and mall], alt: "highway 17 with gas station and mall", caption: none)


          + Make a table of values showing how far Marlene has driven on Highway 17 and how far she is from the mall.
+ Make a graph of Marlene’s distance from the mall versus the number of miles she has driven on the highway. Which of the basic graphs from Section 2.2 does your graph most resemble?
+ Find a piecewise defined formula that describes Marlene’s distance from the mall as a function of the distance she has driven on the highway.


        
        #solutionbox[
+ Marlene gets closer to the mall for each mile that she has driven on the highway until she has driven 15 miles, and after that she gets farther from the mall.#figure(table(
  columns: 8, align: left, inset: 6pt,
  table.header([Miles on highway], [#math.equation(block: false, alt: "0")[$0$]], [#math.equation(block: false, alt: "5")[$5$]], [#math.equation(block: false, alt: "10")[$10$]], [#math.equation(block: false, alt: "15")[$15$]], [#math.equation(block: false, alt: "20")[$20$]], [#math.equation(block: false, alt: "25")[$25$]], [#math.equation(block: false, alt: "30")[$30$]]),
  [Miles from mall], [#math.equation(block: false, alt: "15")[$15$]], [#math.equation(block: false, alt: "10")[$10$]], [#math.equation(block: false, alt: "5")[$5$]], [#math.equation(block: false, alt: "0")[$0$]], [#math.equation(block: false, alt: "5")[$5$]], [#math.equation(block: false, alt: "10")[$10$]], [#math.equation(block: false, alt: "15")[$15$]],
))
+ We plot the points in the table to obtain the graph shown below. This graph looks like the absolute value function, except that the vertex is the point #math.equation(block: false, alt: "open parenthesis 15 , 0 close parenthesis")[$( 15 , 0 )$] instead of the origin.#figure(figph[graph of distance from mall], alt: "graph of distance from mall", caption: none)
+ Let #math.equation(block: false, alt: "x")[$x$] represent the number of miles on the highway and #math.equation(block: false, alt: "f open parenthesis x close parenthesis")[$f ( x )$] the number of miles from the mall. For #math.equation(block: false, alt: "x")[$x$]-values less than #math.equation(block: false, alt: "15")[$15$], the graph is a straight line with slope #math.equation(block: false, alt: "minus 1")[$− 1$] and #math.equation(block: false, alt: "y")[$y$]-intercept at #math.equation(block: false, alt: "open parenthesis 0 , 15 close parenthesis")[$( 0 , 15 )$], so its equation is #math.equation(block: false, alt: "y equals minus x plus 15")[$" " y = − x + 15$]. Thus,
        #math.equation(block: true, alt: "f open parenthesis x close parenthesis equals minus x plus 15 when 0 less than or equal to x less than 15")[$f ( x ) = − x + 15 " " " " " " " when " " " " " " " 0 ≤ x < 15$]
    On the other hand, when #math.equation(block: false, alt: "x greater than or equal to 15")[$x ≥ 15$], the graph of #math.equation(block: false, alt: "f")[$f$] is a straight line with slope #math.equation(block: false, alt: "1")[$1$] that passes through the point #math.equation(block: false, alt: "open parenthesis 15 , 0 close parenthesis")[$( 15 , 0 )$]. The point-slope form of this line is
        #math.equation(block: true, alt: "y equals 0 plus 1 open parenthesis x minus 15 close parenthesis")[$y = 0 + 1 ( x − 15 )$]
    so #math.equation(block: false, alt: "y equals x minus 15")[$y = x − 15$]. Thus,
        #math.equation(block: true, alt: "f open parenthesis x close parenthesis equals x minus 15 when x greater than or equal to 15")[$f ( x ) = x − 15 " " " " " " " when " " " " " " " x ≥ 15$]
    Combining the two pieces, we obtain
        #math.equation(block: true, alt: "f open parenthesis x close parenthesis equals open brace minus x plus 15, when 0 less than or equal to x less than 15; x minus 15, when x greater than or equal to 15")[$f ( x ) = \{ − x + 15 " " " " " " & "when " " " " " " " 0 ≤ x < 15 \ x − 15 " " " " " " & "when " " " " " " " x ≥ 15$]
    The graph of #math.equation(block: false, alt: "f open parenthesis x close parenthesis")[$f ( x )$] is a part of the graph of #math.equation(block: false, alt: "y equals vertical bar x minus 15 vertical bar")[$" " y = | x − 15 |$]. If we think of the highway as a portion of the real number line, with Marlene’s on-ramp located at the origin, then the outlet mall is located at #math.equation(block: false, alt: "15")[$15$]. Marlene's coordinate as she drives along the highway is #math.equation(block: false, alt: "x")[$x$], and the distance from Marlene to the mall is given by #math.equation(block: false, alt: "f open parenthesis x close parenthesis equals vertical bar x minus 15 vertical bar")[$" " f ( x ) = | x − 15 |$].
]
]


      
        #figure(figph[graph of distance from mall], alt: "graph of distance from mall", caption: none)


        + Use the graph above from the previous example  to determine how far Marlene has driven when she is within 5 miles of the mall. Write and solve an absolute value inequality to verify your answer. #linebreak() Inequality: \_\_\_\_\_#math.equation(block: false, alt: "less than")[$<$] \_\_\_\_\_ Note: Use "abs(x)" to get #math.equation(block: false, alt: "vertical bar x vertical bar")[$| x |$]. #linebreak() Solution: \_\_\_\_\_
+ Write and solve an absolute value inequality to determine how far Marlene has driven when she is at least #math.equation(block: false, alt: "10")[$10$] miles from the mall. #linebreak() Inequality: \_\_\_\_\_#math.equation(block: false, alt: "greater than or equal to")[$≥$] \_\_\_\_\_ Note: Use "abs(x)" to get #math.equation(block: false, alt: "vertical bar x vertical bar")[$| x |$]. #linebreak() Solution: \_\_\_\_\_


        You may enter your solutions using inequality symbols or using interval notation. When using inequalities, type in "\<=" for #math.equation(block: false, alt: "less than or equal to")[$≤$], and type in "\>=" for #math.equation(block: false, alt: "greater than or equal to")[$≥$]. When using interval notation, type "inf" for #math.equation(block: false, alt: "∞")[$upright(∞)$] and type "U" (an upper case letter u) for the union symbol #math.equation(block: false, alt: "union")[$∪$].


        #solutionbox[
+ #math.equation(block: false, alt: "vertical bar x minus 15 vertical bar less than 5")[$| x − 15 | < 5$]; #math.equation(block: false, alt: "10 less than x less than 20")[$" " 10 < x < 20$] #linebreak() A graph is below.
+ #math.equation(block: false, alt: "vertical bar x minus 15 vertical bar greater than or equal to 10")[$| x − 15 | ≥ 10$]; #math.equation(block: false, alt: "x less than or equal to 5")[$" " x ≤ 5 " "$] or #math.equation(block: false, alt: "x greater than or equal to 25")[$" " x ≥ 25$] #linebreak() A graph is below.
]


        Graphs are below.


        + #figure(figph[absolute value inequality], alt: "absolute value inequality", caption: none)
+ #figure(figph[absolute value inequality], alt: "absolute value inequality", caption: none)


      
      
        + Use the graph above from the previous example  to determine how far Marlene has driven when she is within 5 miles of the mall. Write and solve an absolute value inequality to verify your answer.
+ Write and solve an absolute value inequality to determine how far Marlene has driven when she is at least #math.equation(block: false, alt: "10")[$10$] miles from the mall.


        #solutionbox[
+ #math.equation(block: false, alt: "vertical bar x minus 15 vertical bar less than 5")[$| x − 15 | < 5$]; #math.equation(block: false, alt: "10 less than x less than 20")[$" " 10 < x < 20$]#figure(figph[absolute value inequality], alt: "absolute value inequality", caption: none)
+ #math.equation(block: false, alt: "vertical bar x minus 15 vertical bar greater than or equal to 10")[$| x − 15 | ≥ 10$]; #math.equation(block: false, alt: "x less than or equal to 5")[$" " x ≤ 5 " "$] or #math.equation(block: false, alt: "x greater than or equal to 25")[$" " x ≥ 25$]#figure(figph[absolute value inequality], alt: "absolute value inequality", caption: none)
]


      
    
    
      
=== Measurement Error


      If you weigh a sample in chemistry lab, the scale’s digital readout might show #math.equation(block: false, alt: "6.0")[$6.0$] grams. But it is unlikely that the sample weighs #emph[exactly] #math.equation(block: false, alt: "6")[$6$] grams; there is always some error in measured values.


      Because the scale shows the weight as #math.equation(block: false, alt: "6.0")[$6.0$] grams, we know that the true weight of the sample must be between #math.equation(block: false, alt: "5.95")[$5.95$] grams and #math.equation(block: false, alt: "6.05")[$6.05$] grams: If the weight were less than #math.equation(block: false, alt: "5.95")[$5.95$] grams, the scale would round down to #math.equation(block: false, alt: "5.9")[$5.9$] grams, and if the weight were more than #math.equation(block: false, alt: "6.05")[$6.05$] grams, the scale would round up to #math.equation(block: false, alt: "6.1")[$6.1$] grams. We should report the mass of the sample as #math.equation(block: false, alt: "6 plus or minus 0.05")[$6 ± 0.05$] grams, which tells the reader that the error in the measurement is no more than #math.equation(block: false, alt: "0.05")[$0.05$] grams.


      We can also describe this measurement error, or #strong[error tolerance], using an absolute value inequality. Because the measured mass #math.equation(block: false, alt: "m")[$m$] can be no more than #math.equation(block: false, alt: "0.05")[$0.05$] from #math.equation(block: false, alt: "6")[$6$], we write


      #math.equation(block: true, alt: "vertical bar m minus 6 vertical bar less than or equal to 0.05")[$| m − 6 | ≤ 0.05$]
      Note that the solution of this inequality is #math.equation(block: false, alt: "5.95 less than or equal to m less than or equal to 6.05")[$5.95 ≤ m ≤ 6.05$].


      #examplebox("Example 6")[][
+ The specifications for a computer chip state that its thickness in millimeters must satisfy #math.equation(block: false, alt: "vertical bar t minus 0.023 vertical bar less than 0.001")[$" " | t − 0.023 | < 0.001$]. What are the acceptable values for the thickness of the chip?
+ The safe dosage of a new drug is between #math.equation(block: false, alt: "250")[$250$] and #math.equation(block: false, alt: "450")[$450$] milligrams, inclusive. Write the safe dosage as an error tolerance involving absolute values.


        
        #solutionbox[
+ The error tolerance can also be stated as #math.equation(block: false, alt: "t equals 0.023 plus or minus 0.001")[$t = 0.023 ± 0.001$] millimeters, so the acceptable values are between #math.equation(block: false, alt: "0.022")[$0.022$] and #math.equation(block: false, alt: "0.024")[$0.024$] millimeters.
+ The safe dosage #math.equation(block: false, alt: "d")[$d$] satisfies #math.equation(block: false, alt: "250 less than or equal to d less than or equal to 450")[$250 ≤ d ≤ 450$], as shown below.#figure(figph[safe dosage], alt: "safe dosage", caption: none)

 #linebreak() The center of this interval is #math.equation(block: false, alt: "350")[$350$], and the endpoints are each #math.equation(block: false, alt: "100")[$100$] units from the center. Thus, the safe values are within #math.equation(block: false, alt: "100")[$100$] units of #math.equation(block: false, alt: "350")[$350$], or
        #math.equation(block: true, alt: "vertical bar d minus 350 vertical bar less than or equal to 100")[$| d − 350 | ≤ 100$]
]
]


      
        Which statement is false?


        \_\_\_\_\_


        #solutionbox[
"The solutions of #math.equation(block: false, alt: "vertical bar 3 x plus 12 vertical bar greater than 6")[$| 3 x + 12 | > 6$] form a closed interval" is a false statement.
]


      
      
        Which statement is false?


        + The equation #math.equation(block: false, alt: "vertical bar 3 x plus 12 vertical bar equals 0")[$" " | 3 x + 12 | = 0 " "$] has one solution.
+ The statement #math.equation(block: false, alt: "vertical bar T minus 0.1 vertical bar less than 0.05")[$" " | T − 0.1 | < 0.05 " "$] describes an error tolerance of 0.05 units.
+ #math.equation(block: false, alt: "vertical bar m x plus b vertical bar less than k")[$" " | m x + b | < k " "$] is equivalent to #math.equation(block: false, alt: "minus k less than m x plus b less than k")[$" " − k < m x + b < k$].
+ The solutions of #math.equation(block: false, alt: "vertical bar 3 x plus 12 vertical bar greater than 6")[$" " | 3 x + 12 | > 6 " "$] form a closed interval.


      
      
        The temperature, #math.equation(block: false, alt: "T")[$T$], in a laboratory must remain between #math.equation(block: false, alt: "9 degrees C")[$9^(∘) "C"$] and #math.equation(block: false, alt: "12 degrees C")[$12^(∘) "C"$].


        + Write the error tolerance as an absolute value inequality. #linebreak() Inequality: \_\_\_\_\_#math.equation(block: false, alt: "less than")[$<$] \_\_\_\_\_ Note: Use "abs(x)" to get #math.equation(block: false, alt: "vertical bar x vertical bar")[$| x |$].
+ For a special experiment, the temperature in degrees celsius must satisfy #math.equation(block: false, alt: "vertical bar T minus 6.7 vertical bar less than or equal to 0.03")[$| T − 6.7 | ≤ 0.03$]. Give the interval of possible temperatures. #linebreak() Interval: \_\_\_\_\_


        You may enter your solutions using inequality symbols or using interval notation. When using inequalities, type in "\<=" for #math.equation(block: false, alt: "less than or equal to")[$≤$], and type in "\>=" for #math.equation(block: false, alt: "greater than or equal to")[$≥$]. When using interval notation, type "inf" for #math.equation(block: false, alt: "∞")[$upright(∞)$] and type "U" (an upper case letter u) for the union symbol #math.equation(block: false, alt: "union")[$∪$].


        #solutionbox[
+ #math.equation(block: false, alt: "vertical bar T minus 10.5 vertical bar less than 1.5")[$| T − 10.5 | < 1.5$]
+ #math.equation(block: false, alt: "6.67 less than or equal to T less than or equal to 6.73")[$6.67 ≤ T ≤ 6.73$]
]


      
      
        The temperature, #math.equation(block: false, alt: "T")[$T$], in a laboratory must remain between #math.equation(block: false, alt: "9 degrees C")[$9^(∘) "C"$] and #math.equation(block: false, alt: "12 degrees C")[$12^(∘) "C"$].


        + Write the error tolerance as an absolute value inequality.
+ For a special experiment, the temperature in degrees celsius must satisfy #math.equation(block: false, alt: "vertical bar T minus 6.7 vertical bar less than or equal to 0.03")[$" " | T − 6.7 | ≤ 0.03$]. Give the interval of possible temperatures.


        #solutionbox[
+ #math.equation(block: false, alt: "vertical bar T minus 10.5 vertical bar less than 1.5")[$| T − 10.5 | < 1.5$]
+ #math.equation(block: false, alt: "6.67 less than or equal to T less than or equal to 6.73")[$6.67 ≤ T ≤ 6.73$]
]


      
      
        What is error tolerance used for, and why do we need it?


        \_\_\_\_\_


      
      
        What is error tolerance used for, and why do we need it?


      
    
    
      
=== Section Summary


      
        
==== Vocabulary


        Look up the definitions of new terms in the Glossary.


        - Absolute value equation
- Absolute value inequality
- Error tolerance


      
      
        
==== CONCEPTS


        + The absolute value is used to model distance: The #strong[distance] between two points #math.equation(block: false, alt: "x")[$x$] and #math.equation(block: false, alt: "a")[$a$] is given by #math.equation(block: false, alt: "vertical bar x minus a vertical bar")[$| x − a |$].
+ #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
#emph[Absolute Value Equations]

The equation

#math.equation(block: true, alt: "vertical bar a x plus b vertical bar equals c open parenthesis c greater than 0 close parenthesis")[$| a x + b | = c #hide($b r e a k$) ( c > 0 )$]is equivalent to

#math.equation(block: true, alt: "a x plus b equals c or a x plus b equals minus c")[$a x + b = c " " " " " or " " " " " a x + b = − c$]
]
+ #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
#emph[Absolute Value Inequalities]

Suppose the solutions of the equation #math.equation(block: false, alt: "vertical bar a x plus b vertical bar equals c")[$| a x + b | = c$] are #math.equation(block: false, alt: "r")[$r$] and #math.equation(block: false, alt: "s")[$s$], with #math.equation(block: false, alt: "r less than s")[$r < s$]. Then

+ The solutions of #math.equation(block: false, alt: "vertical bar a x plus b vertical bar less than c")[$| a x + b | < c$] are 
			        #math.equation(block: true, alt: "r less than x less than s")[$r < x < s$]
+ The solutions of #math.equation(block: false, alt: "vertical bar a x plus b vertical bar greater than c")[$| a x + b | > c$] are
			        #math.equation(block: true, alt: "x less than r or x greater than s")[$x < r " " " " " or " " " " " x > s$]
]
+ The #strong[error tolerance] #math.equation(block: false, alt: "e")[$e$] in a measurement #math.equation(block: false, alt: "M")[$M$] can be expressed as #math.equation(block: false, alt: "vertical bar x minus M vertical bar less than e")[$| x − M | < e$], or as #math.equation(block: false, alt: "x equals M plus or minus e")[$x = M ± e$]. Both indicate that #math.equation(block: false, alt: "M minus e less than x less than M plus e")[$M − e < x < M + e$].


      
      
        
==== STUDY QUESTIONS


        + Write a function that models the distance between #math.equation(block: false, alt: "x")[$x$] and a fixed point #math.equation(block: false, alt: "k")[$k$] on the number line.
+ For what values of #math.equation(block: false, alt: "c")[$c$] does the equation #math.equation(block: false, alt: "vertical bar a x plus b vertical bar equals c")[$| a x + b | = c$] have one solution? No solution?
+ If you know that the solutions of #math.equation(block: false, alt: "vertical bar a x plus b vertical bar less than c")[$| a x + b | < c$] are #math.equation(block: false, alt: "minus 3 less than x less than 6")[$− 3 < x < 6$], what are the solutions of #math.equation(block: false, alt: "vertical bar a x plus b vertical bar greater than c")[$| a x + b | > c$]?
+ What is the center of the interval #math.equation(block: false, alt: "open bracket 220 , 238 close bracket")[$[ 220 , 238 ]$]?
+ What is the center of the interval #math.equation(block: false, alt: "open bracket a , b close bracket")[$[ a , b ]$]?


      
      
        
==== SKILLS


        Practice each skill in the Homework problems listed.


        + Use absolute value notation to write statements about distance: \#1–8
+ Use graphs to solve absolute value equations and inequalities: \#9–12
+ Solve absolute value equations: \#13–24
+ Solve absolute value inequalities: \#25–40
+ Express error tolerances using absolute value notation: \#41–48
+ Analyze absolute value functions: \#49–56
+ Model problems about distance using the absolute value function: \#57–60


      
    
    
      
=== Homework 2.5


      In Problems 1–8,


      + Use absolute value notation to write each expression as an equation or an inequality. (It may be helpful to restate each sentence using the word #emph[distance].)
+ Illustrate the solutions on a number line.


      
        #math.equation(block: false, alt: "x")[$x$] is six units from the origin.


        
          + #math.equation(block: false, alt: "vertical bar x vertical bar equals 6")[$| x | = 6$]
+ #figure(figph[number line], alt: "number line", caption: none)


        
      
      
        #math.equation(block: false, alt: "a")[$a$] is seven units from the origin.


      
      
        The distance from #math.equation(block: false, alt: "p")[$p$] to #math.equation(block: false, alt: "minus 3")[$− 3$] is five units.


        
          + #math.equation(block: false, alt: "vertical bar p plus 3 vertical bar equals 5")[$| p + 3 | = 5$]
+ #figure(figph[number line], alt: "number line", caption: none)


        
      
      
        The distance from #math.equation(block: false, alt: "q")[$q$] to #math.equation(block: false, alt: "minus 7")[$− 7$] is two units.


      
      
        #math.equation(block: false, alt: "t")[$t$] is within three units of #math.equation(block: false, alt: "6")[$6$].


        
          + #math.equation(block: false, alt: "vertical bar t minus 6 vertical bar less than 3")[$| t − 6 | < 3$]
+ #figure(figph[number line], alt: "number line", caption: none)


        
      
      
        #math.equation(block: false, alt: "w")[$w$] is no more than one unit from #math.equation(block: false, alt: "minus 5")[$− 5$].


      
      
        #math.equation(block: false, alt: "b")[$b$] is at least #math.equation(block: false, alt: "0.5")[$0.5$] unit from #math.equation(block: false, alt: "minus 1")[$− 1$].


        
          + #math.equation(block: false, alt: "vertical bar b plus 1 vertical bar greater than or equal to 0.5")[$| b + 1 | ≥ 0.5$]
+ #figure(figph[number line], alt: "number line", caption: none)


        
      
      
        #math.equation(block: false, alt: "m")[$m$] is more than #math.equation(block: false, alt: "0.1")[$0.1$] unit from #math.equation(block: false, alt: "8")[$8$].


      
      
        Graph #math.equation(block: false, alt: "y equals vertical bar x plus 3 vertical bar")[$y = | x + 3 |$]. Use your graph to solve the following equations and inequalities.


        + #math.equation(block: false, alt: "vertical bar x plus 3 vertical bar equals 2")[$| x + 3 | = 2$]
+ #math.equation(block: false, alt: "vertical bar x plus 3 vertical bar less than or equal to 4")[$| x + 3 | ≤ 4$]
+ #math.equation(block: false, alt: "vertical bar x plus 3 vertical bar greater than 5")[$| x + 3 | > 5$]


        
          #figure(figph[absolute value graph], alt: "absolute value graph", caption: none)


          + #math.equation(block: false, alt: "x equals minus 5")[$x = − 5$] or #math.equation(block: false, alt: "x equals minus 1")[$x = − 1$]
+ #math.equation(block: false, alt: "minus 7 less than or equal to x less than or equal to 1")[$− 7 ≤ x ≤ 1$]
+ #math.equation(block: false, alt: "x less than minus 8")[$x < − 8$] or #math.equation(block: false, alt: "x greater than 2")[$x > 2$]


        
      
      
        Graph #math.equation(block: false, alt: "y equals vertical bar x minus 2 vertical bar")[$y = | x − 2 |$]. Use your graph to solve the following equations and inequalities.


        + #math.equation(block: false, alt: "vertical bar x minus 2 vertical bar equals 5")[$| x − 2 | = 5$]
+ #math.equation(block: false, alt: "vertical bar x minus 2 vertical bar less than 8")[$| x − 2 | < 8$]
+ #math.equation(block: false, alt: "vertical bar x minus 2 vertical bar greater than or equal to 4")[$| x − 2 | ≥ 4$]


      
      
        Graph #math.equation(block: false, alt: "y equals vertical bar 2 x minus 8 vertical bar")[$y = | 2 x − 8 |$]. Use your graph to solve the following equations and inequalities.


        + #math.equation(block: false, alt: "vertical bar 2 x minus 8 vertical bar equals 0")[$| 2 x − 8 | = 0$]
+ #math.equation(block: false, alt: "vertical bar 2 x minus 8 vertical bar equals minus 2")[$| 2 x − 8 | = − 2$]
+ #math.equation(block: false, alt: "vertical bar 2 x minus 8 vertical bar less than minus 6")[$| 2 x − 8 | < − 6$]


        
          #figure(figph[absolute value graph], alt: "absolute value graph", caption: none)


          + #math.equation(block: false, alt: "x equals 4")[$x = 4$]
+ No solution
+ No solution


        
      
      
        Graph #math.equation(block: false, alt: "y equals vertical bar 4 x plus 8 vertical bar")[$y = | 4 x + 8 |$]. Use your graph to solve the following equations and inequalities.


        + #math.equation(block: false, alt: "vertical bar 4 x plus 8 vertical bar equals 0")[$| 4 x + 8 | = 0$]
+ #math.equation(block: false, alt: "vertical bar 4 x plus 8 vertical bar less than 0")[$| 4 x + 8 | < 0$]
+ #math.equation(block: false, alt: "vertical bar 4 x plus 8 vertical bar greater than minus 3")[$| 4 x + 8 | > − 3$]


      
      For Problems 13–24, solve.


      
        #math.equation(block: true, alt: "vertical bar 2 x minus 1 vertical bar equals 4")[$| 2 x − 1 | = 4$]


        
          #math.equation(block: false, alt: "x equals the fraction minus 3 over 2")[$x = display(frac(− 3, 2))$] or #math.equation(block: false, alt: "x equals the fraction 5 over 2")[$x = display(frac(5, 2))$]


        
      
      
        #math.equation(block: true, alt: "vertical bar 3 x minus 1 vertical bar equals 5")[$| 3 x − 1 | = 5$]


      
      
        #math.equation(block: true, alt: "0 equals vertical bar 7 plus 3 q vertical bar")[$0 = | 7 + 3 q |$]


        
          #math.equation(block: true, alt: "q equals the fraction minus 7 over 3")[$q = display(frac(− 7, 3))$]


        
      
      
        #math.equation(block: true, alt: "vertical bar minus 11 minus 5 t vertical bar equals 0")[$| − 11 − 5 t | = 0$]


      
      
        #math.equation(block: true, alt: "4 equals the fraction vertical bar b plus 2 vertical bar over 3")[$4 = display(frac(| b + 2 |, 3))$]


        
          #math.equation(block: false, alt: "b equals minus 14")[$b = − 14$] or #math.equation(block: false, alt: "b equals 10")[$b = 10$]


        
      
      
        #math.equation(block: true, alt: "6 vertical bar n plus 2 vertical bar equals 9")[$6 | n + 2 | = 9$]


      
      
        #math.equation(block: true, alt: "vertical bar 2 open parenthesis w minus 7 close parenthesis vertical bar equals 1")[$| 2 ( w − 7 ) | = 1$]


        
          #math.equation(block: false, alt: "w equals the fraction 13 over 2")[$w = display(frac(13, 2))$] or #math.equation(block: false, alt: "w equals the fraction 15 over 2")[$w = display(frac(15, 2))$]


        
      
      
        #math.equation(block: true, alt: "2 equals vertical bar the fraction a minus 4 over 5 vertical bar")[$2 = | display(frac(a − 4, 5)) |$]


      
      
        #math.equation(block: true, alt: "vertical bar c minus 2 vertical bar plus 3 equals 1")[$| c − 2 | + 3 = 1$]


        
          No solution


        
      
      
        #math.equation(block: true, alt: "5 equals 4 minus vertical bar h plus 3 vertical bar")[$5 = 4 − | h + 3 |$]


      
      
        #math.equation(block: true, alt: "minus 7 equals vertical bar 2 m plus 3 vertical bar")[$− 7 = | 2 m + 3 |$]


        
          No solution


        
      
      
        #math.equation(block: true, alt: "vertical bar 5 r minus 3 vertical bar equals minus 2")[$| 5 r − 3 | = − 2$]


      
      For Problems 25–36, solve.


      
        #math.equation(block: true, alt: "vertical bar 2 x plus 6 vertical bar less than 3")[$| 2 x + 6 | < 3$]


        
          #math.equation(block: true, alt: "the fraction minus 9 over 2 less than x less than the fraction minus 3 over 2")[$display(frac(− 9, 2)) < x < display(frac(− 3, 2))$]


        
      
      
        #math.equation(block: true, alt: "vertical bar 5 minus 3 x vertical bar less than or equal to 1")[$| 5 − 3 x | ≤ 1$]


      
      
        #math.equation(block: true, alt: "7 less than or equal to vertical bar 3 minus 2 d vertical bar")[$7 ≤ | 3 − 2 d |$]


        
          #math.equation(block: false, alt: "d less than or equal to minus 2")[$d ≤ − 2 " "$] or #math.equation(block: false, alt: "d greater than or equal to 5")[$" " d ≥ 5$]


        
      
      
        #math.equation(block: true, alt: "10 less than vertical bar 3 r plus 2 vertical bar")[$10 < | 3 r + 2 |$]


      
      
        #math.equation(block: true, alt: "vertical bar 6 s plus 15 vertical bar greater than minus 3")[$| 6 s + 15 | > − 3$]


        
          All real numbers


        
      
      
        #math.equation(block: true, alt: "vertical bar 8 b minus 12 vertical bar less than minus 4")[$| 8 b − 12 | < − 4$]


      
      
        #math.equation(block: true, alt: "vertical bar t minus 1.5 vertical bar less than 0.1")[$| t − 1.5 | < 0.1$]


        
          #math.equation(block: true, alt: "1.4 less than t less than 1.6")[$1.4 < t < 1.6$]


        
      
      
        #math.equation(block: true, alt: "vertical bar z minus 2.6 vertical bar less than or equal to 0.1")[$| z − 2.6 | ≤ 0.1$]


      
      
        #math.equation(block: true, alt: "vertical bar T minus 3.25 vertical bar greater than or equal to 0.05")[$| T − 3.25 | ≥ 0.05$]


        
          #math.equation(block: false, alt: "T less than or equal to 3.2")[$T ≤ 3.2 " "$] or #math.equation(block: false, alt: "T greater than or equal to 3.3")[$" " T ≥ 3.3$]


        
      
      
        #math.equation(block: true, alt: "vertical bar P minus 0.6 vertical bar greater than 0.01")[$| P − 0.6 | > 0.01$]


      
      
        #math.equation(block: true, alt: "minus 1 greater than or equal to vertical bar the fraction n minus 3 over 2 vertical bar")[$− 1 ≥ | display(frac(n − 3, 2)) |$]


        
          No solution


        
      
      
        #math.equation(block: true, alt: "minus 0.1 less than or equal to vertical bar 9 open parenthesis p plus 2 close parenthesis vertical bar")[$− 0.1 ≤ | 9 ( p + 2 ) |$]


      
      In Problems 37–40, give an interval of possible values for the measurement.


      
        The length, #math.equation(block: false, alt: "l")[$l$], of a rod is given by #math.equation(block: false, alt: "vertical bar l minus 4.3 vertical bar less than 0.001")[$| l − 4.3 | < 0.001$], in centimeters.


        
          #math.equation(block: true, alt: "4.299 less than l less than 4.301")[$4.299 < l < 4.301$]


        
      
      
        The mass, #math.equation(block: false, alt: "m")[$m$], of the device shall be #math.equation(block: false, alt: "vertical bar m minus 450 vertical bar less than 4")[$| m − 450 | < 4$], in grams.


      
      
        The candle will burn for #math.equation(block: false, alt: "t")[$t$] minutes, where #math.equation(block: false, alt: "vertical bar t minus 300 vertical bar less than or equal to 50")[$| t − 300 | ≤ 50$].


        
          #math.equation(block: true, alt: "250 less than or equal to t less than or equal to 350")[$250 ≤ t ≤ 350$]


        
      
      
        The ramp will have angle of inclination #math.equation(block: false, alt: "α")[$α$], and #math.equation(block: false, alt: "vertical bar α minus 10 degrees vertical bar less than or equal to 0.5 degrees")[$| α − 10^(∘) | ≤ 0.5^(∘)$].


      
      In Problems 41–44, write the error tolerance using absolute values.


      
        The chemical compound must be maintained at a temperature, #math.equation(block: false, alt: "T")[$T$], between #math.equation(block: false, alt: "4.7 degrees")[$4.7^(∘)$] and #math.equation(block: false, alt: "5.3 degrees")[$5.3^(∘)$]C.


        
          #math.equation(block: true, alt: "vertical bar T minus 5 vertical bar less than 0.3")[$| T − 5 | < 0.3$]


        
      
      
        The diameter, #math.equation(block: false, alt: "d")[$d$], of the hole shall be in the range of
#math.equation(block: false, alt: "24.98")[$24.98$] to #math.equation(block: false, alt: "25.02")[$25.02$] centimeters.


      
      
        The subject will receive a dosage #math.equation(block: false, alt: "D")[$D$] from #math.equation(block: false, alt: "95")[$95$] to #math.equation(block: false, alt: "105")[$105$] milligrams of the drug.


        
          #math.equation(block: true, alt: "vertical bar D minus 100 vertical bar less than or equal to 5")[$| D − 100 | ≤ 5$]


        
      
      
        The pendulum swings out and back in a time period #math.equation(block: false, alt: "t")[$t$]
between #math.equation(block: false, alt: "0.9995")[$0.9995$] and #math.equation(block: false, alt: "1.0005")[$1.0005$] seconds.


      
      
        An electrical component of a high-tech sensor requires #math.equation(block: false, alt: "0.25")[$0.25$] ounce of gold. Assume that the actual amount of gold used, #math.equation(block: false, alt: "g")[$g$], is not in error by more than #math.equation(block: false, alt: "0.001")[$0.001$] ounce. Write an absolute value inequality for the possible error and show the possible values of #math.equation(block: false, alt: "g")[$g$] on a number line.


        
          #math.equation(block: true, alt: "vertical bar g minus 0.25 vertical bar less than or equal to 0.001")[$| g − 0.25 | ≤ 0.001$]


          #figure(figph[number line], alt: "number line", caption: none)


        
      
      
        In a pasteurization process, milk is to be irradiated for #math.equation(block: false, alt: "10")[$10$] seconds. The actual period #math.equation(block: false, alt: "t")[$t$] of irradiation cannot be off by more than #math.equation(block: false, alt: "0.8")[$0.8$] second. Write an absolute value inequality for the possible error and show the possible values of #math.equation(block: false, alt: "t")[$t$] on a number line.


      
      
        In a lab assignment, a student reports that a chemical reaction required #math.equation(block: false, alt: "200")[$200$] minutes to complete. Let #math.equation(block: false, alt: "t")[$t$] represent the actual time of the reaction.


        + Write an absolute value inequality for #math.equation(block: false, alt: "t")[$t$], assuming that the student rounded his answer to the nearest #math.equation(block: false, alt: "100")[$100$] minutes. Give the smallest and largest possible value for #math.equation(block: false, alt: "t")[$t$].
+ Write an absolute value inequality for #math.equation(block: false, alt: "t")[$t$], assuming that the student rounded his answer to the nearest minute. Give the smallest and largest possible value for #math.equation(block: false, alt: "t")[$t$].
+ Write an absolute value inequality for #math.equation(block: false, alt: "t")[$t$], assuming that the student rounded his answer to the nearest #math.equation(block: false, alt: "0.1")[$0.1$] minute. Give the smallest and largest possible value for #math.equation(block: false, alt: "t")[$t$].


        
          What is the shortest time that would round to #math.equation(block: false, alt: "200")[$200$] minutes? The greatest time?


        
        
          + #math.equation(block: false, alt: "vertical bar t minus 200 vertical bar less than 50")[$| t − 200 | < 50$], #math.equation(block: false, alt: "150 less than or equal to t less than 250")[$" " 150 ≤ t < 250$]
+ #math.equation(block: false, alt: "vertical bar t minus 200 vertical bar less than 0.5")[$| t − 200 | < 0.5$], #math.equation(block: false, alt: "199.5 less than or equal to t less than 200.5")[$" " 199.5 ≤ t < 200.5$]
+ #math.equation(block: false, alt: "vertical bar t minus 200 vertical bar less than 0.05")[$| t − 200 | < 0.05$], #math.equation(block: false, alt: "199.95 less than or equal to t less than 200.05")[$" " 199.95 ≤ t < 200.05$]


        
      
      
        An espresso machine has a square metal plate. The side of the plate is #math.equation(block: false, alt: "2 plus or minus 0.01")[$2 ± 0.01$] cm.


        + Write an absolute value inequality for the length of the side, #math.equation(block: false, alt: "s")[$s$]. Give the smallest and largest possible value for #math.equation(block: false, alt: "s")[$s$].
+ Compute the smallest and largest possible area of the plate, including units.
+ Write an absolute value inequality for the area, #math.equation(block: false, alt: "A")[$A$].


      
      
        + Write the piecewise definition for #math.equation(block: false, alt: "vertical bar 3 x minus 6 vertical bar")[$| 3 x − 6 |$].
+ Use your answer to part (a) to write two inequalities that together are equivalent to #math.equation(block: false, alt: "vertical bar 3 x minus 6 vertical bar less than 9")[$| 3 x − 6 | < 9$].
+ Solve the inequalities in part (b) and check that the solutions agree with the solutions of #math.equation(block: false, alt: "vertical bar 3 x minus 6 vertical bar less than 9")[$| 3 x − 6 | < 9$].
+ Show that #math.equation(block: false, alt: "vertical bar 3 x minus 6 vertical bar less than 9")[$| 3 x − 6 | < 9$] is equivalent to the compound inequality #math.equation(block: false, alt: "minus 9 less than 3 x minus 6 less than 9")[$− 9 < 3 x − 6 < 9$].


        
          + #math.equation(block: false, alt: "vertical bar 3 x minus 6 vertical bar equals open brace minus open parenthesis 3 x minus 6 close parenthesis, if x less than 2; 3 x minus 6, if x greater than or equal to 2")[$| 3 x − 6 | = \{ − ( 3 x − 6 ) & "if " x < 2 \ 3 x − 6 & "if " x ≥ 2$]
+ #math.equation(block: false, alt: "minus open parenthesis 3 x minus 6 close parenthesis less than or equal to 9")[$− ( 3 x − 6 ) ≤ 9$], #math.equation(block: false, alt: "3 x minus 6 less than 9")[$" " 3 x − 6 < 9$]
+ #math.equation(block: false, alt: "minus 1 less than x less than 5")[$− 1 < x < 5$]
+ The solutions are the same.


        
      
      
        + Write the piecewise definition for #math.equation(block: false, alt: "vertical bar 3 x minus 6 vertical bar")[$| 3 x − 6 |$].
+ Use your answer to part (a) to write two inequalities that together are equivalent to #math.equation(block: false, alt: "vertical bar 3 x minus 6 vertical bar greater than 9")[$| 3 x − 6 | > 9$].
+ Solve the inequalities in part (b) and check that the solutions agree with the solutions of #math.equation(block: false, alt: "vertical bar 3 x minus 6 vertical bar greater than 9")[$| 3 x − 6 | > 9$].
+ Show that #math.equation(block: false, alt: "vertical bar 3 x minus 6 vertical bar greater than 9")[$| 3 x − 6 | > 9$] is equivalent to the compound inequality #math.equation(block: false, alt: "3 x minus 6 less than minus 9")[$3 x − 6 < − 9 " "$] or #math.equation(block: false, alt: "3 x minus 6 greater than 9")[$" " 3 x − 6 > 9$].


      
      
        + Write the piecewise definition for #math.equation(block: false, alt: "vertical bar 2 x plus 5 vertical bar")[$| 2 x + 5 |$].
+ Use your answer to part (a) to write two inequalities that together are equivalent to #math.equation(block: false, alt: "vertical bar 2 x plus 5 vertical bar greater than 7")[$| 2 x + 5 | > 7$].
+ Solve the inequalities in part (b) and check that the solutions agree with the solutions of #math.equation(block: false, alt: "vertical bar 2 x plus 5 vertical bar greater than 7")[$| 2 x + 5 | > 7$].
+ Show that #math.equation(block: false, alt: "vertical bar 2 x plus 5 vertical bar greater than 7")[$| 2 x + 5 | > 7$] is equivalent to the compound inequality #math.equation(block: false, alt: "2 x plus 5 less than minus 7")[$2 x + 5 < − 7 " "$] or #math.equation(block: false, alt: "2 x plus 5 greater than 7")[$" " 2 x + 5 > 7$].


        
          + #math.equation(block: false, alt: "vertical bar 2 x plus 5 vertical bar equals open brace minus open parenthesis 2 x plus 5 close parenthesis, if x less than the fraction minus 5 over 2; 2 x plus 5, if x greater than or equal to the fraction minus 5 over 2")[$| 2 x + 5 | = \{ − ( 2 x + 5 ) & "if " x < display(frac(− 5, 2)) \ 2 x + 5 & "if " x ≥ display(frac(− 5, 2))$]
+ #math.equation(block: false, alt: "minus open parenthesis 2 x plus 5 close parenthesis greater than 7")[$− ( 2 x + 5 ) > 7$], #math.equation(block: false, alt: "2 x plus 5 greater than 7")[$" " 2 x + 5 > 7$]
+ #math.equation(block: false, alt: "x less than minus 6")[$x < − 6$] or #math.equation(block: false, alt: "x greater than 1")[$" " x > 1$]
+ The solutions are the same.


        
      
      
        + Write the piecewise definition for #math.equation(block: false, alt: "vertical bar 2 x plus 5 vertical bar")[$| 2 x + 5 |$].
+ Use your answer to part (a) to write two inequalities that together are equivalent to #math.equation(block: false, alt: "vertical bar 2 x plus 5 vertical bar less than 7")[$| 2 x + 5 | < 7$].
+ Solve the inequalities in part (b) and check that the solutions agree with the solutions of #math.equation(block: false, alt: "vertical bar 2 x plus 5 vertical bar less than 7")[$| 2 x + 5 | < 7$].
+ Show that #math.equation(block: false, alt: "vertical bar 2 x plus 5 vertical bar less than 7")[$| 2 x + 5 | < 7$] is equivalent to the compound inequality #math.equation(block: false, alt: "minus 7 less than 2 x plus 5 less than 7")[$− 7 < 2 x + 5 < 7$].


      
      For Problems 53–56, graph the function and answer the questions.


      
        #math.equation(block: true, alt: "f open parenthesis x close parenthesis equals vertical bar x plus 4 vertical bar plus vertical bar x minus 4 vertical bar")[$f ( x ) = | x + 4 | + | x − 4 |$]


        + Using your graph, write a piecewise formula for #math.equation(block: false, alt: "f open parenthesis x close parenthesis")[$f ( x )$].
+ Experiment by graphing #math.equation(block: false, alt: "g open parenthesis x close parenthesis equals vertical bar x plus p vertical bar plus vertical bar x minus q vertical bar")[$g ( x ) = | x + p | + | x − q |$] for different positive values of #math.equation(block: false, alt: "p")[$p$] and #math.equation(block: false, alt: "q")[$q$]. Make a conjecture about how the graph depends on #math.equation(block: false, alt: "p")[$p$] and #math.equation(block: false, alt: "q")[$q$].
+ Write a piecewise formula for #math.equation(block: false, alt: "g open parenthesis x close parenthesis equals vertical bar x plus p vertical bar plus vertical bar x minus q vertical bar")[$g ( x ) = | x + p | + | x − q |$].


        
          #figure(figph[piecewise linear], alt: "piecewise linear", caption: none)


          + #math.equation(block: false, alt: "f open parenthesis x close parenthesis equals open brace minus 2 x ,, x less than minus 4; 8 ,, minus 4 less than or equal to x less than or equal to 4; 2 x ,, x greater than 4")[$f ( x ) = \{ − 2 x , & x < − 4 \ 8 , & − 4 ≤ x ≤ 4 \ 2 x , & x > 4$]
+ The graphs looks like like a trough. The middle horizontal section is #math.equation(block: false, alt: "y equals p plus q")[$y = p + q$] for #math.equation(block: false, alt: "minus p less than or equal to x less than or equal to q")[$− p ≤ x ≤ q$], the left side, #math.equation(block: false, alt: "x less than minus p")[$x < − p$], has slope #math.equation(block: false, alt: "minus 2")[$− 2$] and the right side, #math.equation(block: false, alt: "x greater than q")[$x > q$], has slope #math.equation(block: false, alt: "2")[$2$].
+ #math.equation(block: false, alt: "g open parenthesis x close parenthesis equals open brace minus 2 x plus q minus p ,, x less than minus p; p plus q ,, minus p less than or equal to x less than or equal to q; 2 x plus p minus q ,, x greater than q")[$g ( x ) = \{ − 2 x + q − p , & x < − p \ p + q , & − p ≤ x ≤ q \ 2 x + p − q , & x > q$]


        
      
      
        #math.equation(block: true, alt: "f open parenthesis x close parenthesis equals vertical bar x plus 4 vertical bar minus vertical bar x minus 4 vertical bar")[$f ( x ) = | x + 4 | − | x − 4 |$]


        + Using your graph, write a piecewise formula for #math.equation(block: false, alt: "f open parenthesis x close parenthesis")[$f ( x )$].
+ Experiment by graphing #math.equation(block: false, alt: "g open parenthesis x close parenthesis equals vertical bar x plus p vertical bar minus vertical bar x minus q vertical bar")[$g ( x ) = | x + p | − | x − q |$] for different positive values of #math.equation(block: false, alt: "p")[$p$] and #math.equation(block: false, alt: "q")[$q$]. Make a conjecture about how the graph depends on #math.equation(block: false, alt: "p")[$p$] and #math.equation(block: false, alt: "q")[$q$].


      
      
        #math.equation(block: true, alt: "f open parenthesis x close parenthesis equals vertical bar x plus 4 vertical bar plus vertical bar x vertical bar plus vertical bar x minus 4 vertical bar")[$f ( x ) = | x + 4 | + | x | + | x − 4 |$]


        + Using your graph, write a piecewise formula for #math.equation(block: false, alt: "f open parenthesis x close parenthesis")[$f ( x )$].
+ What is the minimum value of #math.equation(block: false, alt: "f open parenthesis x close parenthesis")[$f ( x )$]?
+ If #math.equation(block: false, alt: "p , q greater than or equal to 0")[$p , q ≥ 0$], what is the minimum value of  #math.equation(block: false, alt: "g open parenthesis x close parenthesis equals vertical bar x plus p vertical bar plus vertical bar x vertical bar plus vertical bar x minus q vertical bar")[$g ( x ) = | x + p | + | x | + | x − q |$]?


        
          #figure(figph[piecewise linear], alt: "piecewise linear", caption: none)


          + #math.equation(block: false, alt: "f open parenthesis x close parenthesis equals open brace minus 3 x ,, x less than minus 4; minus x plus 8 ,, minus 4 less than or equal to x less than or equal to 0; x plus 8 ,, 0 less than x less than 4; 3 x ,, x greater than or equal to 4")[$f ( x ) = \{ − 3 x , & x < − 4 \ − x + 8 , & − 4 ≤ x ≤ 0 \ x + 8 , & 0 < x < 4 \ 3 x , & x ≥ 4$]
+ #math.equation(block: false, alt: "8")[$8$]
+ #math.equation(block: false, alt: "p plus q")[$p + q$]


        
      
      
        #math.equation(block: true, alt: "f open parenthesis x close parenthesis equals vertical bar x plus 4 vertical bar minus vertical bar x vertical bar plus vertical bar x minus 4 vertical bar")[$f ( x ) = | x + 4 | − | x | + | x − 4 |$]


        + Using your graph, write a piecewise formula for #math.equation(block: false, alt: "f open parenthesis x close parenthesis")[$f ( x )$].
+ What is the minimum value of #math.equation(block: false, alt: "f open parenthesis x close parenthesis")[$f ( x )$]?
+ If #math.equation(block: false, alt: "p , q greater than or equal to 0")[$p , q ≥ 0$], what is the minimum value of  #math.equation(block: false, alt: "g open parenthesis x close parenthesis equals vertical bar x plus p vertical bar minus vertical bar x vertical bar plus vertical bar x minus q vertical bar")[$g ( x ) = | x + p | − | x | + | x − q |$]?


      
      Problems 57–60 use the absolute value function to model distance. Use the strategy outlined in Problems 57 and 58 to solve Problems 59 and 60.


      
        A small pottery is setting up a workshop to produce mugs. Three machines are located on a long table, as shown in the figure. The potter must use each machine once in the course of producing a mug. Let #math.equation(block: false, alt: "x")[$x$] represent the coordinate of the potter's station.


        #figure(figph[three machines on number line], alt: "three machines on number line", caption: none)


        + Write expressions for the distance from the potter's station to each of the machines.
+ Write a function that gives the sum of the distances from the potter's station to the three machines.
+ Graph your function for #math.equation(block: false, alt: "minus 20 less than or equal to x less than or equal to 30")[$− 20 ≤ x ≤ 30$]. Where should the potter stand in order to minimize the distance she must walk to the machines?


        
          + #math.equation(block: false, alt: "vertical bar x plus 12 vertical bar")[$| x + 12 |$], #math.equation(block: false, alt: "vertical bar x plus 4 vertical bar")[$| x + 4 |$], #math.equation(block: false, alt: "vertical bar x minus 24 vertical bar")[$| x − 24 |$]
+ #math.equation(block: false, alt: "f open parenthesis x close parenthesis equals vertical bar x plus 12 vertical bar plus vertical bar x plus 4 vertical bar plus vertical bar x minus 24 vertical bar")[$f ( x ) = | x + 12 | + | x + 4 | + | x − 24 |$]
+ #figure(figph[piecewise linear], alt: "piecewise linear", caption: none)

 #linebreak() At #math.equation(block: false, alt: "x")[$x$]-coordinate #math.equation(block: false, alt: "minus 4")[$− 4$]


        
      
      
        Suppose the pottery in Problem 57 adds a fourth machine to the procedure for producing a mug, located at #math.equation(block: false, alt: "x equals 16")[$x = 16$] in the figure.


        + Write and graph a new function for the sum of the potter's distances to the four machines.
+ Where should the potter stand now to minimize the distance she has to walk while producing a mug?


      
      
        Richard and Marian are moving to Parkville to take jobs after they graduate. The main road through Parkville runs east and west, crossing a river in the center of town. Richard's job is located #math.equation(block: false, alt: "10")[$10$] miles east of the river on the main road, and Marian's job is #math.equation(block: false, alt: "6")[$6$] miles west of the river. There is a health club they both like located #math.equation(block: false, alt: "2")[$2$] miles east of the river. If they plan to visit the health club every workday, where should Richard and Marian look for an apartment to minimize their total daily driving distance?


        
          #math.equation(block: false, alt: "2")[$2$] miles east of the river


        
      
      
        Romina's Bakery has just signed contracts to provide baked goods for three new restaurants located on Route 28 outside of town. The Coffee Stop is #math.equation(block: false, alt: "2")[$2$] miles north of town center, Sneaky Pete's is #math.equation(block: false, alt: "8")[$8$] miles north, and the Sea Shell is #math.equation(block: false, alt: "12")[$12$] miles south. Romina wants to open a branch bakery on Route 28 to handle the new business, with separate days for each restaurant's weekly delivery. Where should she locate the bakery in order to minimize the distance she must drive for deliveries?