copyright © 2007 pearson education, inc. slide 2-1


Chapter 2: Analysis of Graphs of Functions

2.1 Graphs of Basic Functions and Relations; Symmetry

2.2 Vertical and Horizontal Shifts of Graphs

2.3 Stretching, Shrinking, and Reflecting Graphs

2.4 Absolute Value Functions: Graphs, Equations, Inequalities, and Applications

2.5 Piecewise-Defined Functions

2.6 Operations and Composition


2.4 Absolute Value Functions: Graphs, Equations, Inequalities, and Applications

• Recall:

• Use this concept to define the absolute value of a function f :

• Technology Note: The command abs(x) is used by some graphing calculators to find absolute value.

0 if 0 if

)(xxxx

xxf

0)( if )( 0)( if )(

)(xfxfxfxf

xf

.

.


2.4 The Graph of y = | f (x)|

• To graph the function the graph is the same as for values of that are nonnegative and reflected across the x-axis for those that are negative.

• The domain of is the same as the domain of f, while the range of will be a subset of

• Example

Give the domain and range of

Solution

,)(xfy )(xfy )(xf

)(xfy )(xfy ).,0[

.3)4( and 3)4( 22 xyxy

).,0[ is 3)4( of range thewhile),,3[ is 3)4(

of range The ).,( isfunction each ofdomain The22

xyxy


Use the graph of to sketch the graph of Give the domain and range of each function.

Solution

2.4 Sketch the Graph of y = | f (x)| Given y = f (x)

)(xfy .)(xfy

).,0[ is )( of range the while),,3[ is )(

of range The ).,( is functionsboth ofdomain The

xfyxfy


2.4 Properties of Absolute Value

For all real numbers a and b:

1.

2.

3.

4.

Example Consider the following sequence of transformations.

baab

)0(||

|| b

b

a

b

a

aa )inequality triangle(the baba

. ofon translatia as Rewrite 112 xyxy

2.out Factor 2

112

xy

11 112 Property 1 2

2 2y x y x


2.4 Comprehensive Graph

• We are often interested in absolute value functions of the form

where the expression inside the absolute value bars is linear. We will solve equations and inequalities involving such functions.

• The comprehensive graph of will include all intercepts and the lowest point on the “V-shaped” graph.

, )( baxxf

)( baxxf


2.4 Equations and Inequalities Involving Absolute Value

Example Solve

Analytic Solution

For to equal 7, 2x + 1 must be 7 units from 0 on the number line. This can only happen when

Graphing Calculator Solution

.712 x

12 x.712or712 x x

712or 712 xx82or 62 xx

4 or 3 xx}.3,4{ isset solution The

.3or 4when

intersect 7 and

12 graphs The

2

1

xx

y

xy


Let k be a positive number.

1. To solve solve the compound equation

2. To solve solve the compound inequality

3. To solve solve the three-part inequality

Inequalities involving are solved similarly, using the equalitypart of the symbol as well.

2.4 Solving Absolute Value Equations and Inequalities

,kbax

,kbax

,kbax

or

.or kbaxkbax

.or kbaxkbax

.kbaxk


2.4 Solving Absolute Value Inequalities Analytically

Solve the inequalities .712 (b) and ,712 (a) xx

7127

line.number on the 0 from units 7 than less

ist number tha arepresent must 12 expression The (a)

x

x

part.each from 1Subtract 628 x2.by part each Divide 34 x

).3,4( interval theisset solution The

(b) The expression 2 1 must represent a number that is

than 7 units from 0 on either side of the number line.

2 1 7 or 2 1 7

x

more

x x

82 or 62 xx4 or 3 xx

).,3()4,( interval theisset solution The


2.4 Solving Absolute Value Inequalities Graphically

Solve the previous equations graphically by letting and

and find all points of intersection.

(a) The graph of lies below the graph of for

x-values between –4 and 3, supporting the solution set (–4,3).

• The graph of lies above the graph of for

x-values greater than 3 or less than –4, confirming the analytic result.

121

xy7

2y

121

xy 72y

121

xy 72y


2.4 Solving Special Cases of Absolute Value Equations and Inequalities

Solve Analytically

(a) Because the absolute value of an expression is never negative, the equation has no solution. The solution set is Ø.

(b) Using similar reasoning as in part (a), the absolute value of an expression will never be less than –5. The solution set is Ø.

(c) Because absolute value will always be greater than or equal to 0, the absolute value of an expression will always be greater than –5. The solution set is

Graphical Solution

The graphical solution is seen from the

graphing of

553 (c) 553 (b) 553 (a) xxx

).,(

.5 and 5321

yxy


2.4 Solving |ax + b| = |cx + d| Analytically

Example

Solve

To solve the equation analytically, solve the compound equation

dcxbax

.dcxb axd or cxbax )(

6 2 3 analytically.x x

).32(6or 326

if satisfied isequation The

xxxx326or 9 xxx

33 x

1x

The solution set is { 1,9}.


2.4 Solving |ax + b| = |cx + d| Graphically

Solve

Let The equation is equivalent to

so graph and find the x-intercepts. From the graph below, we see that they are –1 and 9, supporting the analytic solution.

6 2 3 graphically.x x

. 32 and 621

xyxy21

yy ,0

21 yy 326

3 xxy


2.4 Solving Inequalities Involving Two Absolute Value Expressions

Solve each inequality graphically.

Solution

(a) The inequality In the previous example, note that the graph of is below the x-axis in the interval

(b) The inequality is satisfied by the closed interval

326 (b) 326 (a) xxxx

.0or ,0 toequivalent is 32121 yyyyy

3y

).,9()1,(

03y

].9,1[


2.4 Solving an Equation Involving a Sum of Absolute Values

Solve graphically by the intersection-of-graphs method.SolutionLet

The points of intersection of the graphs have x-coordinates –9 and 7.To verify these solutions, we substitute them into the equation.

Therefore, the solution set is

1635 xx

.16 and 3521 yxxy

}.7,9{

164124123757 :7Let

161241243)9(59)( :9Let

x

x


2.4 An Application Involving Absolute Value

The inequality describes the range of average monthly temperatures x in degrees Fahrenheit for Spokane, Washington. Solve this inequality, and interpret the result.

48 21x

21 48 21

21 48 48 48 21 48

27 69

x

x

x

This means that the average monthly temperature ranges from through The average monthly temperatures are always within 21º of 48ºF. See the graphical representation below.

27 F

F.69

copyright © 2007 pearson education, inc. slide 2-1

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