copyright © 2007 pearson education, inc. slide 2-1
TRANSCRIPT
Copyright © 2007 Pearson Education, Inc. Slide 2-1
Copyright © 2007 Pearson Education, Inc. Slide 2-2
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
Copyright © 2007 Pearson Education, Inc. Slide 2-3
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
.
.
Copyright © 2007 Pearson Education, Inc. Slide 2-4
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
Copyright © 2007 Pearson Education, Inc. Slide 2-5
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
Copyright © 2007 Pearson Education, Inc. Slide 2-6
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
Copyright © 2007 Pearson Education, Inc. Slide 2-7
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
Copyright © 2007 Pearson Education, Inc. Slide 2-8
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
Copyright © 2007 Pearson Education, Inc. Slide 2-9
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
Copyright © 2007 Pearson Education, Inc. Slide 2-10
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
Copyright © 2007 Pearson Education, Inc. Slide 2-11
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
Copyright © 2007 Pearson Education, Inc. Slide 2-12
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
Copyright © 2007 Pearson Education, Inc. Slide 2-13
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}.
Copyright © 2007 Pearson Education, Inc. Slide 2-14
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
Copyright © 2007 Pearson Education, Inc. Slide 2-15
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[
Copyright © 2007 Pearson Education, Inc. Slide 2-16
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
Copyright © 2007 Pearson Education, Inc. Slide 2-17
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