Transcript sec 2.4

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
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Slide 2-2
2.4 Absolute Value Functions: Graphs,
Equations, Inequalities, and Applications
• Recall:
 x if x  0
f ( x)  x  
 x if x  0 .
• Use this concept to define the absolute value of a function f:
f ( x)



f ( x) if f ( x)  0
f ( x) if f ( x)  0 .
• Technology Note: The command abs(x) is used by some
graphing calculators to find absolute value.
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Slide 2-3
2.4 The Graph of y = | f (x)|
• To graph the function y  f (x) , the graph is the same as y  f (x)
for values of f (x) that are nonnegative and reflected across the x-axis
for those that are negative.
• The domain of y  f (x) is the same as the domain of f, while the
range of y  f (x) will be a subset of [0, ).
• Example
Give the domain and range of y  ( x  4)  3 and y  ( x  4)  3 .
2
2
Solution The domain of each function is (  ,  ). The range of
2
y  ( x  4 )  3 is [ 3,  ), while the range of y  ( x  4 ) 2  3 is [ 0,  ).
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2.4 Sketch the Graph of y = | f (x)| Given
y = f (x)
Use the graph of y  f (x) to sketch the graph of y  f (x) .
Give the domain and range of each function.
Solution
The domain of both functions is (, ). The range of
y  f ( x) is [3, ), while the range of y  f ( x) is [0, ).
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2.4 Properties of Absolute Value
For all real numbers a and b:
1. ab  a  b
a
2.
3.
4.
b

|a|
(b  0)
|b|
a  a
a  b  ab
(the triangle inequality )
Example Consider the following sequence of transformations.
y  2 x  11
Rewrite as a translati on of y  x .
 11
y  2 x  
2

Factor out 2.
11
y  2  x
2
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Property 1

11
y 2 x
2
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2.4 Comprehensive Graph
• We are often interested in absolute value functions
of the form
f ( x)  ax  b ,
where the expression inside the absolute value
bars is linear. We will solve equations and
inequalities involving such functions.
• The comprehensive graph of f ( x)  ax  b will
include all intercepts and the lowest point on the
“V-shaped” graph.
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Slide 2-7
2.4 Equations and Inequalities Involving
Absolute Value
Example Solve 2 x  1  7.
Analytic Solution
For 2 x  1 to equal 7, 2x + 1 must be 7 units from 0 on the number
line. This can only happen when 2 x  1  7 or 2 x  1  7.
2x  1  7
2x  6
x3
The solution set is {4,3}.
or
or
or
2 x  1  7
2 x  8
x  4
Graphing Calculator Solution
The graphs y  2 x  1
1
and y  7 intersect
2
when x  4 or x  3.
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Slide 2-8
2.4 Solving Absolute Value Equations and
Inequalities
Let k be a positive number.
1.
To solve ax  b  k , solve the compound equation
ax  b  k
2.
ax  b  k .
To solve ax  b  k , solve the compound inequality
ax  b  k
3.
or
or
ax  b  k .
To solve ax  b  k , solve the three-part inequality
 k  ax  b  k .
Inequalities involving  or  are solved similarly, using the equality
part of the symbol as well.
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Slide 2-9
2.4 Solving Absolute Value Inequalities
Analytically
Solve the inequalities (a) 2 x  1  7, and (b) 2 x  1  7.
(a) The expression 2 x  1 must represent a number that is
less than 7 units from 0 on the number line.
 7  2x  1  7
 8  2x  6
Subtract1 from each part.
4 x 3
Divide each part by 2.
The solution set is the interval (4,3).
(b) The expression 2x  1 must represent a number that is
more than 7 units from 0 on either side of the number line.
2x  1  7
or
2 x  1  7
2x  6
or
2 x  8
x3
or
x  4
The solution set is the interval (,4)  (3, ).
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2.4 Solving Absolute Value Inequalities
Graphically
Solve the previous equations graphically by letting y  2 x  1 and
y  7 and find all points of intersection.
1
2
(a)
•
The graph of y  2 x  1 lies below the graph of y  7 for
x-values between –4 and 3, supporting the solution set (–4,3).
The graph of y  2 x  1 lies above the graph of y  7 for
x-values greater than 3 or less than –4, confirming the analytic
result.
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1
2
1
2
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2.4 Solving Special Cases of Absolute
Value Equations and Inequalities
Solve Analytically
(a) 3 x  5  5
(a)
(b)
(c)
(b) 3 x  5  5
(c) 3 x  5  5
Because the absolute value of an expression is never negative, the
equation has no solution. The solution set is Ø.
Using similar reasoning as in part (a), the absolute value of an
expression will never be less than –5. The solution set is Ø.
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 y1  3 x  5 and y 2  5.
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Slide 2-12
2.4 Solving |ax + b| = |cx + d| Analytically
To solve the equation ax  b  cx  d analytically, solve the
compound equation
ax  b  cx  d or ax  b  (cx  d ).
Example
Solve x  6
 2 x  3 analytically.
The equation is satisfied if
x  6  2 x  3 or
9x
or
x  6  (2 x  3).
x  6  2 x  3
3x  3
x  1
The solution set is {  1,9}.
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Slide 2-13
2.4 Solving |ax + b| = |cx + d| Graphically
Solve x  6  2 x  3 graphically.
Let y  x  6 and y  2 x  3 . The equation y  y is equivalent to
y  y  0, so graph y  x  6  2 x  3 and find the x-intercepts.
From the graph below, we see that they are –1 and 9, supporting the
analytic solution.
1
1
1
2
2
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3
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2.4 Solving Inequalities Involving Two
Absolute Value Expressions
Solve each inequality graphically.
(a) x  6  2 x  3
(b) x  6  2 x  3
Solution
(a) The inequality y  y is equivalent to y  y  0, or y  0.
In the previous example, note that the graph of y is
below the x-axis in the interval (,1)  (9, ).
1
2
1
2
3
3
(b) The inequality y  0 is satisfied by the closed interval
[1,9].
3
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Slide 2-15
2.4 Solving an Equation Involving a Sum
of Absolute Values
Solve x  5  x  3  16 graphically by the intersection-of-graphs method.
Solution
Let y  x  5  x  3 and y  16.
1
2
The points of intersection of the graphs have x-coordinates –9 and 7.
To verify these solutions, we substitute them into the equation.
Let x  9 :
(9)  5  (9)  3   4   12  4  12  16
Let x  7 :
7  5  7  3  12  4  12  4  16
Therefore, the solution set is {9,7}.
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Slide 2-16
2.4 An Application Involving Absolute
Value
The inequality x  48  21 describes the range of average monthly
temperatures x in degrees Fahrenheit for Spokane, Washington. Solve
this inequality, and interpret the result.
21  x  48  21
21  48  x  48  48  21  48
27  x  69
This means that the average monthly temperature ranges from 27 F
through 69 F. The average monthly temperatures are always
within 21º of 48ºF. See the graphical representation below.

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