Transcript Document

Chapter 2
Linear
Functions and
Equations
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2.5
♦
♦
♦
Absolute Value
Equations and
Inequalities
Evaluate and graph the absolute value function
Solve absolute value equations
Solve absolute value inequalities
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Absolute Value Function
The graph of y = |x|.
V-shaped
Cannot be represented by
single linear function
x
x 
x
if x  0
if x  0
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Absolute Value Function
Alternate Formula
x  x
2
for all real numbers x
That is, regardless of whether a real number x is
positive or negative, the expression x 2 equals the
absolute value of x.
Examples:
y  y
2
x  1
2
 x 1
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2x 
2
 2x
4
Example: Analyzing the graph of
y = |ax + b|
For the linear function f, graph y = f (x) and
y = |f (x)| separately. Discuss how the absolute
value affects the graph of f.
f(x) = –2x + 4
(For continuity of the solution, it appears
completely on the next slide.)
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Example: Analyzing the graph of
y = |ax + b|
The graph of y = |–2x + 4| is a reflection of f
across the x-axis when y = –2x + 4 is below the
x-axis.
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Absolute Value Equations
Solutions to |x| = k with k > 0 are given by
x = ±k.
Solutions to |ax + b| = k are given by
ax + b = ±k.
These concepts can be illustrated visually.
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Absolute Value Equations
Two solutions |ax + b| = k, for k > 0
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Absolute Value Equations
One solution |ax + b| = k, for k = 0
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Absolute Value Equations
No solution |ax + b| = k, for k < 0
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Absolute Value Equations
Let k be a positive number. Then
|ax + b| = k is equivalent to ax + b = ±k.
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Example: Solving an equation with
technology
Solve the equation |2x + 5| = 2 graphically,
numerically, and symbolically.
Solution
Graph Y1 = abs(2X + 5) and Y2 = 2
Solutions: –3.5, –1.5
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Example: Solving an equation with
technology
Table Y1 = abs(2x + 5) and Y2 = 2
Solutions to y1 = y2 are –3.5 and –1.5.
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Example: Solving an equation with
technology
2x  5  2
Symbolic:
2x  5  2
2x  5  2
2x  3
3
x
2
2x  5  2
2x  7
7
x
2
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Absolute Value Inequalities
Solutions |ax + b| = k labeled s1 and s2 and the
graph of y = |ax + b| is below y = k between s1 and
s2 or when s1 < x < s2. Solution to |ax + b| < k is in
green.
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Absolute Value Inequalities
Solutions |ax + b| = k labeled s1 and s2 and the
graph of y = |ax + b| is above y = k to left of s1 and
right of s2 or x < s1 or x >s2. Solution to |ax + b| > k
is in green.
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Absolute Value Inequalities
Let solutions to |ax + b| = k be s1 and s2, where s1
< s2 and k > 0.
1.
|ax + b| < k is equivalent to s1 < x < s2.
2.
|ax + b| > k is equivalent to x < s1 or
x > s 2.
Similar statements can be made for inequalities
involving ≤ or ≥.
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Example: Solving inequalities involving
absolute values symbolically
Solve the inequality |2x – 5| ≤ 6. Write the solution
set in interval notation.
Solution
Solve |2x – 5| = 6 or 2x – 5 = ±6
2x  5  6 or 2x  5  6
2x  11
x  1
1
11
x
x
2
2
 1 11
1
11
Solution set:  2  x  2 , or   2 , 2 


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Absolute Value Inequalities
Alternative Method
Let k be a positive number.
1. |ax + b| < k is equivalent to –k < ax + b < k.
2. |ax + b| > k is equivalent to ax + b < –k or
ax + b > –k.
Similar statements can be made for inequalities
involving ≤ or ≥.
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Example: Using an alternative
method
Solve the inequality |4 – 5x | ≤ 3. Write your
answer in interval notation.
Solution
|4 – 5x| ≤ 3 is equivalent to the three-part
inequality
3  4  5x  3
7  5x  1
7
1
x
5
5
1 7
In interval notation, solution is  , .
5 5 
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