Absolute Value Inequalities §2.5, Functions §3.6
Fall 2013 - Math 1010
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Roadmap
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§2.5 - Review of absolute value inequalities.
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§3.6 - Functions: Relations, Functions
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§3.6 - Evaluating Functions, Finding Domains and Ranges
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§2.5 - Absolute Value Inequality
Write down and solve the absolute value inequality.
|2x − 3| ≤ 6
Be prepared to demonstrate your method and answer on the chalkboard.
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§2.5 - Absolute Value Inequality
Write down and solve the absolute value inequality.
1
|y − 15| > 5
3
Be prepared to demonstrate your method and answer on the chalkboard.
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§2.5 - Absolute Value Inequality
Write down and solve the absolute value inequality.
−5|x| ≥ 1000
Be prepared to demonstrate your method and answer on the chalkboard.
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§2.5 - Absolute Value Inequality
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Identify if the inequality makes sense. (ie −|x| > 2 will have no
solution, −|x| < 2 will have solutions.)
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Identify when to use ”and” (ie |x| < 2) and when to use ”or” (ie
|x| > 2).
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Split the absolute value inequality into two inequalties without
absolute values.
|x + 1| > 4
becomes
x +1>4
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and x + 1 < −4
Tip: Avoid negating the algebraic expression from inside the absolute
value.
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§3.6 - Functions
Our study of functions will come in parts: defining a function, seeing why
functions stand out from all relations, using the notation of functions, and
finding associated domains and ranges of functions.
On Monday we described what parts we’ve taken from graphing. Here is a
sample of popular choices: points / order pairs / (x, y ); intercepts; shapes
- lines, curves, V -shapes; trends; axes; distance; slope between two points
Many of the above ideas are useful when learning about functions.
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§3.6 - Points / Ordered Pairs
Thinking about points is a good place to start. Here is a relation table:
x
0
1
2
3
0
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y
1
3
5
5
3
point
(0,1)
(1,3)
(2,5)
(3.5)
(0.3)
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§3.6 - Points / Ordered Pairs
Thinking about points is a good place to start. Here is a relation table:
x
0
1
2
3
0
y
1
3
5
5
3
point
(0,1)
(1,3)
(2,5)
(3.5)
(0.3)
The set of x values is the domain. The set of y values is the range.
Domain = {0, 1, 2, 3}, Range = {1, 3, 5}
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§3.6 - Points / Ordered Pairs
Thinking about points is a good place to start. Here is a relation table:
x
0
1
2
3
0
y
1
3
5
5
3
point
(0,1)
(1,3)
(2,5)
(3.5)
(0.3)
The set of x values is the domain. The set of y values is the range.
Domain = {0, 1, 2, 3}, Range = {1, 3, 5}
This relation cannot be a function.
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§3.6 - Functions
A function is a rule that pairs each element of a domain to exactly one
element of the range.
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§3.6 - Functions
A function is a rule that pairs each element of a domain to exactly one
element of the range.
Note:
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No two ordered pairs have the same first component and different
second components. That is, (2, 4) and (2, 18) cannot be ordered
pairs of a function.
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§3.6 - Functions
A function is a rule that pairs each element of a domain to exactly one
element of the range.
Note:
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No two ordered pairs have the same first component and different
second components. That is, (2, 4) and (2, 18) cannot be ordered
pairs of a function.
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The rule is a process that turns one number into another number. For
instance, ”the squares of all real numbers” is the rule that turns 1
into 1, 1.1 into 1.21, 1.2 into 1.44, and so on.
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§3.6 - Functions
A function is a rule that pairs each element of a domain to exactly one
element of the range.
Note:
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No two ordered pairs have the same first component and different
second components. That is, (2, 4) and (2, 18) cannot be ordered
pairs of a function.
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The rule is a process that turns one number into another number. For
instance, ”the squares of all real numbers” is the rule that turns 1
into 1, 1.1 into 1.21, 1.2 into 1.44, and so on.
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It is okay to have two or more ordered pairs with different first
components and the same second component. That is, (2, 6) and
(40, 6) and (−19, 6) may be ordered pairs of a function.
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§3.6 - Functions
A function is a rule that pairs each element of a domain to exactly one
element of the range.
Algebraically this is represented with an equation. For the examples below
take x as the independent variable and y as the dependent variable.
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§3.6 - Functions
A function is a rule that pairs each element of a domain to exactly one
element of the range.
Algebraically this is represented with an equation. For the examples below
take x as the independent variable and y as the dependent variable.
1) Is y 2 − x = 0 a function?
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§3.6 - Functions
A function is a rule that pairs each element of a domain to exactly one
element of the range.
Algebraically this is represented with an equation. For the examples below
take x as the independent variable and y as the dependent variable.
1) Is y 2 − x = 0 a function?
2) Is x 2 − y = 0 a function?
(Math 1010)
M 1010 §3.6
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§3.6 - Functions
A function is a rule that pairs each element of a domain to exactly one
element of the range.
Algebraically this is represented with an equation. For the examples below
take x as the independent variable and y as the dependent variable.
1) Is y 2 − x = 0 a function?
2) Is x 2 − y = 0 a function?
3) Is −4x + 8y = 5 a function?
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§3.6 - Functions
A function is a rule that pairs each element of a domain to exactly one
element of the range.
Algebraically this is represented with an equation. For the examples below
take x as the independent variable and y as the dependent variable.
1) Is y 2 − x = 0 a function?
2) Is x 2 − y = 0 a function?
3) Is −4x + 8y = 5 a function?
1) No, 2) Yes, 3) Yes
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§3.6 - Function Notation
When writing the rule of a function in equation form, it is convenient to
keep track of input and output pairs and to give the function a name.
This is all done with function notation.
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§3.6 - Function Notation
When writing the rule of a function in equation form, it is convenient to
keep track of input and output pairs and to give the function a name.
This is all done with function notation.
Example: y = (x + 3)3 becomes f (x) = (x + 3)3
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§3.6 - Function Notation
When writing the rule of a function in equation form, it is convenient to
keep track of input and output pairs and to give the function a name.
This is all done with function notation.
Example: y = (x + 3)3 becomes f (x) = (x + 3)3
• The name of the function is f .
• f (x) is read, ”f-of-x,” or, ”the value of f at x.”
Evaluating a function is done by subsituting an x value into the function.
Example: Evaluate the value of f (x) = (x + 3)3 at x = −5.
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§3.6 - Function Notation
When writing the rule of a function in equation form, it is convenient to
keep track of input and output pairs and to give the function a name.
This is all done with function notation.
Example: y = (x + 3)3 becomes f (x) = (x + 3)3
• The name of the function is f .
• f (x) is read, ”f-of-x,” or, ”the value of f at x.”
Evaluating a function is done by subsituting an x value into the function.
Example: Evaluate the value of f (x) = (x + 3)3 at x = −5.
f (−5) = ((−5) + 3)3 = (−2)2 = −8.
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§3.6 Functions
Are some values of x not allowed for
f (x) =
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1
?
x − 10
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§3.6 Functions
Are some values of x not allowed for
f (x) =
1
?
x − 10
Are some values of x not allowed for
f (x) =
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√
x?
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§3.6 - Implied Domains and Ranges
Implied domains are the set of real values for x that yield real values for
f (x). The rule for the function implies the domain. Exclude:
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sets of values for x that result in division by zero.
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sets of values for x that result in taking the square-roots of negative
numbers.
Otherwise, explicit domains very directly state which x values to exclude
when the rule of the function is given.
Example: f (x) = x − 7, x 6= 7. The domain is {x|x 6= 7} or ’all x except
7.’
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§3.6 Implied Domain and Range
Find the implied domain of
f (x) =
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√
5x − 2
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§3.6 Implied Domain and Range
Find the implied domain of
f (x) =
√
5x − 2
The inequality is 5x − 2 ≥ 0
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§3.6 Implied Domain and Range
Find the implied domain of
f (x) =
√
5x − 2
The inequality is 5x − 2 ≥ 0
Solved: x ≥ 25 . All real numbers x such that x ≥
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5
is the domain.
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§3.6 Explicit Domain and Range
The domain of f given below is all real numbers. What is the rule?
(
−x, if x ≤ 0
f (x) =
x,
if x > 0
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§3.6 Explicit Domain and Range
The domain of f given below is all real numbers. What is the rule?
(
x + 8, if x < 2
f (x) =
6 − 3x, if x ≥ 2
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Assignment
Assignment:
For Monday:
1. Exercises from §3.6 due Monday, September 30.
2. Exam # 1: Chapter 3 & Cumulative Chapters 1 - 2, October 2
3. Read section 3.7.
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