Chapter 2

Functions and Graphs

Section 1

Functions

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Learning Objectives for Section 2.1Functions

The student will be able to do point-by-point plotting of equations in two variables.

The student will be able to give and apply the definition of a function.

The student will be able to identify domain and range of a function.

The student will be able to use function notation. The student will be able to solve applications.

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Graphing Equations

If you are not familiar with a graph’s “family”, then use point-by-point plotting. (i.e. make an x-y table)• However, this is a very tedious process.

Knowing a graph’s family, will help you determine its basic shape.

Knowing a graph’s basic shape and the transformations on its parent, will help you graph it without making an x-y table.• This will be reviewed in tomorrow’s lesson.

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Families and Shapes

Line

Parabola “V-shaped”

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Families and Shapes

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Functions

A relation (set of ordered pairs) represents a function if for each x, there is only one y.

The set of all x’s is called the domain, and the set of all corresponding y’s is called the range.

Which of these relations is a function?• {(1, 3), (4, 9), (7, 15), (10, 21)}• {(2, 4), (-2, 4), (3, 9), (-3, 0)}• {(16, 4), (16, -4), (9, 3), (9, -3)}• Answer: The first two are functions.

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Vertical Line Test for a Function

If you have the graph of an equation, you can easily determine if it is the graph of a function by doing the vertical line test.

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Vertical Line Test for a Function(continued)

This graph fails the vertical line test, so it’s not a function.

This graph passes the vertical line test, so it is a function.

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Function Notation

The following notation is used to describe functions. The variable y will now be called f (x).

This is read as “ f of x” and simply means the y coordinate of the function corresponding to a given x value.

can now be expressed as

2 2y x

2( ) 2f x x

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Function Evaluation

Consider our function Evaluate:

• f (–3)

(-3)2 – 2 = 7• f(a)

a2 - 2• f(2x)

(2x)2 – 2 = 4x2 – 2• f(x + h)

(x + h)2 – 2 = x2 + 2xh + h2 – 2

2( ) 2f x x

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More Examples

f (6 h) 3(6 h) 2 18 3h 2

16 3h

f (2) 3(2) 2 4 2

f (x) 3x 2

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Domain of a Function

The domain of a function refers to all the possible values of x that produce a valid y.

The domain can be determined from the equation of the function or from its graph.

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Finding the Domain of a Function

¿𝐷𝑜𝑚𝑎𝑖𝑛 :

𝑦=(𝑥− 4)3 𝑦=−(𝑥+1)2 −3 𝑦=|𝑥−2|+5

If a function does not contain a square root or a denominator then its domain is all reals (-, )

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Determining Domain

If a function contains a square root or a denominator containing x, its domain will be restricted.

The next few examples show how to determine the restricted domain.

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Finding the Domain of a Function

( ) 3 2f x x 3 𝑥− 2≥ 0

𝑥≥23

Functions with square roots:Set the expression inside the

square root 0 and solve for x to determine the domain.

[ 23

, ∞ )𝐷𝑜𝑚𝑎𝑖𝑛 :

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Finding the Domain of a Function

Example: Find the domain of the function

1( ) 4

2f x x

12𝑥− 4 ≥ 0

𝑥≥ 8

[ 8 , ∞ )𝐷𝑜𝑚𝑎𝑖𝑛 :

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Finding the Domain of a Function

Functions with x in the denominator:• Set the denominator 0 and solve for x

to determine what x cannot be equal to.

1( )

3 5f x

x

3 𝑥−5≠ 0

𝑥≠53

(− ∞ ,53 )∪( 5

3, ∞ )𝐷𝑜𝑚𝑎𝑖𝑛 :

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Finding the Domain of a Function

𝑥≠ 3

𝑓 (𝑥 )= 4

√2 𝑥+5

Find each domain:

𝑓 (𝑥 )= 72 (𝑥− 3 ) (𝑥+2 ) 2 𝑥+5>0

𝑥>−52

(− ∞ , −2 )∪ (−2,3 )∪ (3 ,∞ )

(− 52

, ∞ )𝐷𝑜𝑚𝑎𝑖𝑛 :

𝐷𝑜𝑚𝑎𝑖𝑛 :

𝑥−3 ≠ 0 𝑥+2≠ 0𝑥≠ −2

−2 3

𝑓 (𝑥 )= 7

2 𝑥2−2 𝑥−12

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Business Analysis

Types of relations involving business applications:• Total Costs = fixed costs + variable costs

C = a + bx (linear relation)• Price-Demand function = the price for which an item

should be sold when you know the demand

p = m – nx (linear relation)• Price-Supply function (similar to above)• Revenue = number of items sold price per item

R = xp = x(m – nx) (quadratic relation)• Profit = Revenue – Cost

P = x(m – nx) – (a + bx) (quadratic relation)

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Mathematical Modeling

The price-demand function for a company is given by

where x represents the number of items and p(x) represents the price of the item.

A) Determine the revenue function.

B) Find the revenue generated if 50 items are sold.

C) What is the domain of the revenue function?

( ) 1000 5 , 0 100p x x x

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Solution

A) Revenue = Quantity Price R(x) = x ∙ p = x(1000 – 5x)

R(x) = 1000x – 5x2

B) When 50 items are sold, we set x = 50:

C) The domain of the function is the same as the domain for the price-demand function (which was given):

0 100x

𝑅 (𝑥 )=1000 (50)− 5(50)2

𝑅 (50 )=$37,500

𝑜𝑟 [0 ,100 ]

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Break-Even and Profit-Loss Analysis

Any manufacturing company has costs C and revenues R. They determine the following:

• If R < C loss • If R = C break even • If R > C profit

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Example of Profit-Loss Analysis

A company manufactures notebook computers. Its marketing research department has determined that the data is modeled by the price-demand functionp(x) = 2,000 – 60x, when 1 < x < 25,

(x is in thousands, p(x) is in dollars).

A) What is the price per computer when the demand is 20 thousand computers?

B) What is the company’s revenue function and what is its domain?

C) How much revenue is generated for 20 thousand computers?

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Answer to Revenue Problem

B) Revenue = Quantity Price

The domain of this function is the same as the domain of the price-demand function, which is [1, 25] (in thousands.)

𝑅 (𝑥 )=𝑥 ∙𝑝 (𝑥)𝑅 (𝑥 )=𝑥 (2000 −60 𝑥)𝑅 (𝑥 )=2000 𝑥− 60𝑥2

A ¿𝑝 (20 )=2000 −60 (20)𝑝 (20 )=$ 800𝑝𝑒𝑟𝑛𝑜𝑡𝑒𝑏𝑜𝑜𝑘(when the demand is 20,000)

C ¿ 𝑅 (20 )=2000 (20 ) −60 (20)2

𝑅 (20 )=16000 (𝑖𝑛 h𝑡 𝑜𝑢𝑠𝑎𝑛𝑑𝑠𝑜𝑓 𝑑𝑜𝑙𝑙𝑎𝑟𝑠 )The revenue is $ 16,000,000 for 20,000 notebooks .

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Profit Problem

The financial department for the company in the preceding problem has established the following cost function for producing and selling x thousand notebook computers:

C(x) = 4,000 + 500x

x is in thousands, C(x) is in thousands of dollars

A) Write a profit function for producing and selling x thousand notebook computers, and indicate the domain of this function.

B) Does the company make a profit/loss if 20 thousand notebooks are made and sold?

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Answer to Profit Problem

A) Since Profit = Revenue – Cost, and our revenue function from the preceding problem was R(x) = 2000x – 60x2,

P(x) = R(x) – C(x) = 2000x – 60x2 – (4000 + 500x) = –60x2 + 1500x – 4000.

The domain of this function is the same as the domain of the original price-demand function, 1< x < 25 (x is in thousands of notebooks)

B) P(20) = 2000 (in thousands of dollars)

The profit is $2,000,000 when 20,000 notebooksare made and sold.

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5000

Thousand dollars

Thousand notebooks

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