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Chapter 2
Functions and
Graphs
Section 1
Functions
Learning Objectives for Section
2.1
Functions
 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
𝑦=𝑥
Line
𝑦 = 𝑥2
Parabola
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𝑦 = |𝑥|
“V-shaped”
4
Families and Shapes
𝑦 = 𝑥3
𝑦= 𝑥
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𝑦=
3
𝑥
5
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.
y  x 2
2
can now be expressed as
f ( x)  x  2
2
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Function Evaluation
 Consider our function f ( x)  x  2
 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
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More Examples
f (x)  3x  2
f (2)  3(2)  2  4  2
f (6  h)  3(6  h)  2  18  3h  2
 16  3h
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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
If a function does not contain a square root or a
denominator then its domain is all reals (-, )
𝑦 = (𝑥 − 4
3
𝑦 = −(𝑥 + 1
2
−3
𝑦 = 𝑥−2 +5
𝐷𝑜𝑚𝑎𝑖𝑛: (−∞, ∞
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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
Functions with square roots:
Set the expression inside the
square root  0 and solve for x to
determine the domain.
3𝑥 − 2 ≥ 0
f ( x)  3x  2
2
𝑥≥
3
2
𝐷𝑜𝑚𝑎𝑖𝑛:
,∞
3
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Finding the Domain of a Function
 Example: Find the domain of the function
1
𝑥−4≥0
2
𝑥 ≥8
1
f ( x) 
x4
2
𝐷𝑜𝑚𝑎𝑖𝑛:
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8,∞
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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
f ( x) 
3x  5
3𝑥 − 5 ≠ 0
5
𝑥≠
3
𝐷𝑜𝑚𝑎𝑖𝑛: (−∞,
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5
5
∪
,∞
3
3
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Finding the Domain of a Function
Find each domain:
7
𝑓 𝑥 = 2
2𝑥 − 2𝑥 − 12
7
𝑓 𝑥 =
2 𝑥−3 𝑥+2
𝑥−3≠0
𝑥≠3
𝑥+2≠0
𝑥 ≠ −2
−2
3
𝐷𝑜𝑚𝑎𝑖𝑛:
(−∞,−2 ∪ −2,3 ∪ (3, ∞
Barnett/Ziegler/Byleen Business Calculus 12e
𝑓 𝑥 =
4
2𝑥 + 5
2𝑥 + 5 > 0
5
𝑥>−
2
5
𝐷𝑜𝑚𝑎𝑖𝑛: − , ∞
2
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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 = xp = 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
p( x)  1000  5 x,
0  x  100
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?
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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:
𝑅 𝑥 = 1000(50 − 5(50
2
𝑅 50 = $37,500
C) The domain of the function is the same as the domain for
the price-demand function (which was given):
0  x  100
Barnett/Ziegler/Byleen Business Calculus 12e
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 function
p(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
A 𝑝 20 = 2000 − 60(20
𝑝 20 = $800 𝑝𝑒𝑟 𝑛𝑜𝑡𝑒𝑏𝑜𝑜𝑘 (when the demand is 20,000
B) Revenue = Quantity  Price
𝑅 𝑥 = 𝑥 ∙ 𝑝(𝑥
𝑅 𝑥 = 𝑥(2000 − 60𝑥
𝑅 𝑥 = 2000𝑥 − 60𝑥 2
The domain of this function is the same as the domain of the
price-demand function, which is [1, 25] (in thousands.)
C 𝑅 20 = 2000 20 − 60(20
2
𝑅 20 = 16000 (𝑖𝑛 𝑡ℎ𝑜𝑢𝑠𝑎𝑛𝑑𝑠 𝑜𝑓 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)
5000
B) P(20) = 2000 (in thousands of dollars)
Thousand
dollars
The profit is $2,000,000 when 20,000 notebooks
are made and sold.
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Thousand
notebooks
25
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