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Week-2 B

The document covers the basics of functions and graphing equations, including how to plot points and determine if an equation specifies a function using the vertical line test. It explains function notation, evaluation, and the concept of domain, providing examples related to price-demand functions and profit-loss analysis for companies. Additionally, it includes mathematical modeling to calculate revenue and profit based on given functions.

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0% found this document useful (0 votes)
18 views29 pages

Week-2 B

The document covers the basics of functions and graphing equations, including how to plot points and determine if an equation specifies a function using the vertical line test. It explains function notation, evaluation, and the concept of domain, providing examples related to price-demand functions and profit-loss analysis for companies. Additionally, it includes mathematical modeling to calculate revenue and profit based on given functions.

Uploaded by

ecebozalan
Copyright
© © All Rights Reserved
We take content rights seriously. If you suspect this is your content, claim it here.
Available Formats
Download as PDF, TXT or read online on Scribd
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MAT1005

Week 2(B)

Chapter 1
Functions and Graphs
Section 1
Functions

Copyright © 2015, 2011, and 2008 Pearson Education, Inc. 1


Graphing an Equation

 To sketch the graph an equation in x and y, we need to find


ordered pairs that solve the equation and plot the ordered
pairs on a grid. This process is called point-by-point
plotting.

For example, let’s plot the graph of the equation

y  x 2
2

Copyright © 2015, 2011, and 2008 Pearson Education, Inc. 2


Graphing an Equation:
Making a Table of Ordered Pairs

 Make a table of ordered x y


pairs that satisfy the
–3 (–3)2+2 = 11
equation
–2 (–2)2+2 = 6
y  x 2
2
–1 (–1)2+2 = 3
0 (0)2+2 = 2
1 (1)2+2 = 3
2 (2)2+2 = 6

Copyright © 2015, 2011, and 2008 Pearson Education, Inc. 3


(x, y)
(–3,11)
(–2, 6)
(–1, 3)
(0, 2)
(1, 3)
(2, 6)

Copyright © 2015, 2011, and 2008 Pearson Education, Inc. 4


Graphing an Equation:
Plotting the points

 Next, plot the points and connect them with a smooth


curve. You may need to plot additional points to see the
pattern formed.
y = x2 - 2

Copyright © 2015, 2011, and 2008 Pearson Education, Inc. 5


Functions

 The previous graph is the graph of a function. The idea of


a function is this: a correspondence between two sets D
and R such that to each element of the first set, D, there
corresponds one and only one element of the second set,
R.
 The first set is called the domain, and the set of
corresponding elements in the second set is called the
range.
For example, the cost of a pizza (C) is related to the size of
the pizza. A 10 inch diameter pizza costs $9.00, while a 16
inch diameter pizza costs $12.00.

Copyright © 2015, 2011, and 2008 Pearson Education, Inc. 6


Function Definition

 You can visualize a function by the following diagram which


shows a correspondence between two sets: D, the domain of
the function, gives the diameter of pizzas, and R, the range of
the function gives the cost of the pizza.

10 9.00
12 10.00
16
12.00

domain D range R

Copyright © 2015, 2011, and 2008 Pearson Education, Inc. 7


Functions Specified by Equations

 If in an equation in two variables, we get exactly one


output (value for the dependent variable) for each input
(value for the independent variable), then the equation
specifies a function. The graph of such a function is just
the graph of the specifying equation.

 If we get more than one output for a given input, the


equation does not specify a function.

Copyright © 2015, 2011, and 2008 Pearson Education, Inc. 8


Functions Specified by Equations

 Consider the equation that was graphed on a previous slide

y  x 2 2
–2
Input:
x = –2

Process:
(–2,2) is an 2 2
2 square (–2),
then subtract 2
ordered pair of
the function.
Output:
2 result is 2

Copyright © 2015, 2011, and 2008 Pearson Education, Inc. 9


Vertical Line Test for a Function

If you have the graph of an equation, there is an easy


way to determine if it is the graph of an function. It is
called the vertical line test which states that:
An equation specifies a function if each vertical line in
the coordinate system passes through at most one
point on the graph of the equation.
If any vertical line passes through two or more points
on the graph of an equation, then the equation does
not specify a function.

Copyright © 2015, 2011, and 2008 Pearson Education, Inc. 10


Vertical Line Test for a Function
(continued)

This graph is not the graph of a


function because you can draw a
vertical line which crosses it
twice.

This is the graph of a


function because any vertical
line crosses only once.

Copyright © 2015, 2011, and 2008 Pearson Education, Inc. 11


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.
Our previous equation
y  x 2 2

can now be expressed as

f ( x)  x  2 2

Copyright © 2015, 2011, and 2008 Pearson Education, Inc. 12


Function Evaluation

 Consider our function f ( x)  x  2 2

 What does f (–3) mean?

Copyright © 2015, 2011, and 2008 Pearson Education, Inc. 13


Function Evaluation

 Consider our function f ( x)  x 2  2

 What does f (–3) mean?


Replace x with the value –3 and evaluate the expression
f (3)  (3)2  2
 The result is 7 . This means that the point (–3,7) is on the
graph of the function.

Copyright © 2015, 2011, and 2008 Pearson Education, Inc. 14


Some Examples

1.
f (x)  3x  2
f (2)  3(2)  2  4  2

f (a)  3(a)  2

f (6  h)  3(6  h)  2  18  3h  2
 16  3h
Copyright © 2015, 2011, and 2008 Pearson Education, Inc. 15
Domain of a Function

 Consider f ( x)  3 x  2
f (0)  ?
f (0)  3(0)  2  2
which is not a real number.
 Question: for what values of x is the function defined?

Copyright © 2015, 2011, and 2008 Pearson Education, Inc. 16


Domain of a Function

 Answer: f ( x)  3 x  2
is defined only when the radicand (3x – 2) is equal to
or greater than zero. This implies that

2
x
3

Copyright © 2015, 2011, and 2008 Pearson Education, Inc. 17


Domain of a Function
(continued)

 Therefore, the domain of our function is the set of real


numbers that are greater than or equal to 2/3.

 Example: Find the domain of the function


1
f ( x)  x4
2

Copyright © 2015, 2011, and 2008 Pearson Education, Inc. 18


Domain of a Function
(continued)

 Therefore, the domain of our function is the set of real


numbers that are greater than or equal to 2/3.

 Example: Find the domain of the function


1
f ( x)  x4
2
 Answer:
x x  8 , [8, )
Copyright © 2015, 2011, and 2008 Pearson Education, Inc. 19
Domain of a Function:
Another Example

1
 Find the domain of f ( x) 
3x  5

Copyright © 2015, 2011, and 2008 Pearson Education, Inc. 20


Domain of a Function:
Another Example

1
 Find the domain of f ( x) 
3x  5

 In this case, the function is defined for all values of x


except where the denominator of the fraction is zero. This
means all real numbers x except 5/3.

Copyright © 2015, 2011, and 2008 Pearson Education, Inc. 21


Mathematical Modeling

The price-demand function for a company is given by

p(x)  1000  5x, 0  x  100

where x represents the number of items and p(x) represents the


price of the item. Determine the revenue function and find the
revenue generated if 50 items are sold.

Copyright © 2015, 2011, and 2008 Pearson Education, Inc. 22


Solution

Revenue = Price ∙ Quantity, so


R(x)= p(x) ∙ x = (1000 – 5x) ∙ x
When 50 items are sold, x = 50, so we will evaluate the
revenue function at x = 50:
R(50)  1000  5  50  50  750  50  37,500
The domain of the function has already been specified. We
are told that
0  x  100

Copyright © 2015, 2011, and 2008 Pearson Education, Inc. 23


Break-Even and Profit-Loss
Analysis

 Any manufacturing company has costs C and revenues R.


 The company will have a loss if R < C, will break even
if R = C, and will have a profit if R > C.
 Costs include fixed costs such as plant overhead, etc. and variable
costs, which are dependent on the number of items produced.
C = a + bx
(x is the number of items produced)

Copyright © 2015, 2011, and 2008 Pearson Education, Inc. 24


Break-Even and Profit-Loss
Analysis
(continued)
 Price-demand functions, usually determined by financial
departments, play an important role in profit-loss analysis.
p = m – nx
(x is the number of items than can be sold at $p per item.)
 The revenue function is
R = (number of items sold) ∙ (price per item)
= xp = x(m – nx)
 The profit function is
P = R – C = x(m – nx) – (a + bx)

Copyright © 2015, 2011, and 2008 Pearson Education, Inc. 25


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

Copyright © 2015, 2011, and 2008 Pearson Education, Inc. 26


Answer to Revenue Problem

Since Revenue = Price ∙ Quantity,

R( x)  x  p( x)  x  (2000  60 x)  2000 x  60 x 2
The domain of this function is the same as the domain
of the price-demand function, which is 1 ≤ x ≤ 25 (in
thousands.)

Copyright © 2015, 2011, and 2008 Pearson Education, Inc. 27


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

Copyright © 2015, 2011, and 2008 Pearson Education, Inc. 28


Answer to Profit Problem

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 (in
thousands.) 5000

Thousand dollars

25
Thousand notebook computers

Copyright © 2015, 2011, and 2008 Pearson Education, Inc. 29

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