Chabot Mathematics
§2.1 Intro to
Functions
Bruce Mayer, PE
Licensed Electrical & Mechanical Engineer
BMayer@ChabotCollege.edu
Chabot College Mathematics
1
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-04_Sec_2-1_Fcn_Intro.ppt
Review §
1.6
MTH 55
 Any QUESTIONS About
• §1.6 → Exponent Rules & Properties
 Any QUESTIONS About HomeWork
• §1.6 → HW-02
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-04_Sec_2-1_Fcn_Intro.ppt
Ordered Pair Defined
 An ordered pair (a, b) is said to satisfy
an equation with variables a and b if,
when a is substituted for x and b is
substituted for y in the equation, the
resulting statement is true.
 An ordered pair that satisfies an
equation is called a solution of the
equation
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-04_Sec_2-1_Fcn_Intro.ppt
Ordered Pair Dependency
 Frequently, the numerical values of the
variable y can be determined by
assigning appropriate values to the
variable x. For this reason,
y is sometimes referred to as the
dependent variable
and x as the
independent variable.
• i.e., if we KNOW x,
we can CALCULATE y
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-04_Sec_2-1_Fcn_Intro.ppt
Mathematical RELATION
 Any set of ordered pairs is called a
relation. The set of all first
components is called the domain
of the relation, and the set of all
SECOND components is called the
RANGE of the relation
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-04_Sec_2-1_Fcn_Intro.ppt
Example  Domain & Range
 Find the Domain and Range of the
relation:
• { (Titanic, $600.8), (Star Wars IV, $461.0),
(Shrek 2, $441.2), (E.T., $435.1),
(Star Wars I, $431.1),
(Spider-Man, $403.7)}
 SOLUTION
• The DOMAIN is the set of all first
components, or {Titanic, Star Wars IV,
Shrek 2, E.T., Star Wars I, Spider-Man}
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-04_Sec_2-1_Fcn_Intro.ppt
Example  Domain & Range
 Find the Domain and Range for the
relation:
• { (Titanic, $600.8), (Star Wars IV, $461.0),
(Shrek 2, $441.2), (E.T., $435.1),
(Star Wars I, $431.1),
(Spider-Man, $403.7)}
 SOLUTION
• The RANGE is the set of all
second components, or {$600.8, $461.0,
$441.2, $435.1, $431.1, $403.7)}.
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-04_Sec_2-1_Fcn_Intro.ppt
FUNCTION Defined
 A function which “takes” a set X
to a set Y is a relation in which
each element of X corresponds
to ONE, and ONLY ONE,
element of Y.
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-04_Sec_2-1_Fcn_Intro.ppt
Functional Correspondence
 A relation may be defined by a
correspondence diagram, in which an arrow
points from each domain element to the
element or elements in the range that
correspond to it.
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-04_Sec_2-1_Fcn_Intro.ppt
Example  Is Relation a Fcn?

Determine whether the relations that
follow are functions. The domain of
each relation is the family consisting of
Malcolm (father), Maria (mother),
Ellen (daughter), and Duane (son).
1. For the relation defined by the following
diagram, the range consists of the ages
of the four family members, and each
family member corresponds to that family
member’s age.
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-04_Sec_2-1_Fcn_Intro.ppt
Example  Is Relation a Fcn?
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-04_Sec_2-1_Fcn_Intro.ppt
Example  Is Relation a Fcn?
1. SOLUTION: The relation IS a
FUNCTION, because each element
in the domain corresponds to
exactly ONE element in the range.
•
For a function, it IS permissible for the
same range element to correspond to
different domain elements. The set of
ordered pairs that define this relation is
{(Malcolm, 36), (Maria, 32), (Ellen, 11),
(Duane, 11)}.
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-04_Sec_2-1_Fcn_Intro.ppt
Example  Is Relation a Fcn?
2. For the relation defined by the diagram
on the next slide, the range consists of
the family’s home phone number, the
office phone numbers for both Malcolm
and Maria, and the
cell phone number
for Maria. Each family
member corresponds
to all phone numbers
at which that family
member can be reached.
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-04_Sec_2-1_Fcn_Intro.ppt
Example  Is Relation a Fcn?
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-04_Sec_2-1_Fcn_Intro.ppt
Example  Is Relation a Fcn?
2. SOLUTION: The relation is NOT a
function, because more than one
range element corresponds to the
same domain element. For example,
both an office ph. number and a home
ph. number correspond to Malcolm.
•
The set of ordered pairs that define this
relation is {(Malcolm, 220-307-4112),
(Malcolm, 220-527-6277 ), (MARIA, 220527-6277), (MARIA, 220-416-5204),
(MARIA, 220-433-8195), (Ellen, 220-5276277), (Duane, 220-527-6277)}.
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-04_Sec_2-1_Fcn_Intro.ppt
Function Notation
 Typically use single letters such as f, F, g, G,
h, H, and so on as the name of a function.
 For each x in the domain of f, there
corresponds a unique y in its range. The
number y is denoted by f(x) read as “f of x”
or “f at x”.
 We call f(x) the value of f at the number x
and say that f assigns the f(x) value to y.
• Since the value of y depends on the given value
of x, y is called the dependent variable
and x is called the independent variable.
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-04_Sec_2-1_Fcn_Intro.ppt
Function Forms
 Functions can be described by:
• A
Table
x
y
• A
Graph
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-04_Sec_2-1_Fcn_Intro.ppt
Function Forms
 Functions are MOST OFTEN
described by:
2
• An EQUATION
 NOTE: f(x) ≠ “f times x”
• f(x) indicates
EVALUATION of the
function AT the
INDEPENDENT
variable-value of x
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yx
f x   x
2
y  x  6x  8
2
g x   x  6x  8
2
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-04_Sec_2-1_Fcn_Intro.ppt
Evaluating a Function
 Let g be the function defined by the
equation  y = g(x) = x2 – 6x + 8
 Evaluate each function value:
 1
b. g 2 
c. g  
a. g 3
 2
e. g x  h 
d. g a  2 
 SOLUTION
a. g 3  3  6 3  8  1
2
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-04_Sec_2-1_Fcn_Intro.ppt
Evaluating a Function
 Evaluate fcn  y = g(x) = x2 – 6x + 8
 1
b. g 2 
c. g  
 2
e. g x  h 
d. g a  2 
 SOLUTION
b. g 2   2   6 2   8  24
2
2
21
 1  1
 1
c. g       6    8 
 2  2
 2
4
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-04_Sec_2-1_Fcn_Intro.ppt
Evaluating a Function
 Evaluate fcn  y = g(x) = x2 – 6x + 8
e. g x  h 
d. g a  2 
 SOLUTION
2
d. g a  2   a  2   6 a  2   8
 a 2  4a  4  6a  12  8
 a  2a
2
e. g x  h   x  h   6 x  h   8
2
 x 2  2xh  h 2  6x  6h  8
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-04_Sec_2-1_Fcn_Intro.ppt
Example  is an EQN a FCN??
 Determine whether each equation
determines y as a function of x.
a. 6x2 – 3y = 12
b. y2 – x2 = 4
 SOLUTION a.

any
value
of
x
6x 2  3y  12
corresponds to
6x 2  3y  3y  12  12  3y  12
6x 2  12  3y
2x 2  4  y
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ONE value of y
so it DOES
define y as a
function of x
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-04_Sec_2-1_Fcn_Intro.ppt
Example  is an EQN a FCN??
 Determine whether each equation
determines y as a function of x.
a. 6x2 – 3y = 12
b. y2 – x2 = 4
 SOLUTION b.
 TWO values of y
2
2
y x 4
correspond to the
y x x 4x
2
2
2
2
y x 4
2
2
y x 4
2
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same value of x so
the expression does
NOT define y as a
function of x.
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-04_Sec_2-1_Fcn_Intro.ppt
Implicit Domain
 If the domain of a function that is
defined by an equation is not
explicitly specified, then we take
the domain of the function to be the
LARGEST SET OF REAL
NUMBERS that result in REAL
NUMBERS AS OUTPUTS.
• i.e., DEFAULT Domain is all x’s that
produce VALID Functional RESULTS
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-04_Sec_2-1_Fcn_Intro.ppt
Example  Find the Domain
 Find the DOMAIN of each function.
1
a. f x  
b. g x   x
2
1 x
1
c. h x  
d. P t   2t  1
x 1
 SOLUTION
a. f is not defined when the denominator is 0.
1−x2 ≠ 0 → Domain: {x|x ≠ −1 and x ≠ 1}
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-04_Sec_2-1_Fcn_Intro.ppt
Example  Find the Domain
 SOLUTION b. g x   x
• The square root of a negative number is
not a real number and is thus excluded
from the domain
x NONnegative → Domain: {x|x ≥ 0}, [0, ∞)
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-04_Sec_2-1_Fcn_Intro.ppt
Example  Find the Domain
1
 SOLUTION c. h x  
x 1
• The square root of a negative number is
not a real number and is excluded from the
domain, so x − 1 ≥ 0. Thus have x ≥ 1
• However, the denominator must ≠ 0, and it
does = 0 when x = 1. So x = 1 must be
excluded from the domain as well
DeNom NONnegative-&-NONzero →
Domain: {x|x > 1}, (1, ∞)
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-04_Sec_2-1_Fcn_Intro.ppt
Example  Find the Domain
 SOLUTION d. P t   2t  1
• Any real number substituted for t yields a
unique real number.
NO UNDefinition →
Domain: {t|t is a real number}, or (−∞, ∞)
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-04_Sec_2-1_Fcn_Intro.ppt
Function Equality
 Two functions f and g are equal if
and only if:
1. f and g have the same domain
• and
2. f(x) = g(x) for all x in the domain.
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-04_Sec_2-1_Fcn_Intro.ppt
WhiteBoard Work
 Problems From
§2.1 Exercise Set
• 18, 26

P2.1-26
Functional
Relationships
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x
f(x)
g(x)
-2
6
0
-1
3
4
0
-1
1
1
-4
-3
2
0
-6
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-04_Sec_2-1_Fcn_Intro.ppt
All Done for Today
Some
Statin
Drugs
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-04_Sec_2-1_Fcn_Intro.ppt
Chabot Mathematics
Appendix
r  s  r  s r  s 
2
2
Bruce Mayer, PE
Licensed Electrical & Mechanical Engineer
BMayer@ChabotCollege.edu
–
Chabot College Mathematics
32
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-04_Sec_2-1_Fcn_Intro.ppt