Chabot Mathematics
§1.1 Intro to
Functions
Bruce Mayer, PE
Licensed Electrical & Mechanical Engineer
BMayer@ChabotCollege.edu
Chabot College Mathematics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-01_sec_1-1_Fcn_Intro.pptx
§1.1 Learning Goals
 Identify the domain of a function, and
evaluate a function from an equation
 Gain familiarity with piecewise-defined
functions
 Introduce and illustrate functions used
in economics
 Form and use composite functions in
applied problems
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BMayer@ChabotCollege.edu • MTH15_Lec-01_sec_1-1_Fcn_Intro.pptx
ReCall the Ordered-Pair
 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; e.g.,
3,16
Eqn  y  x 2  7 Satisfied by
?
2
as
16  3  7 16  9  7  16 
 An ordered pair that satisfies an
equation is called a solution of the eqn
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-01_sec_1-1_Fcn_Intro.pptx
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 • MTH15_Lec-01_sec_1-1_Fcn_Intro.pptx
Mathematical RELATION
 Any SET of ordered pairs is called a
relation.
 The set of all first components is
called the domain of the relation,
 The set of all SECOND
components is called the RANGE
of the relation
Chabot College Mathematics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-01_sec_1-1_Fcn_Intro.pptx
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}
Chabot College Mathematics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-01_sec_1-1_Fcn_Intro.pptx
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)}.
Chabot College Mathematics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-01_sec_1-1_Fcn_Intro.pptx
FUNCTION Defined
 A function is a “Rule” which
“takes” a set X to a set Y, and 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 • MTH15_Lec-01_sec_1-1_Fcn_Intro.pptx
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.
Chabot College Mathematics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-01_sec_1-1_Fcn_Intro.pptx
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.
Chabot College Mathematics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-01_sec_1-1_Fcn_Intro.pptx
Example  Is Relation a Fcn?
Chabot College Mathematics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-01_sec_1-1_Fcn_Intro.pptx
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)}.
Chabot College Mathematics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-01_sec_1-1_Fcn_Intro.pptx
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.
Chabot College Mathematics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-01_sec_1-1_Fcn_Intro.pptx
Example  Is Relation a Fcn?
Chabot College Mathematics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-01_sec_1-1_Fcn_Intro.pptx
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)}.
Chabot College Mathematics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-01_sec_1-1_Fcn_Intro.pptx
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 • MTH15_Lec-01_sec_1-1_Fcn_Intro.pptx
Function Forms
 Functions can be described by:
• A
Table
x
y
• A
Graph
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-01_sec_1-1_Fcn_Intro.pptx
Function Forms
 Math Functions are MOST
OFTEN described by:
• An EQUATION
– The Eqn can be
used to MAKE a
Table or Graph
 NOTE: f(x) ≠ “f times x”
• f(x) indicates
EVALUATION of the
function AT the
INDEPENDENT
variable-value of x
Chabot College Mathematics
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yx
2
f x   x
2
y  x  6x  8
2
g x   x  6x  8
2
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-01_sec_1-1_Fcn_Intro.pptx
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
Chabot College Mathematics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-01_sec_1-1_Fcn_Intro.pptx
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
Chabot College Mathematics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-01_sec_1-1_Fcn_Intro.pptx
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
Chabot College Mathematics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-01_sec_1-1_Fcn_Intro.pptx
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
Chabot College Mathematics
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ONE value of y
so it DOES
define y as a
function of x
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-01_sec_1-1_Fcn_Intro.pptx
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
Chabot College Mathematics
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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 • MTH15_Lec-01_sec_1-1_Fcn_Intro.pptx
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 • MTH15_Lec-01_sec_1-1_Fcn_Intro.pptx
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}
Chabot College Mathematics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-01_sec_1-1_Fcn_Intro.pptx
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, ∞)
Chabot College Mathematics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-01_sec_1-1_Fcn_Intro.pptx
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, ∞)
Chabot College Mathematics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-01_sec_1-1_Fcn_Intro.pptx
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 • MTH15_Lec-01_sec_1-1_Fcn_Intro.pptx
Composite Functions
 In the real world, functions
frequently occur in which some
quantity depends on a variable that,
in turn, depends on yet another
variable.
 Functions such as these are called
COMPOSITE FUNCTIONS
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-01_sec_1-1_Fcn_Intro.pptx
Composing a Function
 Composition with sets A & B by fcns g & f
g
f
g ( x)  3x  1
A
B
1
3
7
4
10
22
h
Chabot College Mathematics
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1
f ( x)  x  3
2
C
−1
2
8
h(x) = ?
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-01_sec_1-1_Fcn_Intro.pptx
g ( x)  3x  1
A
Composing a Function
f ( x) 
B
4
-1
3
10
2
7
22
8
 From The Diagram notice that since
f takes the output from g we can
combine f and g to get a function h:
f (g (x)) = f (3x + 1)
1
 (3x  1)  3
2
3
5
 x
2
2
3
5
 This Yields an eqn for h: h( x)  x  .
2
2
31
C
1
h
Chabot College Mathematics
1
x 3
2
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-01_sec_1-1_Fcn_Intro.pptx
h(x) = ?
Composing a Function
g ( x)  3x  1
f ( x) 
1
x 3
2
A
B
1
4
-1
3
10
2
7
22
8
h
C
3
5
h( x )  x  .
2
2
h(x) = ?
 The function h is the composition of
f and g and is denoted f○g (read “the
composition of f and g”, or “f composed
with g”, or “f circle g”).
Chabot College Mathematics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-01_sec_1-1_Fcn_Intro.pptx
COMPOSITION OF FUNCTIONS
 If f and g are two functions, the
composition of function f with function g
is written as f○g and is defined by the
equation
 f og x   f g x ,
 The function where the domain of f○g
consists of those values x in the domain
of g for which g(x) is in the domain of f
Chabot College Mathematics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-01_sec_1-1_Fcn_Intro.pptx
COMPOSITION OF FUNCTIONS
 Graphically the f○g Domain Chain
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-01_sec_1-1_Fcn_Intro.pptx
COMPOSITION OF FUNCTIONS
 Conceptually the f○g Operation Chain
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-01_sec_1-1_Fcn_Intro.pptx
Example  Evaluate Composites
 Given: f x   x 3 and g x   x  1.
 Find Each of the Following
a.  f og 1
b. g o f 1
c.  f o f 1
d. g og 1
 Solution a.a.  f og 1  f g 1
 f 2 
Chabot College Mathematics
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 23
8
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-01_sec_1-1_Fcn_Intro.pptx
Example  Evaluate Composites
 Solution b.
b. g o f 1  g  f 1
f x   x 3 and g x   x  1.
 g 1  1  1  2
 Solution c.  f o f 1  f  f 1
 f 1  1  1
3
 Solution d.
d. g og 1  g g 1
 g 0   0  1  1
Chabot College Mathematics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-01_sec_1-1_Fcn_Intro.pptx
Example  Fcn Composition
 Given f(x) = 4x and g(x) = x2 + 2, find
f
g  ( x) and  g f  ( x).
 SOLUTION
f
g  ( x)  f ( g ( x)) = f (x2 + 2)
= 4(x2 + 2)
= 4x2 + 8
Chabot College Mathematics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-01_sec_1-1_Fcn_Intro.pptx
Example  Fcn Composition
 Given f(x) = 4x and g(x) = x2 + 2, find
f
g  ( x) and  g f  ( x).
 SOLUTION
g
f  ( x)  g ( f ( x)) = g(4x)
= (4x)2 + 2
= 16x2 + 2
 This example shows
that in general  f g  ( x)   g f  ( x).
Chabot College Mathematics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-01_sec_1-1_Fcn_Intro.pptx
Example  Fcn Composition
 Given: f x   2x  1 and g x   x  3.
2
 Find Each Composite Function
a.  f og x 
b. g o f x 
c.  f o f x 
a.  f og x   f g x 

 2 x

 3 1
 f x2  3
 Solution a.
2
 2x  6  1
2
 2x  5
2
Chabot College Mathematics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-01_sec_1-1_Fcn_Intro.pptx
Example  Fcn Composition
 Given: f x   2x  1 and g x   x  3.
2
 Solution
 f og
x  b.b. g o f x  c.  f o f x 
b. g o f x   g  f x 
 g 2x  1
2
2
 2x  1  3  4x  4x  2
Chabot College Mathematics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-01_sec_1-1_Fcn_Intro.pptx
Example  Fcn Composition
 Given: f x   2x  1 and g x   x  3.
2
g o f Solution
x  c.c.  f o f x 
c.  f o f x   f  f x 
 f 2x  1
 2 2x  1  1  4 x  3
Chabot College Mathematics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-01_sec_1-1_Fcn_Intro.pptx
Example  Composite Domain
1
Let f x   x  1 and g x   .
 Given:
x
a. Find  f og 1.
b. Find g o f 1.
c. Find  f og x  and its domain.
d. Find g o f x  and its domain.
 Solution
a.  f og 1  f g 1
a.
 f 1  1  1  0
Chabot College Mathematics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-01_sec_1-1_Fcn_Intro.pptx
Example  Composite Domain
1
Let f x   x  1 and g x   .
 Given:
x
 Solution
b. g o f 1  g  f 1
b.
 g 0  not defined
 1 1
 Soln
c.  f og x   f g x   f     1
 x x
c.
• Domain: (−∞, 0)U(0, ∞) or {x|x ≠ 0}
Chabot College Mathematics
44
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-01_sec_1-1_Fcn_Intro.pptx
Example  Composite Domain
1
Let f x   x  1 and g x   .
 Given:
x
1
 Soln
d. g o f x   g  f x   g x  1 
x 1
d.
• Domain: (−∞, −1)U(−1, ∞) or {x|x ≠ −1}
Chabot College Mathematics
45
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-01_sec_1-1_Fcn_Intro.pptx
DEcomposing a Function
Let H x  
 Given:
1
.
2x  1
 Show that each of the following
provides a DEcomposition of H(x)
2
a. Express H x  as f g x ,
1
where f x  
and g x   2x 2  1.
x
b. Express H x  as f g x ,
1
where f x   and g x   2x 2  1.
x
Chabot College Mathematics
46
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-01_sec_1-1_Fcn_Intro.pptx
Decomposing a Function
Let
 Solution: a. Express H x  as f g x ,
1
H x  
.
1
2
2x  1 where f x  

and g x   2x 2  1.
x

a. f g x   f 2x  1

2
1
2x  1
 H x 
Chabot College Mathematics
47
2
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-01_sec_1-1_Fcn_Intro.pptx
Decomposing a Function
Let
 Solution: b. Express H x  as f g x ,
1
H x  
.
1
2
2x  1
where f x  
b. f g x   f

Chabot College Mathematics
48
x
and g x   2x 2  1.
 2x  1
2
1
2x  1
 H x 
2
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-01_sec_1-1_Fcn_Intro.pptx
WhiteBoard Work
 Problems From §1.1 Exercise Set
• 37, 65
Chabot College Mathematics
49
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-01_sec_1-1_Fcn_Intro.pptx
All Done for Today
Some
Statin
Drugs
Chabot College Mathematics
50
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-01_sec_1-1_Fcn_Intro.pptx
Chabot Mathematics
Appendix
r  s  r  s r  s 
2
2
Bruce Mayer, PE
Licensed Electrical & Mechanical Engineer
BMayer@ChabotCollege.edu
–
Chabot College Mathematics
51
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-01_sec_1-1_Fcn_Intro.pptx
Chabot College Mathematics
52
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-01_sec_1-1_Fcn_Intro.pptx
P1-37
Chabot College Mathematics
53
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-01_sec_1-1_Fcn_Intro.pptx
Chabot College Mathematics
54
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-01_sec_1-1_Fcn_Intro.pptx
Chabot College Mathematics
55
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-01_sec_1-1_Fcn_Intro.pptx
Chabot College Mathematics
56
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-01_sec_1-1_Fcn_Intro.pptx
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.
Chabot College Mathematics
57
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-01_sec_1-1_Fcn_Intro.pptx