§2.1 Intro to Functions Chabot Mathematics Bruce Mayer, PE

Chabot Mathematics

§2.1 Intro to

Functions

Bruce Mayer, PE

Licensed Electrical & Mechanical Engineer

BMayer@ChabotCollege.edu

Bruce Mayer, PE

BMayer@ChabotCollege.edu • MTH55_Lec-04_Sec_2-1_Fcn_Intro.ppt

Chabot College Mathematics

1

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


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; e.g.,

• (7, −17) is a solution to: y = x 2 − 66

3

 −17 = 7 2 − 66 = 49 − 66 = −17

Chabot College Mathematics Bruce Mayer, PE


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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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


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


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


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


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


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


Example





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Bruce Mayer, PE


Example




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 different domain elements tto correspond the same range element .

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


Example




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


Example





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Bruce Mayer, PE


Example




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.

15

• The set of ordered pairs that define this relation is {( Malcolm , 220-307-4112),

( Malcolm , 220-527-6277 ), (MARIA, 220-

527-6277), (MARIA, 220-416-5204),

(MARIA, 220-433-8195), (Ellen, 220-527-

6277), (Duane, 220-527-6277)}.

Bruce Mayer, PE Chabot College Mathematics


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 .

16

• 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.

Bruce Mayer, PE Chabot College Mathematics


Function Forms

 Functions can be described by:

• A

Table y x

• A

Graph


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Bruce Mayer, PE


Function Forms

 Functions are MOST OFTEN described by:

• An EQUATION y  x 2

 

 x 2

 NOTE: f ( x ) ≠ “f times x”

• f ( x ) indicates

EVALUATION of the function AT the

INDEPENDENT variable-value of x y  x 2  6 x  8

   x 2  6 x  8


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Bruce Mayer, PE


Finding a Function Value

 Let g be the function defined by the equation



y = g(x) = x 2 – 6x + 8

 Determine each function value: a. d. g



 

 2

 b. e. g



 

 h

 c. g



 1

2 



 SOLUTION a. g

   3 2  6

   8   1


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Bruce Mayer, PE


Determine a Function Result

 Find fcn Value



y = g(x) = x 2 – 6x + 8 c. g



 1

2 

 d.



 2

 b. g

  e. g x  h



 SOLUTION b. g

    2

2  6  2  8  24 c. g



 1

2 




20





 1

2 

 2

 6



 1

2 



 8 

21

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Bruce Mayer, PE


Evaluating a Function

 Evaluate fcn



y = g(x) = x 2 – 6x + 8 d.

  2

 e. g x  h



 SOLUTION d. g

 a  2

   a  2

 2  6

 a  2

  8

 a 2  4 a  4  6 a  12  8

 a 2  2 a e. g

 x  h





 x  h

 2  6

 x  h



 8


21

 x 2  2 xh  h 2  6 x  6 h  8

Bruce Mayer, PE


Example

 is an EQN a FCN??

 Determine whether each equation determines y as a function of x .

a. 6x 2 – 3y = 12 b. y 2 – x 2 = 4



6 x 2

SOLUTION a.

6 x 2  3 y  12

 3 y  3 y  12  12  3 y  12

 any value of x corresponds to only ONE value of y , so the Eqn

6 x 2

2 x 2

 12  3 y

 4  y

DOES define y as a function of x


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Bruce Mayer, PE


Example

 is an EQN a FCN??

 Determine whether each equation determines y as a function of x .

a. 6 x 2 – 3 y = 12 b. y 2 – x 2 = 4

 SOLUTION b.

y 2  x 2  4 y 2  x 2  x 2  4  x 2 y 2  x 2  4 y   x 2  4

 TWO values of y correspond to the same value of x so the expression does

NOT define y as a function of x .


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Bruce Mayer, PE


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., the DEFAULT Domain is all x ’s that produce VALID Functional RESULTS


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Bruce Mayer, PE


Example



Find the Domain

 Find the DOMAIN of each function.

a.

 



1

1  x 2 b.

   x c.

  

1 x  1

 SOLUTION d.

 

 2 t  1 a. f is not defined when the denominator is 0.

1−x 2 ≠ 0 → Domain: {x|x ≠ −1 and x ≠ 1}


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Bruce Mayer, PE


Example



Find the Domain

 SOLUTION b.

   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


Example



Find the Domain

 SOLUTION c.

 



1 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


Example



Find the Domain

 SOLUTION d.

   2 t  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


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


WhiteBoard Work

 Problems From

§2.1 Exercise Set

• 18, 26

 P2.1-26

Functional

Relationships x

-2

-1

0

1

2


30 f(x) g(x)

6

3

-1

-4

0

-3

-6

0

4

1

Bruce Mayer, PE


All Done for Today

Some

Statin

Drugs


31

Bruce Mayer, PE


Chabot Mathematics

Appendix

r

2  s

2 

 r

 s

 r

 s



Bruce Mayer, PE

Licensed Electrical & Mechanical Engineer

BMayer@ChabotCollege.edu

–


32

Bruce Mayer, PE


§2.1 Intro to Functions Chabot Mathematics Bruce Mayer, PE

Chabot Mathematics

§2.1 Intro to

Functions

A function which “takes” a set

to a set

is a relation in which each element of

corresponds to ONE, and ONLY ONE, element of

.

Some

Statin

Drugs

Chabot Mathematics

Appendix

Related documents

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Support

§2.1 Intro to Functions Chabot Mathematics Bruce Mayer, PE

Chabot Mathematics

§2.1 Intro to

Functions

A function which “takes” a set

to a set

is a relation in which each element of

corresponds to ONE, and ONLY ONE, element of

.

Some

Statin

Drugs

Chabot Mathematics

Appendix

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