Transcript 2.3 Introduction to Functions

2.3 Introduction to Functions
Definition of a Relation
A relation is any set of ordered pairs.
The set of all first components of the
ordered pairs is called the domain of
the relation, and the set of all second
components is called the range of the
relation.
Definition of a Function
A function is a correspondence between
two sets X and Y that assings to each
element x of set X exactly one element y
of Y .
Domain and Range
For each element x in X , the
corresponding element y in Y is called
the value of the function at x. The set X
is called the domain of the function, and
the set of all function values, Y , is called
the range of the function.
Ex 1: Determine whether each
relation is a function.
a. {(4,5), (6, 7), (8,8)}
b. {(5, 6), (4, 7), (6, 6), (6, 7)}
Solution We begin by making a figure
for each relation that shows set X , the
domain, and set Y , the range.
Solution for part (a)
X
Y
4
6
8
5
7
8
The figure shows that every
element in the domain
corresponds to exactly one
element in the range.
Domain Range
No two ordered pairs in the given relation have
the same first component different second
components. Thus, the relation is a function
Solution for part (b)
X
Y
4
5
6
6
7
The figure shows that 6
corresponds to both 6 and 7.
Domain Range
If any element in the domain corresponds to
more than one element in the range, the
relation is not a function, Thus, the relation
is not a function.
Practice Exercises
Determine whether each relation is a
function. Give the domain and range
for each relation.
1. {(7, 7), (5, 5), (3, 3), (0, 0)}
2. {(4,1), (5,1), (4, 2)}
Answers
1. Domain {7, 5, 3, 0}
Range {7, 5, 3, 0}
Given relation is a function.
2. Domain {4,5}
Range {1, 2}
Given relation is not a function.
Function Notation
y  f ( x)
The variable x is called the independent
variable, because it can be assigned any
of the permissible numbers from the domain.
The variable y is called the dependent
variable, because its value depends on x.
Function Notation
The special notation f ( x), read "f of x"
or "f at x," represents the value of the
function at the number x.
The notation f ( x) does not mean
"f times x."
Ex 2: Determine whether each
equation defines y as a function of x.
1. x  y  25
2. x  y  25
2
2
Solution
Solve each equation for y in terms of
x. If two or more values of y can be
obtained for a given x, the equation
is not a function.
1.
x  y  25
y  25  x
Solution continued
From this last equation we can see that
for each value of x, there is one and
only one value of y. Thus, the equation
defines y as a function of x.
Solution of 2 x  y  25
2
2
y  25  x
2
2
y   25  x
2
The  in this last equation shows that for
certain values of x (all values between  5 and
5), there are two values of y. For this reason
the equation does not define y as a function of x.
Practice Exercises
Determine whether each equation
difines y as a function of x.
1. x  y  25
2
2. 4 x  y
2
Answers
1. The equation defines a function.
2. The equation does not define a
function.
Ex 3: Evaluating a Function
If f ( x)  x  10 x  3, evaluate:
a. f (1) b. f ( x  2) c. f ( x)
2
Solution
We substitute  1, x  2, and  x
for x in the definition of f .
Solution part a.
We find f (1) by substituting  1 for
x in the equation f ( x)  x  10 x  3.
2
f ( 1)  ( 1 ) 10( 1 )  3
2
 1  10  3  8
Thus f (1)  8.
Solution part b.
We find f ( x  2) by substituting x  2 for
x in the equation f ( x)  x  10 x  3.
2
f
(x  2)  (x  2)
2
10(x  2)  3
f ( x)  x  4 x  4 10 x  20  3
2
f ( x)  x  6x 19
2
Solution part c.
We find f ( x) by substituting  x for
x in the equation f ( x)  x  10 x  3.
2
2

x

x
f( )( )
10( x )  3
f ( x)  x  10x  3
2
Practice Exercise
Evaluate the function h( x)  x  x  1
at the given values of the independent
variable and simplify.
a. h(2) b. h(1) c. h( x) d. h(3a )
3
Answer
a. 7
b. 1
3
c.  x  x  1
d. 27a  3a  1
3
Piecewise Functions
A function defined by two (or more)
equations over a specified domain
is called a piecewise function.
Ex 4: Evaluating a Piecewise
Function
Evaluate the piecewise function at the
given values of the independent variable.
 6 x  1 if x  0
f ( x)  
7 x  3 if x  0
a. f ( 3) b. f (0) c. f (4)
Solution part a
To find f (3), we let x  3.
Because  3 is less than 0, we use
the first line of the piecewise function.
f ( x)  6 x  1
This is the function's
equation for x  0.
f (3)  6(3)  1  19
Solution part b
To find f (0), we let x  0.
Because 0 is equal to 0, we use
the second line of the piecewise function.
f ( x)  7 x  3
This is the function's
equation for x  0.
f (0)  7(0)  3  3
Solution part c
To find f (4), we let x  4.
Because 4 is greater than 0, we use
the second line of the piecewise function.
f ( x)  7 x  3
This is the function's
equation for x  0.
f (4)  7(4)  3  31
Practice Exercise
Evaluate the piecewise function at the
given values of the independent variable.
if x  5
x  5
g ( x)  
 ( x  5) if x  5
a. g (0) b. g ( 6) c. g ( 5)
Answer
a. g (0)  5
b. g (6)  1
c. g (5)  0
Finding a Function’s Domain
If a function f does not model data or
verbal conditions, its domain is the
largest set of real numbers for which the
value of f ( x) is a real number.
Finding a Function’s Domain
Exclude from a function's domain real
numbers that cause division by zero
and real numbers that result in an even
root of a negative number.
Ex 5: Find the domain of each
function
a. f ( x)  8 x  5 x  2
2
b. g ( x) 
x5
c. h( x)  x  2
2
Solution part a
The function f ( x)  8 x  5 x  2 contains
neither division nor an even root.
The domain of f is the set of all real numbers.
2
Solution part b
2
The function g ( x) 
contains division.
x5
Because division by zero is undefined, we
must exclude from the domain values of x
that cause x  5 to be 0. Thus x cannot equal
to  5. The domain of function g is {x | x  5}.
Solution part c
The function h( x)  x  2 contains an even
root. Because only non-negative numbers
have real square roots, the quantity under the
radical sign, x  2 must be greater than or
equal to 0. Thus, x  2  0 or x  2
Therefore the domain of h is {x | x  2}
or the interval [2, ).
Practice Exercises
Find the domain of each function:
12 x
1. h( x)  2
x  36
1
2. f ( x) 
x2
Answers
1. {x | x  6, x  6}
2. {x | x  2} or (  2, )