Transcript Definition of a Relation A relation is any set of ordered pairs.

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.
Example:Analyzing U.S. Mobile-Phone Bills as a Relation
Find the domain and range of the relation
{(1994, 56.21), (1995, 51.00), (1996, 47.70), (1997, 42.78), (1998, 39.43)}
Solution The domain is the set of all first components. Thus, the domain is
{1994,1995,1996,1997,1998}.
The range is the set of all second components. Thus, the range is
{56.21, 51.00, 47.70, 42.78, 39.43}.
Definition of a Function
A function is a correspondence between two sets X and Y that
assigns to each element x of set X exactly one element y of set
Y. 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.
Example: Determining Whether a Relation is a
Function
Determine whether each relation is a function.
a. {(1, 6), (2, 6), (3, 8), (4, 9)}
b. {(6,1),(6,2),(8,3),(9,4)}
Solution We begin by making a figure for each relation that shows set X,
the domain, and set Y, the range, shown below.
(a)
(b)
1
2
3
4
6
8
9
Domain
Range
Figure (a) shows that every element in the domain corresponds
to exactly one element in the range. No two ordered pairs in the
given relation have the same first component and different
second components. Thus, the relation is a function.
6
8
9
1
2
3
4
Figure (b) shows that 6 corresponds to both 1 and 2. This
relation is not a function; two ordered pairs have the same first
component and different second components.
Domain
Range
When is a relation a function?
Determine whether each relation is a function.
• S = {(1,2), (3,4),(5,6),(7,8)}
Each first component is unique, therefore S is a
function
•
T = {(1,2), (3,4),(6,5),(1,5)}
Note that the first component in the first pair is the
same as the first component in the second pair,
therefore T is not a function.
Function Notation
When an equation represents a function, the function is often named by a
letter such as f, g, h, F, G, or H. Any letter can be used to name a function.
Suppose that f names a function. Think of the domain as the set of the
function's inputs and the range as the set of the function's outputs. The input is
represented by x and the output by f (x). The special notation f(x), read "f of x"
or "f at x," represents the value of the function at the number x.
If a function is named f and x represents the independent variable, the notation
f (x) corresponds to the y-value for a given x. Thus, f (x) = 4 - x2 and y = 4 - x2
define the same function. This function may be written as
y = f (x) = 4 - x2.
Example: Evaluating a Function
If f (x) = x2 + 3x + 5, evaluate:
a. f (2)
b. f (x + 3)
c. f (-x)
Solution We substitute 2, x + 3, and -x for x in the definition of f. When
replacing x with a variable or an algebraic expression, you might find it
helpful to think of the function's equation as
f (x) = x2 + 3x + 5.
a. We find f (2) by substituting 2 for x in the equation.
f (2) = 22 + 3 • 2 + 5 = 4 + 6 + 5 = 15
Thus, f (2) = 15.
more
Example: Evaluating a Function
If f (x) = x2 + 3x + 5, evaluate:
a. f (2)
b. f (x + 3)
c. f (-x)
Solution
b. We find f (x + 3) by substituting x + 3 for x in the equation.
f (x + 3) = (x + 3)2 + 3(x + 3) + 5
Equivalently,
f (x + 3) = (x + 3)2 + 3(x + 3) + 5
= x2 + 6x + 9 + 3x + 9 + 5
= x2 + 9x + 23.
Square x + 3 and
distribute 3 throughout
the parentheses.
more
Example: Evaluating a Function
If f (x) = x2 + 3x + 5, evaluate:
a. f (2)
b. f (x + 3)
c. f (-x)
Solution
c. We find f (-x) by substituting -x for x in the equation.
f (-x) = (-x)2 + 3(-x) + 5
Equivalently,
f (-x) = (-x)2 + 3(-x) + 5
= x2 –3x + 5.
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.
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.
Example: Finding the Domain of a Function
Normally it is safe to say the domain of a function is all real numbers.
However, there are 2 conditions which must be considered: 1)division by
zero and 2)even roots of negative numbers. Consider the following
functions and find the domain of each function:
a. f ( x )  x 2  7 x
b. g ( x ) 
6x
x2  9
c. h ( x ) 
3 x  12
Solution
a. The function f (x) = x2 – 7x contains neither division nor an even root.
The domain of f is the set of all real numbers.
6x
b. The function g ( x )  2
contains division. Because division by 0 is
x 9
undefined, we must exclude from the domain values of x that cause
x2 – 9 to be 0. Thus, x cannot equal –3 or 3. The domain of function g is
{x | x =/ -3, x =/ 3}.
more
Example: Finding the Domain of a Function
Continuing…
a. f ( x )  x 2  7 x
b. g ( x ) 
6x
x2  9
c. h ( x ) 
3 x  12
Solution
c. The function h ( x )  3 x  12 contains an even root. Because only
nonnegative numbers have real square roots, the quantity under the
radical sign, 3x + 12, must be greater than or equal to 0.
3x + 12 > 0
3x > -12
x > -4
The domain of h is { x | x > -4} or the interval [-4, oo).
Problems
Evaluate each function for the given values.
1.
F(x) = 3x + 7
a.
2.
F(4)
F(x) =
a.
f(16)
b. F(x+1)
6 
c. F(-x)
25  r
b. f(-24)
c.f(25-2x)
Exercises page 160, numbers 1-7 odd, 21 – 31 odd, 51 – 71 odd.
Definition of the Difference Quotient
The expression
f ( x  h)  f ( x)
,h  0
h
is called the difference quotient.
Example: Evaluating the Difference Quotient
For
a.
f ( x)  x  3x  5
2
f ( x  h)
, find and simplify:
b.
f ( x  h)  f ( x)
,h  0
h
Repeat for
f ( x)  x  7 x  3
2
.
Exercises pg 160, numbers 33-41 odd.
Piecewise-Defined Functions
Functions that are made up of different pieces
based upon different domains are called
piecewise-defined functions.
Example: Piecewise-defined function
If
x2  3
f ( x)  
5 x  3
a.
f ( 5)
if x  0
if x  0
b. f ( 6 )
, find :