Date:1.2 Functions And Their Properties

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.

The domain is the set of all first components (the x-values). domain:

Find the domain and range of the relation{(1994,56.21), (1995,51.00), (1996,47.70), (1997,42.78), (1998,39.43)}

{1994,1995,1996,1997,1998}

The range is the set of all second components (the y-values). range: {56.21, 51.00, 47.70, 42.78, 39.43}

Definition of a Function: A function is a relation that assigns

to each domain exactly one range.

Solution We begin by making a figure for each relation that shows the domain and the range.

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)}

1234

689

Domain Range

(a) 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 x and different y’s. For every x, there is only one y. Thus, the relation is a function.

689

1234

Domain Range

(b) Figure (b) shows that 6 corresponds to both 1 and 2 (6 has two y’s). This relation is not a function; two ordered pairs have the same x and different y’s.

Determine whether each equation is a function.a. y = 2x + 3 b. x = y2

Consider the values of x and y. For every x is there onlyone y value.

x y x y

-10

1

-10

1

13

5

10

1

Function

x has two y values

Not a Function

x

yc.

y is a function of x for the graphs in (b) and (c).

y is not a function since 2 values of y correspond to an x-value.

y is a function of x y is a function of x

x

yd.

y is not a function since 2 values of y correspond to an x-value.

The Vertical Line Test for Functions•If any vertical line intersects a graph in more than one point, the graph does not define y as a function of x.Use the vertical line test to identify graphs in which y is a function of x.

x

y a.

x

y b.

Finding a Function’s Domain: It’s 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.

f(x) x3 2xx2 1

Find the domain of the function:

x(x2 2)

(x 1)(x 1)

domain :{x : x 1, 1}or , 1 , 1,1 , 1,

1 and -1 need to be excluded from the domain because they cause division by zero.

32)( xxgFind the domain of the function:

2x 30

Numbers that result in an even root of a negative number must be excluded. 2x + 3 must be positive.

domain : x : x 32

or 32

,

2x 3 - 3 - 3

x 32

2 2

For these values an even root will be positive.

Confirm this graphically.Find the range graphically. range : y : y 0

or 0,

h(x) 2x

Find the domain and range of the function:

2x0

x = 0 is not in the domain because a denominator cannot be zero.

domain : x : x 0 or ,0 , 0,

y = 0 is not in the range

Find the domain and range graphically.range : y : y 0 or ,0 , 0,

Confirm the domain and range algebraically.

y = 0 is not in the range by trying to solve when y = 0

2 0

h(x) 20

Solution We begin by setting up a partial table of coordinates.

Graph f (x) = x2 + 1. Use integer values of x from the set {-3, -2, -1, 0, 1, 2, 3}. Use the graph to specify the function's domain and range.

Plot the points.

(-1, 2)f (-1) = (-1)2 + 1 = 2-1

(0, 1)f (0) = 02 + 1 = 10

(-2, 5)f (-2) = (-2)2 + 1 = 5-2

(-3, 10)f (-3) = (-3)2 + 1 = 10-3

(x, y) or (x, f (x))f (x) = x2 + 1x

(3, 10)f (3) = (3)2 + 1 = 103

(2, 5)f (2) = (2)2 + 1 = 52

(1, 2)f (1) = 12 + 1 = 21

Domain: all Reals

10987654321

1 2 3 4-4 -3 -2 -1

Ran

ge: [

1, o

o)

The points on the graph of f have x-coordinates that extend indefinitely far to the left and to the right. Thus, the domain consists of all real numbers, represented by (-∞, ∞).

The points on the graph have y-coordinates that start at 1 and extend indefinitely upward. Thus, the range consists of all real numbers greater than or equal to 1, represented by [1, ∞).

Complete Student CheckpointDetermine whether each relation is a function:a. {(1,2),(3,4),(5,6),(5,8)}

b. {(1,2),(3,4),(6,5),(8,5)}

Solve each equation for y and then determine whether the equation defines y as a function of x. a. 2x y 6 b. x2 y2 1

Not a function

Function

2x 2x y 2x 6

x y -101

864

Function

x2 x2

y2 x2 1

y x2 1x y -100

01

-1Not a function

Smooth, Continuous Graphs

Two important features of the graphs of polynomial functions are that they are smooth and continuous. By smooth, we mean that the graph contains only roundedcurves with no sharp corners. Bycontinuous, we mean that the graph has no breaks and can be drawn without lifting your pencil from the rectangular coordinate system. These ideas are illustrated in the figure.

Smooth rounded corner

Smooth rounded corner

Smooth rounded corner

Smooth rounded corner

x x

y y

This graph has removable discontinuity at x = 1.

It is removable because we could define f(1) in such a way as to plug the hole and make f(x) continuous as 1.

These graphs have discontinuity. They are not continuous everywhere.

-5 -4 -3 -2 -1 1 2 3 4 5

543

1

-1-2-3-4-5

-5 -4 -3 -2 -1 1 2 3 4 5

54321

-1-2-3-4-5

a. b.

Discontinuities

This graph has jump discontinuity.

This discontinuity is not removable because there is more than just a hole at x = 0; the gap makes it impossible to plug.

-5 -4 -3 -2 -1 1 2 3 4 5

54321

-1-2-3-4-5

This graph has discontinuity. It is not continuous everywhere.Discontinuities

This graph has infinite discontinuity.

This discontinuity is definitely not removable at x = -2.

Definitions of Local Maximum

and Local MinimumA function value f(a) is a local maximum of f if there exists an open interval about a such that f(a) > f(x) for all x in the open interval.

-5 -4 -3 -2 -1 1 2 3 4 5

54321

-1-2-3-4-5

A function value f(b) is a local minimum of f if there exists an open interval about b such that f(b) < f(x) for all x in the open interval.

It occurs at (1, 4), the maximum value is 4

It occurs at (-3, -4), the minimum value is -4

Local is also referred to as Relative or Absolute .

Increasing, Decreasing, and Constant Functions

Constantf (x1) = f (x2)

(x1, f (x1))

(x2, f (x2))

Increasingf (x1) < f (x2)

(x1, f (x1))

(x2, f (x2))

Decreasingf (x1) > f (x2)

(x1, f (x1))

(x2, f (x2))

A function is increasing on an interval if for any x1, and x2 in the interval, where x1 < x2, then f (x1) < f (x2).A function is decreasing on an interval if for any x1, and x2 in the interval, where x1 < x2, then f (x1) > f (x2). A function is constant on an interval if for any x1, and x2 in the interval, where x1 < x2, then f (x1) = f (x2).

decreasing on the interval (-∞, 0)increasing on the interval (0, 2) decreasing on the interval (2, ∞)

Describe the increasing, decreasing, or constant behavior of each function whose graph is shown.

-5 -4 -3 -2 -1 1 2 3 4 5

543

1

-1-2-3-4-5

-5 -4 -3 -2 -1 1 2 3 4 5

54321

-1-2-3-4-5

a. b.

Example

The function is defined in two pieces.

constant on the interval (-∞, 0) increasing on the interval [0, ∞)

Lower Bound, Upper Bound, and BoundedThe function is bounded below if

there is some number b that is less than or equal to every number in the range of f. Any such number b is called a lower bound of f.

x

y

The function is bounded above if there is some number b that is less than or equal to every number in the range of f. Any such number b is called a upper bound of f.

bounded below

x

y

not bounded above bounded abovenot bounded below

Lower Bound, Upper Bound, and BoundedThe function is bounded if it is

bounded both above and below.The function is:

bounded not bounded abovenot bounded below

x

y

x

y

Definition of Even and Odd Functions

The function f is an even function if f (-x) = f (x) for all x in the domain of f.

The right side of the equation of an even function does not change if x is replaced with -x.

The function f is an odd function if f (-x) = -f (x) for all x in the domain of f.

Every term in the right side of the equation of an odd function changes sign if x is replaced by -x.

Identify the following function as even, odd, or neither: f(x) = 3x2 - 2.

Use the given function equation to find f(-x). f(-x) = 3(-x) 2-2 = 3x2 - 2The right side of the equation of the given function did not change when we replaced x with -x. Algebraically:

Because f(-x) = f(x) f is an even function.

x

yEven Functions and y-Axis Symmetry

The graph of an even function in which f (-x) = f (x)

is symmetric with respect to the y-axis.

Odd Functions and Origin Symmetry

The graph of an odd function in which f (-x) = - f (x)

is symmetric with respect to the origin.

x

y

Identify the following function as even, odd, or neither: f(x) = 3x2 - 2.

Identify graphically.Because f(x) looks the same on

either side of the y-axis f is an even function.

Determine algebraically whether each of the following functions is even, odd, or neither.

a. f (x) x2 6 b. g(x) 7x3 x

c. h(x) x5 1

f ( x) ( x)2 6 x2 6

doesnot

change

EVEN

g( x) 7( x)3 ( x) 7x3 x

all termschangesigns

ODD

h( x) ( x)5 1 x5 1

all termsdo notchangesigns

NEITHER

Arrow NotationSymbol Meaningx a x approaches a from the right.x a x approaches a from the left.x x approaches infinity; that is, x

increases without bound.x x approaches negative infinity;

that is, x decreases without bound.

The line x a is a vertical asymptote of the graph of a function f if f (x) increases or decreases without bound as x approaches a. f (x) as x a f (x) as x a

f

a

y

x

x = a

fa

y

xx = a

Definition of a Vertical Asymptote

limx a

f (x) or limx a

f (x)

The line x a is a vertical asymptote of the graph of a function f if f (x) increases or decreases without bound as x approaches a.

x = a

f a

y

x

fa

y

x

x = a

f (x) as x a f (x) as x a

Definition of a Vertical Asymptote

limx a

f (x) or limx a

f (x)

Locating Vertical AsymptotesIf f(x) = p(x) / q(x) is a rational function in

which p(x) and q(x) have no common factors and a is a zero of q(x) , the denominator, then x = a is a vertical asymptote of the graph of f.

5 is a zero of x - 5, therefore x = 5 is a vertical asymptote of f

Find the vertical asymptote of this function:

f (x) x 7x 5

Find the vertical asymptotes, if any, of the graph of each rational function.a. b.

x 1 and x 1

g(x)

x 1x2 1

h(x) x 1x2 1

g(x)

x 1x 1 x 1 no vertical asymptotes

The line y = b is a horizontal asymptote of the graph of a function f if f (x) approaches b as x increases or decreases without bound.

f

y

x

y = b

x

y

fy = b

f

y

x

y = b

f(x)b as x f(x)b as x f(x)b as x

Definition of a Horizontal Asymptote

limx

f (x) b

Locating Horizontal AsymptotesLet f be the rational function given by

f (x) an xn an 1x

n 1 ... a1x a0

bm xm bm 1xm 1 ... b1x b0

, an 0,bm 0

The degree of the numerator is n. The degree of the denominator is m.

1. If n<m, the x-axis, or y=0, is the horizontal asymptote of the graph of f.

2. If n=m, the line y = an/bm is the horizontal asymptote of the graph of f.

3. If n>m, the graph of f has no horizontal asymptote.

Find the horizontal asymptotes, if any, of the graph of each rational function.

a. b.

c.

h(x)

9x3

3x2 1 g(x)

9x3x2 1

If n<m, the x-axis, or y = 0

If n>m, no horizontal asymptote

g(x)

9x3x 1

If n=m, then y = a/b, or y = 3

Strategy for Graphing a Rational FunctionSuppose that f(x) = p(x) / q(x) , where p(x) and q(x) are polynomial functions

with no common factors.1. Determine whether the graph of f has symmetry.

f (x) f (x): y-axis symmetry f (x) f (x): origin symmetry

2. Find the y-intercept (if there is one) by evaluating f (0).3. Find the x-intercepts (if there are any) by solving the equation p(x) 0.4. Find any vertical asymptote(s) by solving the equation q (x) 0.5. Find the horizontal asymptote (if there is one) using the rule for determining

the horizontal asymptote of a rational function.6. Plot at least one point between and beyond each x-intercept and vertical

asymptote.7. Use the information obtained previously to graph the function between

and beyond the vertical asymptotes.

• the graph has no symmetry• The y-intercept is (0,-3/10)• The x-intercept is (3/2, 0)• The vertical asymptote is x = -2• The horizontal asymptote is y = 2/5• Test points include (-5, 13/15), (-4, 11/10),

(-3, 9/5), (-7/2, 16/5), (-1, -1), (2, 1/20)

10532)(

xxxf

-5 -4 -3 -2 -1 1 2 3 4 5

54321

-1-2-3-4-5

Sketch the graph of

Identify the intervals on which the function is increasing, decreasing, and constant.x y

-5-4-3 -3-2-1 0 1 2 2 3

f (x) x 3 x 2 5 if x 32x 1 if -3 x 25 if x 2

need to find out value it begins or ends but is not included.

-5-5-3 -1 13555

Constant: , 3 and 2, Increasing: 3,2 3,2

graphing calculator interpretation

Suppose f is an odd function. For each statement below, explain why the statement is possible, necessarily true, or impossible.a. f(3) = 4 and f(-4) = -3

b. f(3) = 4 and f(-3) = -4

c. f(0) = -2 and f(-0) = 2

PossibleFor odd functions f (-x) = - f (x), so if (-3,-4) and (4,3) are in the function this statement is possible.

ImpossibleThe graph of an odd function in which f (-x) = - f (x) is symmetric with respect to the origin. f(0) = 0 for this to be an odd function

Necessarily trueFor odd functions f (-x) = - f (x), so f is an odd function in this statement.

Matching functions using end behaviorMatch the functions to their graphs on page 97.

a. y 3x

x2 1b. y

3x2

x2 1c. y

3x3

x2 1d. y

3x4

x2 1When x is very large, the denominator x2+1 is almost the same as x2, replace this to make simpler functions and compare end behaviors to match with graphs.a. y

3xx2

a. y 3x

a. matches graph (iv)

ending behavior, it approaches

y = 0

b. y 3x2

x2

b. y 3

b. matches graph (iii)

end behavior, it approaches

y = 3

c. y 3x3

x2

c. y 3x

c. matches graph (ii)

end behavior, it approaches

y = 3x

d. y 3x4

x2

d. y 3x2

d. matches graph (i)

end behavior, it approaches

y = 3x2

Functions and Their Properties