Copyright © 2010 Pearson Education, Inc. All rights reserved Sec 12.1 - 1.


Nonlinear Functions, Conic Sections,

and Nonlinear Systems

Chapter 12


12.1

Additional Graphs of Functions;

Composition

Copyright © 2010 Pearson Education, Inc. All rights reserved. Sec 12.1 - 4

12.1 Additional Graphs of Functions; Composition

Objectives

1. Recognize the graphs of the elementary functions defined

by | x |, , and , and graph their translations.

2. Recognize and graph step functions.

3. Find the composition of functions.

x1x


x

y

(0, 0)


The Absolute Value Function

f ( x ) = | x |

x y

0 0

1 1+

2 2+

3 3+

The domain of the absolute value function is (– ∞, ∞) and its range is [0, ∞).


x

y


The Reciprocal Function

f ( x ) = 1x

x y

1 1 3

2 1 2

11

2 1 2

3 1 3

x y

1 1 3

2 1 2

11

2 1 2

3 1 3

– –

– –

– –

– –

– –

For the reciprocal function, the domain and the range are both (– ∞, 0) U (0, ∞).

Notice the vertical asymptote at x = 0 and the

horizontal asymptote at y = 0 .


x

y

(0, 0)


The Square Root Function

f ( x ) = x

x y

00

11

24

The domain of the square root function is [0, ∞). Because the principal

square root is always nonnegative, the range is

also [0, ∞).


x

y

Graph f ( x ) = | x – 3 |.

EXAMPLE 1 Applying a Horizontal Shift


(3, 0)

f ( x ) = | x – 3 |

x y

30

21

12

03

14

This graph is found by shifting f ( x ) = | x | three units to the right.

The domain of this function is (– ∞, ∞) and its range is [0, ∞).


x

y

Graph f ( x ) = – 4.

EXAMPLE 2 Applying a Vertical Shift


This graph of this function is found by shifting

f ( x ) = four units down.

f ( x ) = – 41x

1x

1x

x y

–1 1 3

–2 1 2

–31

2 –3.5

x y

7 1 3

6 1 2

51

2

– –

– –

– –

– – 4.5


x

y

Graph f ( x ) = – 4.

EXAMPLE 2 Applying a Vertical Shift


This graph of this function is found by shifting

f ( x ) = four units down.

f ( x ) = – 41x

1x

1x

The domain is (– ∞, 0) U (0, ∞) and the

range is (– ∞, –4) U (–4, ∞).

Notice the vertical asymptote at x = 0

and the horizontal asymptote at y = –4.


EXAMPLE 3 Applying Both Horizontal and Vertical Shifts


Graph f ( x ) = – 3.x + 2

x

y

x y

–3–2

–12

07

This graph is found by shifting f ( x ) = two units to the left and

three units down.

The domain of this function is [–2, ∞) and its range is [–3, ∞).

x

f ( x ) = – 3x + 2



Greatest Integer Function

The greatest integer function, usually written f (x) = x , is defined as

follows:

x denotes the largest integer that is less than or equal to x.

For example,

9 = 9, –3.8 = –4, 5.7 = 5.


EXAMPLE 4 Graphing the Greatest Integer Function


x

y

Graph f ( x ) = x .

For x ,

if –1 ≤ x < 0, then x = –1;

if 0 ≤ x < 1, then x = 0;

if 1 ≤ x < 2, then x = 1,

and so on.

The appearance of the graph is the

reason that this function is called a

step function.

f ( x ) = x


EXAMPLE 5 Applying a Greatest Integer Function


An overnight delivery service charges $20 for a package weighing up to 4lb.

For each additional pound or fraction of a pound there is an additional charge

of $2. Let D(x) represent the cost to

send a package weighing x pounds.

Graph D(x) for x in the interval (0, 7].

For x in the interval (0, 4], y = 20.



and so on.

y

x

Do

llar

s10

20

30

1 2 3 4 5 6 7

Pounds

For x in the interval (6, 7], y = 26,



Composition of Functions

Composition of Functions

If f and g are functions, then the composite function, or composition,

of g and f is defined by

( g ◦ f )( x ) = g ( f ( x ) )

for all x in the domain of f such that f (x) is in the domain of g.


EXAMPLE 6 Evaluating a Composite Function


Let f (x) = 3x2 + 5 and g(x) = x – 7. Find ( f ◦ g )( 2 ).

( f ◦ g )( 2 ) = f ( g ( 2 ) ) Definition

= f ( 2 – 7 ) Use the rule for g( x ); g( 2 ) = 2 – 7.

= f ( –5 ) Subtract.

= 3( –5 )2 + 5 Use the rule for f ( x ); f ( –5 ) = 3( –5 )2 + 5.

= 80




Let f (x) = 3x2 + 5 and g(x) = x – 7. Now find ( g ◦ f )( 2 ).

( g ◦ f )( 2 ) = g ( f ( 2 ) ) Definition

= g ( 3( 2 )2 + 5 ) Use the rule for f ( x ); f ( 2 ) = 3( 2 )2 + 5.

= g ( 17 ) Square, multiply, and then add.

= 17 – 7 Use the rule for g( x ); g(17) = 17 – 7.

= 10




Let f (x) = 3x2 + 5 and g(x) = x – 7. Notice that ( f ◦ g )( 2 ) ≠ ( g ◦ f )( 2 ).

( g ◦ f )( 2 ) = g [ f ( 2 ) ]

= g ( 3( 2 )2 + 5 )

= g ( 17 )

= 17 – 7

= 10

( f ◦ g )( 2 ) = f [ g ( 2 ) ]

= f ( 2 – 7 )

= f ( –5 )

= 3( –5 )2 + 5

= 80

In general, ( f ◦ g )( 2 ) ≠ ( g ◦ f )( 2 ).


EXAMPLE 7 Finding Composite Functions


Let f (x) = 5x + 1 and g(x) = x2 – 4. Find each of the following.

( f ◦ g )( –3 ) = f [ g ( –3 ) ]

= f ( (–3)2 – 4 ) g(x) = x2 – 4

= f ( 5 )

= 5( 5 ) + 1 f (x) = 5x + 1

= 26

(a) ( f ◦ g )( –3 )


= 5(n2 – 4) + 1 f (x) = 5x + 1

= 5n2 – 19




( f ◦ g )( n ) = f [ g ( n ) ]

= f ( n2 – 4 ) g(x) = x2 – 4

(b) ( f ◦ g )( n )


= (5n + 1)2 – 4 g(x) = x2 – 4

= 25n2 + 10n + 1 – 4




( g ◦ f )( n ) = g [ f ( n ) ]

= g ( 5n + 1 ) f (x) = 5x + 1

(c) ( g ◦ f )( n )

= 25n2 + 10n – 3

Copyright © 2010 Pearson Education, Inc. All rights reserved Sec 12.1 - 1.

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