Copyright 2010 Pearson Education Inc All rights reserved

Copyright © 2010 Pearson Education, Inc. All rights reserved Sec 12. 1 - 1

Chapter 12 Nonlinear Functions, Conic Sections, and Nonlinear Systems Copyright © 2010 Pearson Education, Inc. All rights reserved Sec 12. 1 - 2

12. 1 Additional Graphs of Functions; Composition Copyright © 2010 Pearson Education, Inc. All rights reserved Sec 12. 1 - 3

12. 1 Additional Graphs of Functions; Composition Objectives 1. Recognize the graphs of the elementary functions defined x , and graph their translations. by | x |, 1 x , and 2. Recognize and graph step functions. 3. Find the composition of functions. Copyright © 2010 Pearson Education, Inc. All rights reserved. Sec 12. 1 - 4

12. 1 Additional Graphs of Functions; Composition The Absolute Value Function The domain of the absolute value function is (– ∞, ∞) and its range is [0, ∞). y x y 0 0 +1 1 +2 2 +3 3 Copyright © 2010 Pearson Education, Inc. All rights reserved. x (0, 0) f(x) = |x| Sec 12. 1 - 5

12. 1 Additional Graphs of Functions; Composition The Reciprocal Function For the reciprocal function, the domain and the range are both (– ∞, 0) U (0, ∞). Notice the vertical asymptote at x = 0 and the y horizontal asymptote at y = 0. x y 1 3 1 – 1 1 2 2 – 1 – 2 1 1 – 1 2 – 2 – 1 3 – 3 – 1 3 2 2 3 Copyright © 2010 Pearson Education, Inc. All rights reserved. x f(x) = 1 x Sec 12. 1 - 6

12. 1 Additional Graphs of Functions; Composition The Square Root Function The domain of the square root function is [0, ∞). Because the principal square root is always nonnegative, the range is y also [0, ∞). x y 0 0 1 1 4 2 (0, 0) f(x) = Copyright © 2010 Pearson Education, Inc. All rights reserved. x x Sec 12. 1 - 7

12. 1 Additional Graphs of Functions; Composition EXAMPLE 1 Applying a Horizontal Shift Graph f ( x ) = | x – 3 |. 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 0 3 1 2 2 1 y x 3 0 (3, 0) 4 1 f(x) = |x– 3| Copyright © 2010 Pearson Education, Inc. All rights reserved. Sec 12. 1 - 8

12. 1 Additional Graphs of Functions; Composition EXAMPLE 2 Applying a Vertical Shift Graph f ( x ) = 1 x – 4. This graph of this function is found by shifting 1 y f ( x ) = x four units down. x y 1 3 – 1 – 7 1 2 – 1 – 6 1 – 3 – 1 – 5 2 – 3. 5 – 2 – 4. 5 3 2 Copyright © 2010 Pearson Education, Inc. All rights reserved. x f(x) = 1 – 4 x Sec 12. 1 - 9

12. 1 Additional Graphs of Functions; Composition EXAMPLE 2 Applying a Vertical Shift Graph f ( x ) = 1 x – 4. This graph of this function is found by shifting 1 y f ( x ) = x four units down. 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. x f(x) = 1 – 4 x Copyright © 2010 Pearson Education, Inc. All rights reserved. Sec 12. 1 - 10

12. 1 Additional Graphs of Functions; Composition EXAMPLE 3 Graph f ( x ) = Applying Both Horizontal and Vertical Shifts x + 2 – 3. This graph is found by shifting f ( x ) = three units down. x two units to the left and The domain of this function is [– 2, ∞) and its range is [– 3, ∞). y x y – 2 – 3 2 – 1 x f(x) = 7 x+2 – 3 0 Copyright © 2010 Pearson Education, Inc. All rights reserved. Sec 12. 1 - 11

12. 1 Additional Graphs of Functions; Composition Greatest Integer Function The greatest integer function, usually written f (x) = follows: x , is defined as x denotes the largest integer that is less than or equal to x. For example, 9 = 9, – 3. 8 = – 4, Copyright © 2010 Pearson Education, Inc. All rights reserved. 5. 7 = 5. Sec 12. 1 - 12

12. 1 Additional Graphs of Functions; Composition EXAMPLE 4 Graph f ( x ) = For Graphing the Greatest Integer Function x. x , y if – 1 ≤ x < 0, then x = – 1; if 0 ≤ x < 1, then x = 0; if 1 ≤ x < 2, then x = 1, and so on. f(x) = x x The appearance of the graph is the reason that this function is called a step function. Copyright © 2010 Pearson Education, Inc. All rights reserved. Sec 12. 1 - 13

12. 1 Additional Graphs of Functions; Composition EXAMPLE 5 Applying a Greatest Integer Function For x in the interval (0, 4], y = 20. For x in the interval (4, 5], y = 22. Dollars An overnight delivery service charges $20 for a package weighing up to 4 lb. For each additional pound or fraction of a pound there is an additional charge of $2. Let D(x) represent the cost to y send a package weighing x pounds. 30 Graph D(x) for x in the interval (0, 7]. For x in the interval (5, 6], y = 24. 20 10 For x in the interval (6, 7], y = 26, and so on. 1 2 3 4 5 6 7 x Pounds Copyright © 2010 Pearson Education, Inc. All rights reserved. Sec 12. 1 - 14

12. 1 Additional Graphs 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. Copyright © 2010 Pearson Education, Inc. All rights reserved. Sec 12. 1 - 15

12. 1 Additional Graphs of Functions; Composition EXAMPLE 6 Evaluating a Composite Function Let f (x) = 3 x 2 + 5 and g(x) = x – 7. Find ( 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 Copyright © 2010 Pearson Education, Inc. All rights reserved. Sec 12. 1 - 16

12. 1 Additional Graphs of Functions; Composition EXAMPLE 6 Evaluating a Composite Function Let f (x) = 3 x 2 + 5 and g(x) = x – 7. Now find ( 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 Copyright © 2010 Pearson Education, Inc. All rights reserved. Sec 12. 1 - 17

12. 1 Additional Graphs of Functions; Composition EXAMPLE 6 Evaluating a Composite Function Let f (x) = 3 x 2 + 5 and g(x) = x – 7. Notice that ( f ◦ g )( 2 ) ≠ ( g ◦ f )( 2 ). ( f ◦ g )( 2 ) = f [ g ( 2 ) ] ( g ◦ f )( 2 ) = g [ f ( 2 ) ] = f (2– 7) = g ( 3( 2 )2 + 5 ) = f ( – 5 ) = g ( 17 ) = 3( – 5 )2 + 5 = 17 – 7 = 80 = 10 In general, ( f ◦ g )( 2 ) ≠ ( g ◦ f )( 2 ). Copyright © 2010 Pearson Education, Inc. All rights reserved. Sec 12. 1 - 18

12. 1 Additional Graphs of Functions; Composition EXAMPLE 7 Finding Composite Functions Let f (x) = 5 x + 1 and g(x) = x 2 – 4. Find each of the following. (a) ( f ◦ g )( – 3 ) = f [ g ( – 3 ) ] = f ( (– 3)2 – 4 ) g(x) = x 2 – 4 = f(5) = 5( 5 ) + 1 f (x) = 5 x + 1 = 26 Copyright © 2010 Pearson Education, Inc. All rights reserved. Sec 12. 1 - 19

12. 1 Additional Graphs of Functions; Composition EXAMPLE 7 Finding Composite Functions Let f (x) = 5 x + 1 and g(x) = x 2 – 4. Find each of the following. (b) ( f ◦ g )( n ) = f [ g ( n ) ] = f ( n 2 – 4 ) g(x) = x 2 – 4 = 5(n 2 – 4) + 1 f (x) = 5 x + 1 = 5 n 2 – 19 Copyright © 2010 Pearson Education, Inc. All rights reserved. Sec 12. 1 - 20

12. 1 Additional Graphs of Functions; Composition EXAMPLE 7 Finding Composite Functions Let f (x) = 5 x + 1 and g(x) = x 2 – 4. Find each of the following. (c) ( g ◦ f )( n ) = g [ f ( n ) ] = g ( 5 n + 1 ) f (x) = 5 x + 1 = (5 n + 1)2 – 4 g(x) = x 2 – 4 = 25 n 2 + 10 n + 1 – 4 = 25 n 2 + 10 n – 3 Copyright © 2010 Pearson Education, Inc. All rights reserved. Sec 12. 1 - 21