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

2. Copyright © 2010 Pearson Education, Inc. All rights reserved Sec 12.1 - 2 Nonlinear Functions, Conic Sections, and Nonlinear Systems Chapter 12

3. Copyright © 2010 Pearson Education, Inc. All rights reserved Sec 12.1 - 3 12.1 Additional Graphs of Functions; Composition

4. 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. x 1 x

5. Copyright © 2010 Pearson Education, Inc. All rights reserved. Sec 12.1 - 5 x y (0, 0) 12.1 Additional Graphs of Functions; Composition The Absolute Value Function f ( x ) = | x | x y 0 1 1 + 2 2 + 3 3 + The domain of the absolute value function is (– ∞, ∞ ) and its range is [0, ∞ ).

6. Copyright © 2010 Pearson Education, Inc. All rights reserved. Sec 12.1 - 6 x y 12.1 Additional Graphs of Functions; Composition The Reciprocal Function f ( x ) = 1 x x y 1 1 3 2 1 2 1 1 2 1 2 3 1 3 1 1 3 2 1 2 1 1 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.

7. Copyright © 2010 Pearson Education, Inc. All rights reserved. Sec 12.1 - 7 x y (0, 0) 12.1 Additional Graphs of Functions; Composition 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, ∞ ).

8. Copyright © 2010 Pearson Education, Inc. All rights reserved. Sec 12.1 - 8 x y Graph f ( x ) = | x – 3 |. EXAMPLE 1 Applying a Horizontal Shift 12.1 Additional Graphs of Functions; Composition (3, 0) f ( x ) = | x – 3 | x y 3 0 2 1 1 2 0 3 1 4 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, ∞ ).

9. Copyright © 2010 Pearson Education, Inc. All rights reserved. Sec 12.1 - 9 x y Graph f ( x ) = – 4. EXAMPLE 2 Applying a Vertical Shift 12.1 Additional Graphs of Functions; Composition This graph of this function is found by shifting f ( x ) = four units down. f ( x ) = – 4 1 x 1 x 1 x x y –1 1 3 –2 1 2 –3 1 2 –3.5 x y 7 1 3 6 1 2 5 1 2 –– – – –– –– 4.5

10. Copyright © 2010 Pearson Education, Inc. All rights reserved. Sec 12.1 - 10 x y Graph f ( x ) = – 4. EXAMPLE 2 Applying a Vertical Shift 12.1 Additional Graphs of Functions; Composition This graph of this function is found by shifting f ( x ) = four units down. f ( x ) = – 4 1 x 1 x 1 x 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.

11. Copyright © 2010 Pearson Education, Inc. All rights reserved. Sec 12.1 - 11 EXAMPLE 3 Applying Both Horizontal and Vertical Shifts 12.1 Additional Graphs of Functions; Composition 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 ) = – 3 x + 2

12. Copyright © 2010 Pearson Education, Inc. All rights reserved. Sec 12.1 - 12 12.1 Additional Graphs of Functions; Composition 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.

13. Copyright © 2010 Pearson Education, Inc. All rights reserved. Sec 12.1 - 13 EXAMPLE 4 Graphing the Greatest Integer Function 12.1 Additional Graphs of Functions; Composition 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

14. Copyright © 2010 Pearson Education, Inc. All rights reserved. Sec 12.1 - 14 EXAMPLE 5 Applying a Greatest Integer Function 12.1 Additional Graphs of Functions; Composition 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. For x in the interval (4, 5], y = 22. For x in the interval (5, 6], y = 24. and so on. y x Dollars 10 20 30 1234567 Pounds For x in the interval (6, 7], y = 26,

15. Copyright © 2010 Pearson Education, Inc. All rights reserved. Sec 12.1 - 15 12.1 Additional Graphs of Functions; Composition 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.

16. Copyright © 2010 Pearson Education, Inc. All rights reserved. Sec 12.1 - 16 EXAMPLE 6 Evaluating a Composite Function 12.1 Additional Graphs of Functions; Composition Let f ( x ) = 3 x 2 + 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 + 5Use the rule for f ( x ); f ( –5 ) = 3( –5 ) 2 + 5. = 80

17. Copyright © 2010 Pearson Education, Inc. All rights reserved. Sec 12.1 - 17 EXAMPLE 6 Evaluating a Composite Function 12.1 Additional Graphs of Functions; Composition Let f ( x ) = 3 x 2 + 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 – 7Use the rule for g ( x ); g (17) = 17 – 7. = 10

18. Copyright © 2010 Pearson Education, Inc. All rights reserved. Sec 12.1 - 18 EXAMPLE 6 Evaluating a Composite Function 12.1 Additional Graphs of Functions; Composition Let f ( x ) = 3 x 2 + 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 ).

19. Copyright © 2010 Pearson Education, Inc. All rights reserved. Sec 12.1 - 19 EXAMPLE 7 Finding Composite Functions 12.1 Additional Graphs of Functions; Composition Let f ( x ) = 5 x + 1 and g ( x ) = x 2 – 4. Find each of the following. ( 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 (a) ( f ◦ g )( –3 )

20. Copyright © 2010 Pearson Education, Inc. All rights reserved. Sec 12.1 - 20 = 5( n 2 – 4) + 1 f ( x ) = 5 x + 1 = 5 n 2 – 19 EXAMPLE 7 Finding Composite Functions 12.1 Additional Graphs of Functions; Composition Let f ( x ) = 5 x + 1 and g ( x ) = x 2 – 4. Find each of the following. ( f ◦ g )( n ) = f [ g ( n ) ] = f ( n 2 – 4 ) g ( x ) = x 2 – 4 (b) ( f ◦ g )( n )

21. Copyright © 2010 Pearson Education, Inc. All rights reserved. Sec 12.1 - 21 = (5 n + 1) 2 – 4 g ( x ) = x 2 – 4 = 25 n 2 + 10 n + 1 – 4 EXAMPLE 7 Finding Composite Functions 12.1 Additional Graphs of Functions; Composition Let f ( x ) = 5 x + 1 and g ( x ) = x 2 – 4. Find each of the following. ( g ◦ f )( n ) = g [ f ( n ) ] = g ( 5 n + 1 ) f ( x ) = 5 x + 1 (c) ( g ◦ f )( n ) = 25 n 2 + 10 n – 3