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6-5

Operations with Functions. 6-5. Warm Up. Lesson Presentation. Lesson Quiz. Holt McDougal Algebra 2. Holt Algebra2. x – 3. x – 2. Warm Up Simplify. Assume that all expressions are defined. 1. (2 x + 5) – ( x 2 + 3 x – 2). – x 2 – x + 7. 2. ( x – 3)( x + 1) 2.

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6-5

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  1. Operations with Functions 6-5 Warm Up Lesson Presentation Lesson Quiz Holt McDougal Algebra 2 Holt Algebra2

  2. x – 3 x – 2 Warm Up Simplify. Assume that all expressions are defined. 1. (2x + 5) – (x2 + 3x – 2) –x2 – x + 7 2. (x – 3)(x+ 1)2 x3 – x2 – 5x – 3 3.

  3. Objectives Add, subtract, multiply, and divide functions. Write and evaluate composite functions.

  4. Vocabulary composition of functions

  5. You can perform operations on functions in much the same way that you perform operations on numbers or expressions. You can add, subtract, multiply, or divide functions by operating on their rules.

  6. Example 1A: Adding and Subtracting Functions Given f(x) = 4x2 + 3x – 1 and g(x) = 6x + 2, find each function. (f + g)(x) (f + g)(x) = f(x) + g(x) = (4x2+ 3x – 1) + (6x + 2) Substitute function rules. = 4x2 + 9x + 1 Combine like terms.

  7. Example 1B: Adding and Subtracting Functions Given f(x) = 4x2 + 3x – 1 and g(x) = 6x + 2, find each function. (f – g)(x) (f – g)(x) = f(x) – g(x) = (4x2 + 3x – 1) – (6x + 2) Substitute function rules. Distributive Property = 4x2 + 3x – 1 – 6x – 2 Combine like terms. = 4x2–3x – 3

  8. Check It Out! Example 1a Given f(x) = 5x –6and g(x) = x2 – 5x + 6, find each function. (f + g)(x) (f + g)(x) = f(x) + g(x) = (5x – 6) +(x2– 5x + 6) Substitute function rules. = x2 Combine like terms.

  9. Check It Out! Example 1b Given f(x) = 5x –6and g(x) = x2 – 5x + 6, find each function. (f – g)(x) (f – g)(x) = f(x) – g(x) = (5x –6) – (x2 – 5x + 6) Substitute function rules. Distributive Property = 5x – 6– x2+ 5x – 6 Combine like terms. = –x2+ 10x – 12

  10. When you divide functions, be sure to note any domain restrictions that may arise.

  11. Example 2A: Multiplying and Dividing Functions Given f(x) = 6x2– x –12and g(x) = 2x – 3, find each function. (fg)(x) (fg)(x) = f(x) ●g(x) Substitute function rules. = (6x2 – x – 12) (2x – 3) = 6x2 (2x – 3) – x(2x – 3)– 12(2x – 3) Distributive Property = 12x3 – 18x2 – 2x2+ 3x– 24x + 36 Multiply. = 12x3 – 20x2 – 21x + 36 Combine like terms.

  12. ( ) ( ) f f 3 3 (x) (x) g g 2 2 f(x) 6x2 – x –12 = g(x) 2x – 3 = = = Factor completely. Note that x ≠ . (2x – 3)(3x +4) (2x – 3)(3x + 4) 2x – 3 (2x – 3) = 3x + 4, where x ≠ Example 2B: Multiplying and Dividing Functions Set up the division as a rational expression. Divide out common factors. Simplify.

  13. Check It Out! Example 2a Given f(x) = x + 2 and g(x) = x2 – 4, find each function. (fg)(x) (fg)(x) = f(x)●g(x) = (x + 2)(x2 – 4) Substitute function rules. Multiply. = x3 + 2x2– 4x – 8

  14. g f ( ) ( ) g g(x) (x) (x) = f f(x) x2 – 4 = x + 2 = = (x – 2)(x + 2) (x – 2)(x + 2) x + 2 (x + 2) Check It Out! Example 2b Set up the division as a rational expression. Factor completely. Note that x ≠ –2. Divide out common factors. = x– 2, where x ≠ –2 Simplify.

  15. Another function operation uses the output from one function as the input for a second function. This operation is called the composition of functions.

  16. The order of function operations is the same as the order of operations for numbers and expressions. To find f(g(3)), evaluate g(3) first and then substitute the result into f.

  17. Reading Math The composition (f o g)(x) or f(g(x)) is read “f of g of x.”

  18. Caution! Be careful not to confuse the notation for multiplication of functions with composition fg(x) ≠ f(g(x))

  19. Example 3A: Evaluating Composite Functions Given f(x) = 2xand g(x) = 7 – x, find each value. f(g(4)) Step 1 Find g(4) g(4) = 7 – 4 g(x) = 7 – x = 3 Step 2 Find f(3) f(3) = 23 f(x) = 2x = 8 So f(g(4)) = 8.

  20. Example 3B: Evaluating Composite Functions Given f(x) = 2xand g(x) = 7 – x, find each value. g(f(4)) Step 1 Find f(4) f(4) = 24 f(x) = 2x = 16 Step 2 Find g(16) g(16) = 7 – 16 g(x) = 7 – x. = –9 So g(f(4)) = –9.

  21. Check It Out! Example 3a Given f(x) = 2x – 3 and g(x) = x2, find each value. f(g(3)) Step 1 Find g(3) g(3) = 32 g(x) = x2 = 9 Step 2 Find f(9) f(9) = 2(9) – 3 f(x) = 2x – 3 = 15 So f(g(3)) = 15.

  22. Check It Out! Example 3b Given f(x) = 2x –3and g(x) = x2, find each value. g(f(3)) Step 1 Find f(3) f(3) = 2(3) – 3 f(x) = 2x – 3 = 3 Step 2 Find g(3) g(3) = 32 g(x) = x2 = 9 So g(f(3)) = 9.

  23. You can use algebraic expressions as well as numbers as inputs into functions. To find a rule for f(g(x)), substitute the rule for g into f.

  24. Given f(x) = x2–1and g(x) = , write each composite function. State the domain of each. x 1 – x f(g(x)) = f( ) x x = = ()2– 1 1 – x 1 – x –1 + 2x (1 – x)2 Example 4A: Writing Composite Functions f(g(x)) Substitute the rule g into f. Use the rule for f. Note that x ≠ 1. Simplify. The domain of f(g(x)) is x ≠ 1 or {x|x ≠ 1} because g(1) is undefined.

  25. Given f(x) = x2–1and g(x) = , write each composite function. State the domain of each. (x2 – 1) x = 1 – (x2 – 1) 1 – x = x2– 1 2 – x2 Example 4B: Writing Composite Functions g(f(x)) g(f(x)) = g(x2 – 1) Substitute the rule f into g. Use the rule for g. Simplify. Note that x ≠ . The domain of g(f(x)) is x ≠ or {x|x ≠ } because f( ) = 1 and g(1) is undefined.

  26. Given f(x) = 3x –4and g(x) = + 2 , write each composite. State the domain of each. f(g(x)) = 3( + 2) – 4 = + 6 – 4 = + 2 Check It Out! Example 4a f(g(x)) Substitute the rule g into f. Distribute. Note that x ≥ 0. Simplify. The domain of f(g(x)) is x ≥ 0 or {x|x ≥ 0}.

  27. Given f(x) = 3x –4and g(x) = + 2 , write each composite. State the domain of each. g(f(x)) = 4 4 4 3 3 3 = Note that x ≥ . The domain of g(f(x)) is x ≥ or {x|x ≥ }. Check It Out! Example 4b g(f(x)) Substitute the rule f into g.

  28. Composite functions can be used to simplify a series of functions.

  29. Example 5: Business Application Jake imports furniture from Mexico. The exchange rate is 11.30 pesos per U.S. dollar. The cost of each piece of furniture is given in pesos. The total cost of each piece of furniture includes a 15% service charge. A. Write a composite function to represent the total cost of a piece of furniture in dollars if the cost of the item is c pesos.

  30. D(c) = c 11.30 Example 5 Continued Step 1 Write a function for the total cost in U.S. dollars. P(c) = c + 0.15c = 1.15c Step 2 Write a function for the cost in dollars based on the cost in pesos. Use the exchange rate.

  31. = 1.15 ( ) D(P(c) ) = 1.15 ( ) c 1800 11.30 11.30 Example 5 Continued Step 3 Find the composition D(P(c)). D(P(c)) = 1.15P(c) Substitute P(c) for c. Replace P(c) with its rule. B. Find the total cost of a table in dollars if it costs 1800 pesos. Evaluate the composite function for c = 1800. ≈ 183.19 The table would cost $183.19, including all charges.

  32. Check It Out! Example 5 During a sale, a music store is selling all drum kits for 20% off. Preferred customers also receive an additional 15% off. a. Write a composite function to represent the final cost of a kit for a preferred customer that originally cost c dollars. Step 1 Write a function for the final cost of a kit that originally cost c dollars. Drum kits are sold at 80% of their cost. f(c) = 0.80c

  33. Check It Out! Example 5 Continued Step 2 Write a function for the final cost if the customer is a preferred customer. g(c) = 0.85c Preferred customers receive 15% off.

  34. Check It Out! Example 5 Continued Step 3 Find the composition f(g(c)). f(g(c)) = 0.80(g(c)) Substitute g(c) for c. f(g(c)) = 0.80(0.85c) Replace g(c) with its rule. = 0.68c b. Find the cost of a drum kit at $248 that a preferred customer wants to buy. Evaluate the composite function for c = 248. f(g(c) ) = 0.68(248) The drum kit would cost $168.64.

  35. f g ( ) (x) 2x + 1 Lesson Quiz: Part I Given f(x) = 4x2– 1and g(x) = 2x – 1, find each function or value. 4x2 + 2x – 2 1. (f + g)(x) 2. (fg)(x) 8x3 – 4x2 – 2x + 1 3. 4. g(f(2)) 29

  36. Given f(x) = x2and g(x) = , write each composite function. State the domain of each. Lesson Quiz: Part II f(g(x)) = x – 1; 5. f(g(x)) {x|x ≥ 1} 6. g(f(x)) {x|x ≤ – 1 or x ≥ 1}

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