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

Simplify. Assume that all expressions are defined. 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. x – 3 x – 2

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

Vocabulary composition of functions

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.

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.

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. = 4x2 + 3x – 1 – 6x – 2 Distributive Property = 4x2 – 3x – 3 Combine like terms.

Check It Out! Example 1a Given f(x) = 5x – 6 and 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.

Check It Out! Example 1b Given f(x) = 5x – 6 and 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. = 5x – 6 – x2 + 5x – 6 Distributive Property = –x2 + 10x – 12 Combine like terms.

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

Example 2A: Multiplying and Dividing Functions Given f(x) = 6x2 – x – 12 and 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.

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

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. = x3 + 2x2 – 4x – 8 Multiply.

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

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

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

Example 3A: Evaluating Composite Functions Given f(x) = 2x and 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.

Example 3B: Evaluating Composite Functions Given f(x) = 2x and 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.

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.

Check It Out! Example 3b Given f(x) = 2x – 3 and 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.

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.

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

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

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

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

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

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

Lesson Quiz: Part II Given f(x) = x2 and g(x) = , write each composite function. State the domain of each. 5. f(g(x)) f(g(x)) = x – 1; {x|x ≥ 1} 6. g(f(x)) {x|x ≤ – 1 or x ≥ 1}