Digital Lesson Algebra of Functions

Operations on Functions If f and g are functions and x is an element of the domain of each function, then: ( f + g)(x) = f (x) + g(x) ( f – g)(x) = f (x) – g(x) ( f • g)(x) = f (x) • g(x) Copyright © by Houghton Mifflin Company, Inc. All rights reserved. Operations on Functions

Example:(f+g)(x), (f-g)(x) Example: Given f (x) = x2 and g(x) = 2, find ( f + g)(x) and ( f – g)(x). ( f + g)(x) f(x) = x2 - 2 x y 2 4 ( f + g)(x) = f(x) + g(x) = x2 + 2 g(x) = 2 ( f – g)(x) = f(x) – g(x) = x2 – 2 ( f – g)(x) Copyright © by Houghton Mifflin Company, Inc. All rights reserved. Example:(f+g)(x), (f-g)(x)

Example: Given f (x) = 10 – 3x3 and g(x) = 2x2 + 1, find ( f + g)(2) and ( f – g)(–1) . ( f + g)(x) = (10 – 3x3) + (2x2 + 1) = –3x3 + 2x2 + 11 ( f + g)(2) = –3(2)3 + 2(2)2 + 11 = –5 (f – g)(x) = (10 – 3x3) – (2x2 + 1) = –3x3 – 2x2 + 9 (f – g)(–1) = – 3(–1)3 – 2(–1)2 + 9 = 10 Copyright © by Houghton Mifflin Company, Inc. All rights reserved. Example: Sum of f and g

Example: Product of f and g Example: Given f (x) = x – 2 and g(x) = x + 1, find ( f • g)(x) and ( f • g)(3) and ( f • g)(-1). ( f • g)(x) = (x – 2) • (x + 1) - 2 x y 2 4 ( f • g)(x) = x2 – x – 2 g(x) = x + 1 ( f • g)(3) = ((3) – 2) • ((3) + 1) = (3)2 – (3) – 2 f(x) = x – 2 = 4 ( f • g)(-1) = (-1)2 – (-1) – 2 = 0 Copyright © by Houghton Mifflin Company, Inc. All rights reserved. Example: Product of f and g

Example: Quotient of f and g Example: Given f (x) = 2x2 – 3x + 1 and g(x) = x – 1, find (1) and (2). f(x) = 2x2 – 3x + 1 (1) = 1 x y 2 -2 1 This is not defined, so (1) cannot be determined. (2) 2 g(x) = x – 1 Copyright © by Houghton Mifflin Company, Inc. All rights reserved. Example: Quotient of f and g

A function can be evaluated at the value of another function. Let f and g be two functions. Where g(x) is in the domain of f for all x in the domain of g. The composition of the two functions denoted as (f g), is the function whose value at x is given by (f g)(x) = f [g(x)]. Example: Given f (x) = 2x + 5 and g(x) = x2 + 1, find f [g(–2)]. g(–2) = (–2)2 + 1 = 5 f [g(–2)] = f (5) = 2(5) + 5 = 15 Copyright © by Houghton Mifflin Company, Inc. All rights reserved. Composition function

Example: Given f (x) = 3x2 + 2x – 48 and g(x) = 5x – 14, evaluate the following composite functions: 1. g[ f (4)] f (4) = 3(4)2 + 2(4) – 48 = 8 g[ f (4)] = g (8) = 5(8) – 14 = 26 2. f [ g(3)] g(3) = 5(3) – 14 = 1 f [ g (3)] = f (1) = 3(1)2 + 2(1) – 48 = – 43 Copyright © by Houghton Mifflin Company, Inc. All rights reserved. Example: f[g(-2)]

Example: Given f (x) = 3x – 2 and g(x) = x2 – 2x, When evaluating compositions of functions, the order in which the functions are applied is important. Example: Given f (x) = 3x – 2 and g(x) = x2 – 2x, find ( f ° g)(x) and (g ° f )(x). g(x) = x2 – 2x y - 2 x -4 ( f ° g)(x) = f [g(x)] = 3(x2 – 2x) – 2 = 3x2 – 6x – 2 f (x) = 3x – 2 ( f ° g)(x) Example continued Copyright © by Houghton Mifflin Company, Inc. All rights reserved. Example: f[g(x)]

Example Continued: g[f(x)] g(x) = x2 – 2x y - 2 x -4 ( g ° f )(x) = g [ f (x)] = (3x – 2)2 – 2(3x – 2) = 9x2 – 18x + 8 f (x) = 3x – 2 ( f ° g)(x) = 3x2 – 6x – 2 The graphs of (f ° g )(x) and (g ° f )(x) intersect at only two points. There are only two values for x for which (f ° g )(x) and (g ° f )(x) are equal. (f ° g )(x) and (g ° f )(x) are different functions. Copyright © by Houghton Mifflin Company, Inc. All rights reserved. Example Continued: g[f(x)]