Show that y = 1 2 - x is a self-inverse function.

Given: y = 1 2 - x is a self-inverse function. Let f x = 1 2 - x A function f is self-inverse if it has the property that f ( f ( x ) ) = x , for every x in the domain of f . In other words, f ( x ) = f - 1 x . Now, f f x = f 1 2 - x Substitute 1 2 - x in the place of x in the function f x = 1 2 - x = 1 2 - 1 2 - x = x Therefore, y = 1 2 - x is a self inverse function.

Show that f ( x ) = - x - 2 5 x + 1 is a self-inverse function.

Given f ( x ) = - x - 2 5 x + 1 is a self-inverse function. A function f is self-inverse if it has the property that f ( f ( x ) ) = x , for every x in the domain of f . In other words, f ( x ) = f - 1 x . Now, f f x = f - x - 2 5 x + 1 Substitute - x - 2 5 x + 1 in the place of x in the function f x = - x - 2 5 x + 1 = - - x - 2 5 x + 1 - 2 5 - x - 2 5 x + 1 + 1 = ( x + 2 ) ( 5 x + 1 ) - 2 ( 5 x + 1 ) ( 5 x + 1 ) ( - 5 x - 10 ) ( 5 x + 1 ) + ( 5 x + 1 ) ( 5 x + 1 ) = x + 2 - 10 x - 2 5 x + 1 - 9 5 x + 1 ⇒ f f x = x Therefore, f x = - x - 2 5 x + 1 is a self inverse function.

Find the value of m such that g ( x ) = 2 x - 4 x + m is a self-inverse function.

Given g ( x ) = 2 x - 4 x + m is a self-inverse function. We know that, a function g is self-inverse if it has the property that g ( g ( x ) ) = x , for every x in the domain of g . In other words, g ( x ) = g - 1 x . Now, g g x = x ⇒ g 2 x - 4 x + m = x ⇒ 2 2 x - 4 x + m - 4 2 x - 4 x + m + m = x ⇒ 4 x - 8 x + m - 4 x + m x + m 2 x - 4 x + m + m x + m x + m = x ⇒ - 4 m - 8 2 x - 4 + x m + m 2 = x ⇒ - 4 m - 8 = x 2 x - 4 + x m + m 2 ⇒ - 4 m - 8 = 2 x 2 - 4 x + x 2 m + m 2 ⇒ - 4 m - 8 = ( 2 + m ) x 2 - 4 x + m 2 ⇒ 0 = 2 + m Therefore, the value of m = - 2 .

E M B I B E

Mathematics : Analysis and Approaches Standard Level Course Companion>Chapter 2 - Representing Relationships: Introducing Functions>Exercise 2J

Natasha Awada, Paul La Rondie, Laurie Buchanan and, Jill Stevens Solutions for Chapter: Representing Relationships: Introducing Functions, Exercise 34: Exercise 2J

Author:Natasha Awada, Paul La Rondie, Laurie Buchanan & Jill Stevens

Natasha Awada Mathematics Solutions for Exercise - Natasha Awada, Paul La Rondie, Laurie Buchanan and, Jill Stevens Solutions for Chapter: Representing Relationships: Introducing Functions, Exercise 34: Exercise 2J

Attempt the practice questions on Chapter 2: Representing Relationships: Introducing Functions, Exercise 34: Exercise 2J with hints and solutions to strengthen your understanding. Mathematics : Analysis and Approaches Standard Level Course Companion solutions are prepared by Experienced Embibe Experts.