Find the inverse of the function: f x = sin - 1 x 3 , x ∈ - 3 , 3

Given: f x = sin - 1 x 3 , x ∈ - 3 , 3 The given function is one-one and onto, therefore its inverse can be calculated. y = sin - 1 x 3 ⇒ sin y = x 3 ⇒ 3 sin y = x Replace x by y and y by x in 3 sin y = x we get, ⇒ y = 3 sin x Hence, f - 1 x = 3 sin x

Find the inverse of the function f x = log e x 2 + 3 x + 1 , x ∈ 1 , 3 . Assuming it to be an onto function.

Given: f x = log e x 2 + 3 x + 1 , x ∈ 1 , 3 ⇒ f ' x = 2 x + 3 x 2 + 3 x + 1 ⇒ f ' ( x ) > 0 , ∀ x ∈ 1 , 3 i.e. f x is strictly increasing function. Thus, f x is injective, and it is given that f x is onto. Hence the given function is invertible. Now, let y = f x y = ln x 2 + 3 x + 1 ⇒ e y = x 2 + 3 x + 1 ⇒ x 2 + 3 x + 1 - e y = 0 ⇒ x = - 3 ± 9 - 4 1 - e y 2 ⇒ x = - 3 ± 5 + 4 e y 2 But, x ∈ 1 , 3 (given) ∴ x = - 3 + 5 + 4 e y 2 We replace ' x ' by ' y ' and ' y ' by ' x ' in x = - 3 + 5 + 4 e y 2 we get, y = - 3 + 5 + 4 e x 2 ⇒ f - 1 x = - 3 + 5 + 4 e x 2

Find the inverse of the function: f x = log e x + x 2 + 1 .

Given: f x = log e x + x 2 + 1 Given function is bijective, hence its inverse can be calculated. Let, y = log e x + x 2 + 1 ⇒ e y = x + x 2 + 1 ⇒ e y - x = x 2 + 1 ⇒ e 2 y + x 2 - 2 e y x = 1 + x 2 ⇒ e 2 y - 2 e y x - 1 = 0 ⇒ x = e 2 y - 1 2 e y ⇒ x = 1 2 e y - e - y . . . i Replace ' x ' by ' y ' and ' y ' by ' x ' in equation i , we get ⇒ y = 1 2 e x - e - x ⇒ f - 1 x = 1 2 e x - e - x

Find the inverse of the function: f x = x , x < 1 x 2 , 1 ≤ x ≤ 4 8 x , x > 4

Given: f x = x , x < 1 x 2 , 1 ≤ x ≤ 4 8 x , x > 4 Let y = x ; x < 1 Replace ' x ' by ' y ' and ' y ' by ' x ' , we get ⇒ x = y ; y < 1 ⇒ f - 1 x = x ; x < 1 Again let, y = x 2 ; 1 ≤ x ≤ 4 ⇒ x = y ; 1 ≤ x ≤ 4 Replace ' x ' by ' y ' and ' y ' by ' x ', we get ⇒ y = x ; 1 ≤ y ≤ 16 ⇒ f - 1 x = x ; 1 ≤ x ≤ 16 Again let, y = 8 x ; x > 4 ⇒ x = y 8 ⇒ x = y 2 64 ; x > 4 Replace ' x ' by ' y ' and ' y ' by ' x ' , we get ⇒ y = x 2 64 ; y > 16 ⇒ f - 1 x = x 2 64 ; x > 16 Hence, f - 1 x x ; x < 1 x ; 1 ≤ x ≤ 16 x 2 64 ; x > 16

If the function f : [ 1 , ∞ ) → [ 1 , ∞ ) is defined by f x = 2 x x - 1 , then find f - 1 x .

Given f : [ 1 , ∞ ) → [ 1 , ∞ ) & f x = 2 x x - 1 Given function is bijective, so it's inverse function is exist. Let, y = 2 x x - 1 Taking log of both the sides with the base ' 2 ' ⇒ log 2 y = log 2 2 x x - 1 ⇒ log 2 y = x x - 1 ⇒ x 2 - x - log 2 y = 0 ⇒ x = 1 ± 1 + 4 log 2 y 2 ⇒ f - 1 x = 1 + 1 + 4 log 2 x 2 ; x > 0 .

E M B I B E

Skills in Mathematics Differential Calculus for JEE Main & Advanced>Chapter 3 - Functions>Target Exercise 3.8

Amit M Agarwal Solutions for Chapter: Functions, Exercise 8: Target Exercise 3.8

Author:Amit M Agarwal

Amit M Agarwal Mathematics Solutions for Exercise - Amit M Agarwal Solutions for Chapter: Functions, Exercise 8: Target Exercise 3.8

Attempt the practice questions on Chapter 3: Functions, Exercise 8: Target Exercise 3.8 with hints and solutions to strengthen your understanding. Skills in Mathematics Differential Calculus for JEE Main & Advanced solutions are prepared by Experienced Embibe Experts.