Notes On Inverse of a Bijective Function - CBSE Class 12 Maths
Let f: A → B be a function. If, for an arbitrary x ∈ A we have f(x) = y ∈ B, then the function, g: B → A, given by g(y) = x, where y ∈ B and x ∈ A, is called the inverse function of f. Ex: Define f: A → B such that f(2) = -2, f(½) = -2, f(½) = -½, f(-1) = 1, f(-1/9) = 1/9 Here f one-one and onto. g(-2) = 2, g(-½) = 2, g(-½) = ½, g(1) = -1, g(1/9) = -1/9 g is the inverse of f. A function, f: A → B, is said to be invertible, if there exists a function, g : B → A, such that g o f = IA and f o g = IB. The function, g, is called the inverse of f, and is denoted by f -1. Ex: Let 2 ∈ A.Then gof(2) = g{f(2)} = g(-2) = 2 Let -2 ∈ B.Then fog(-2) = f{g(-2)} = f(2) = -2. Example: Let P = {y ϵ N: y = 3x - 2 for some x ϵN}. Show that a function, f : N → P, defined by f (x) = 3x - 2, is invertible, and find f-1. Solution: Let us consider an arbitrary element, y ϵ P. ⇒ y=3x-2 ⇒ x = (y+2)/3 Let us define g : P → N by g(y) = (y+2)/3 Let x ∈ N. Then g o f (x) = g (f (x)) = g (3x - 2) = (3x-2+2)/3 = x This shows g o f = IN …(i) Let y ∈ P. Then fog (y) = f (g (y)) = f((y+2)/3) = 3((y+2)/3) - 2 = y This shows fog = IP …(ii) Hence, f is invertible and g is the inverse of f. Theorem: Let f : X → Y and g : Y → Z be two invertible (i.e. bijective) functions. Then g o f is also invertible with (g o f)-1 = f -1o g-1. Proof: Given, f and g are invertible functions. To prove that g o f is invertible, with (g o f)-1 = f -1o g-1. It is sufficient to prove that: i. (f -1 o g-1) o (g o f) = IX, and ii. (g o f)o( f -1o g-1) = IZ. Now, ( f -1 o g-1) o (g o f) = {( f -1 o g-1) o g} o f {'.' l o (m o n) = (l o m) o n} = { f -1 o (g-1 o g)} o f = ( f -1oIY)of {'.' g-1og = IY} = f-1of { '.' f-1oI = f-1} = IX Hence, (f -1o g-1)o(g o f) =IX …… ( i) Similarly, (g o f)o( f -1o g-1) ={(g o f) o f--1} o g-1 ={g 0 (f o f--1} o g-1 =(g 0 Ix) o g-1 =g o g-1 = Iz (g o f)o( f -1o g-1) =IZ ……. (ii) From equations (i) and (ii), (g o f)-1 = f -1 o g-1 Hence, the composition of two invertible functions is also invertible.

#### Summary

Let f: A → B be a function. If, for an arbitrary x ∈ A we have f(x) = y ∈ B, then the function, g: B → A, given by g(y) = x, where y ∈ B and x ∈ A, is called the inverse function of f. Ex: Define f: A → B such that f(2) = -2, f(½) = -2, f(½) = -½, f(-1) = 1, f(-1/9) = 1/9 Here f one-one and onto. g(-2) = 2, g(-½) = 2, g(-½) = ½, g(1) = -1, g(1/9) = -1/9 g is the inverse of f. A function, f: A → B, is said to be invertible, if there exists a function, g : B → A, such that g o f = IA and f o g = IB. The function, g, is called the inverse of f, and is denoted by f -1. Ex: Let 2 ∈ A.Then gof(2) = g{f(2)} = g(-2) = 2 Let -2 ∈ B.Then fog(-2) = f{g(-2)} = f(2) = -2. Example: Let P = {y ϵ N: y = 3x - 2 for some x ϵN}. Show that a function, f : N → P, defined by f (x) = 3x - 2, is invertible, and find f-1. Solution: Let us consider an arbitrary element, y ϵ P. ⇒ y=3x-2 ⇒ x = (y+2)/3 Let us define g : P → N by g(y) = (y+2)/3 Let x ∈ N. Then g o f (x) = g (f (x)) = g (3x - 2) = (3x-2+2)/3 = x This shows g o f = IN …(i) Let y ∈ P. Then fog (y) = f (g (y)) = f((y+2)/3) = 3((y+2)/3) - 2 = y This shows fog = IP …(ii) Hence, f is invertible and g is the inverse of f. Theorem: Let f : X → Y and g : Y → Z be two invertible (i.e. bijective) functions. Then g o f is also invertible with (g o f)-1 = f -1o g-1. Proof: Given, f and g are invertible functions. To prove that g o f is invertible, with (g o f)-1 = f -1o g-1. It is sufficient to prove that: i. (f -1 o g-1) o (g o f) = IX, and ii. (g o f)o( f -1o g-1) = IZ. Now, ( f -1 o g-1) o (g o f) = {( f -1 o g-1) o g} o f {'.' l o (m o n) = (l o m) o n} = { f -1 o (g-1 o g)} o f = ( f -1oIY)of {'.' g-1og = IY} = f-1of { '.' f-1oI = f-1} = IX Hence, (f -1o g-1)o(g o f) =IX …… ( i) Similarly, (g o f)o( f -1o g-1) ={(g o f) o f--1} o g-1 ={g 0 (f o f--1} o g-1 =(g 0 Ix) o g-1 =g o g-1 = Iz (g o f)o( f -1o g-1) =IZ ……. (ii) From equations (i) and (ii), (g o f)-1 = f -1 o g-1 Hence, the composition of two invertible functions is also invertible.

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