Notes On Inverse of a Bijective Function - CBSE Class 12 Maths
Let f: A â†’ B be a function. If, for an arbitrary x âˆˆ A we have f(x) = y âˆˆ B, then the function, g: B â†’ A, given by g(y) = x, where y âˆˆ B and x âˆˆ A, is called the inverse function of f. Ex: Define f: A â†’ B such that f(2) = -2, f(Â½) = -2, f(Â½) = -Â½, f(-1) = 1, f(-1/9) = 1/9 Here f one-one and onto. g(-2) = 2, g(-Â½) = 2, g(-Â½) = Â½, g(1) = -1, g(1/9) = -1/9 g is the inverse of f. A function, f: A â†’ B, is said to be invertible, if there exists a function, g : B â†’ 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 âˆˆ A.Then gof(2) = g{f(2)} = g(-2) = 2 Let -2 âˆˆ B.Then fog(-2) = f{g(-2)} = f(2) = -2. Example: Let P = {y Ïµ N: y = 3x - 2 for some x ÏµN}. Show that a function, f : N â†’ P, defined by f (x) = 3x - 2, is invertible, and find f-1. Solution: Let us consider an arbitrary element, y Ïµ P. â‡’ y=3x-2 â‡’ x = (y+2)/3 Let us define g : P â†’ N by g(y) = (y+2)/3 Let x âˆˆ N. Then g o f (x) = g (f (x)) = g (3x - 2) = (3x-2+2)/3 = x This shows g o f = IN â€¦(i) Let y âˆˆ P. Then fog (y) = f (g (y)) = f((y+2)/3) = 3((y+2)/3) - 2 = y This shows fog = IP â€¦(ii) Hence, f is invertible and g is the inverse of f. Theorem: Let f : X â†’ Y and g : Y â†’ 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 â€¦â€¦ ( 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 â€¦â€¦. (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 â†’ B be a function. If, for an arbitrary x âˆˆ A we have f(x) = y âˆˆ B, then the function, g: B â†’ A, given by g(y) = x, where y âˆˆ B and x âˆˆ A, is called the inverse function of f. Ex: Define f: A â†’ B such that f(2) = -2, f(Â½) = -2, f(Â½) = -Â½, f(-1) = 1, f(-1/9) = 1/9 Here f one-one and onto. g(-2) = 2, g(-Â½) = 2, g(-Â½) = Â½, g(1) = -1, g(1/9) = -1/9 g is the inverse of f. A function, f: A â†’ B, is said to be invertible, if there exists a function, g : B â†’ 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 âˆˆ A.Then gof(2) = g{f(2)} = g(-2) = 2 Let -2 âˆˆ B.Then fog(-2) = f{g(-2)} = f(2) = -2. Example: Let P = {y Ïµ N: y = 3x - 2 for some x ÏµN}. Show that a function, f : N â†’ P, defined by f (x) = 3x - 2, is invertible, and find f-1. Solution: Let us consider an arbitrary element, y Ïµ P. â‡’ y=3x-2 â‡’ x = (y+2)/3 Let us define g : P â†’ N by g(y) = (y+2)/3 Let x âˆˆ N. Then g o f (x) = g (f (x)) = g (3x - 2) = (3x-2+2)/3 = x This shows g o f = IN â€¦(i) Let y âˆˆ P. Then fog (y) = f (g (y)) = f((y+2)/3) = 3((y+2)/3) - 2 = y This shows fog = IP â€¦(ii) Hence, f is invertible and g is the inverse of f. Theorem: Let f : X â†’ Y and g : Y â†’ 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 â€¦â€¦ ( 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 â€¦â€¦. (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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