## Inverse of a Function:

Consider a one-one onto function, f: x → y and let y be any arbitrary element of Y. Since f is onto, there exists at least one element x ∈ X such that f(x) = y. Again, as f is one-one, the element x ∈ X is unique. So corresponding to every element y ∈ Y, there exists a unique element x ∈ X and we can define a function g from Y to X i.e., g: Y → X such that g(y) = x for all y ∈ Y. This function g, associated with the function f, is known as the inverse of f and is denoted by f^{-1}.

Thus, if f: X → Y be one-one onto function and if f(x) = y, where x ∈ X and y ∈ Y, then f^{-1}: Y → X defined by f^{-1}(y) = x is called the inverse of the function f(x).

A function whose inverse exists is known as an invertible or an inversible function. Obviously, the domain of f^{-1} = range of f.

*Properties of Invertible Functions:*

- The inverse of a bijection is unique.
- The inverse of a bijection is also a bijection.
- If f: A → B and g: B → C are two bijections, then gof: A → C is a bijection, and (gof)
^{-1}= f^{-1}og^{-1}. - If f: A → B is a bijection and g: B → A is the inverse of f, then fog = I
_{B}and gof = I_{A}, where I_{A}and I_{B}are identity functions on the sets A and B respectively. - If f: A → B and g: B → A be two functions such that gof = I
_{A}and fog = I_{B}, then f and g are bijections and g = f^{-1}.

If f: R to R defined by f(x) = (3x + 7)/2 for all x ∈ R. Find fExample-^{-1}. f(x) = (3x + 7)/2Solution-f is one-one:Let x _{1}, x_{2} ∈ R (domain) such that f(x_{1}) = f(x_{2})⇒ (3x _{1} + 7)/2 = (3x_{2} + 7)/2⇒ 3x _{1} + 7 = 3x_{2} + 7⇒ 3x _{1} = 3x_{2}⇒ x _{1} = x_{2}∴ f is one-one. f is onto:Let y ∈ R (codomain) such that f(x) = y ⇒ (3x + 7)/2 = y ⇒ 3x = 2y – 7 ⇒ x = (2y-7)/3 ∴ f is an on-to function. Since f is one-one as well as onto, so the inverse of f exists. Inverse of f:Let f(x) = y ⇒ (3x + 7)/2 = y ⇒ 3x = 2y – 7 ⇒ x = (2y – 7)/3 ⇒ f ^{-1}(y) = (2x – 7)/3∴ Inverse is (2x – 7)/3 |

If f: A → B be given by f(x) = 2x + 1 and A = {1, 3, 5}, B = {3, 7, 11}, then write f and fExample-^{-1} as sets of ordered pairs. f(x) = 2x + 1Solution-Now, f(1) = 2(1) + 1 = 3 f(3) = 2(3) + 1 = 7 f(5) = 2(5) + 1 = 11 ∴ f = {(1, 3) (3, 7) (5, 11)} and f ^{-1} = {(3, 1) (7, 3) (11, 5)} |

If f: R → R is defined by f(x) = xExample-^{3} – 2 for all x ∈ R, show that f is a bijection. Find f^{-1}(x) and f^{-1}(3).Solution-f is one-one:Let x _{1}, x_{2} ∈ R (domain) such that f(x_{1}) = f(x_{2})⇒ x _{1}^{3} – 2 = x_{2}^{3} – 2⇒ x _{1}^{3} = x_{2}^{3}⇒ x _{1} = x_{2}∴ f is one-one. f is onto:Let y ∈ R (codomain) such that f(x) = y ⇒ x ^{3} – 2 = y⇒ x ^{3} = y + 2⇒ x = (y + 2) ^{1/3}For all values of y ∈ R, x ∈ R (domain). ∴ f is an on-to function. Since f is one-one as well as onto. ∴ f is a bijection and hence invertible. Inverse of f:Let y ∈ R such that, f(x) = y ⇒ x ^{3} – 2 = y⇒ x ^{3} = y + 2⇒ x = (y + 2) ^{1/3}⇒ f ^{-1}(y) = (y + 2)^{1/3}∴ Inverse of f(x) is f ^{-1}(x) = (x + 2)^{1/3}Now, f ^{-1}(-3) = (-3 + 2)^{1/3} = (-1)^{1/3} = -1 |