Let f: X → Y. Now, let f represent a one to one function and y be any element of Y, there exists a unique element x ∈ X such that y = f(x).Then the map
f−1: f[X] →X
That associates to each element is called as the inverse map of f.
The function f(x) = x5 and g(x) = x1/2 have the following property:
f(g(x)) = f(x1/5) = (x1/5)5 = x
g(f(x)) = g(x5) = (x5)1/5 = x
Thus, if two functions f and g satisfy f(g(x)) = x for every x in domain of f , then in such a situation we can say that the function f is the inverse of g and g is the inverse of f .
For finding the inverse of a function,we write down the function y as a function of x i.e. y = f(x) and then solve for x as a function of y.
(i) f: X → Y is defined to be invertible, if g: Y → X such that g o f = I x and f o g = IY.
(ii) If f is a bijective function then only a function f: X → Y is invertible.
(iii) If f: X → Y, g: Y → Z and also h: Z → S are functions, then (h o g) o f = h o (g o f)
Consider f: X → Y and g: Y → Z as two invertible functions. In this case g o f is also invertible with (g o f) –1 = f –1 o g–1.
Example: If f(x) = x2, g(x) = x3 and h(x) = 3x+2. Find out fohog(x).
Solution: h(g(x)) = 3 (x3) + 2 = x + 2
fohog(x) = f [h(g(x))] = (x+2)2
This is the required solution.
Example: Example 2: Find the inverse of the function f(x) = x3, x ∈ R.
Solution: The given function f(x) = x3 is a one to one and onto function defined in the range → R .Therefore, we can find the inverse of this function.
To find the inverse, we need to write down this function as
y = x3
In the above equation, y is an arbitrary element from the range of f. If we solve for x from the above equation, we will get:
x = y1/3
This gives a function g:Y →X. This new function g can be defined as
g(y) = y1/3
This function g is the inverse of the function f since its domain is same as the range of the function f.Since g(y) = y1/2, representing the independent variable with x , we get g(x) = x1/3 = f−1(x).