Consider the function of of the set of children to the set of their mothers. It is obviously a manyone type function as several children may have the same mother. Its inverse however will not exist. In this case we will have a relation in which several ordered pairs of the form (mother, child) may have the same first component.
Consider the functions f_{1}, f_{2}, f_{3}, f_{4}, f_{5} exhibited by the following diagrams:
 f_{1}(oneone into type), f_{1}^{1} doesn't exist.
 f_{2}(oneone onto type), f_{2}^{1} doesn't exist.
 f_{3}(manyone into type), f_{3}^{1} doesn't exist.
 f_{4}(manyone onto type), f_{4}^{1} doesn't exist.
 f_{5}(constant type), f_{5}^{1} doesn't exist.
The above illustrations lead to a conclusion that
A function f: A > B will have its inverse g: B > A if and only if
 f is a oneone function.
 f is an onto function i.e f is bijective.
Definition: If f is a function f: A  >B, then there will exist a function g: B >A if f is a oneone onto function, and such that the range of f is the domain of g and domain of f is the range of g; then g is called the inverse of f and is denoted by f^{1}. Also f is called the inverse of g and is denoted as g^{1}.
Symbolically, a function f: x  >y , then its inverse is represented as f^{1}: y  > x, or if y = f(x), then its inverse is represented as f^{1}(y) = x.
Example: If f: R > R be define by f(x) = x^{3} + 7, find a formula that defines f^{1}.
Solution: Method to find the inverse of a bijective function
 Step 1: Put f(x) = y, where y € B and x €A.
 Step 2: Solve f(x) = y to obtainx in terms of y.
 Step 3: In the relation obtained in step 2, replace x by f^{1} to obtain the inverse of f.
 Step 4: Let f(x) = x^{3} + 7 = y, then x^{3} = y  7
 Step 5: x = (y7)^{1/3} = >f^{1}(y) = (y  7)^{1/3}
Answer: Hence f^{1} : R > R : f^{1}(x) = (x7)^{1/3}for all x € R.
Directions: Solve the following problems. Also write at least 5 examples of your own.
