|Consider the function of of the set of children to the set of their mothers. It is obviously a many-one 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 f1, f2, f3, f4, f5 exhibited by the following diagrams:
The above illustrations lead to a conclusion that
- f1(one-one into type), f1-1 doesn't exist.
- f2(one-one onto type), f2-1 doesn't exist.
- f3(many-one into type), f3-1 doesn't exist.
- f4(many-one onto type), f4-1 doesn't exist.
- f5(constant type), f5-1 doesn't exist.
A function f: A -> B will have its inverse g: B -> A if and only if
Definition: If f is a function f: A - >B, then there will exist a function g: B ->A if f is a one-one 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.
- f is a one-one function.
- f is an onto function i.e f is bijective.
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) = x3 + 7, find a formula that defines f-1.
Solution: Method to find the inverse of a bijective function
Answer: Hence f-1 : R -> R : f-1(x) = (x-7)1/3for all x € R.
- 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) = x3 + 7 = y, then x3 = y - 7
- Step 5: x = (y-7)1/3 = >f-1(y) = (y - 7)1/3
Directions: Solve the following problems. Also write at least 5 examples of your own.