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High School Mathematics - 2
4.7 Inverse of Functions

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:

  1. f1(one-one into type), f1-1 doesn't exist.
  2. f2(one-one onto type), f2-1 doesn't exist.
  3. f3(many-one into type), f3-1 doesn't exist.
  4. f4(many-one onto type), f4-1 doesn't exist.
  5. f5(constant type), f5-1 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
  1. f is a one-one function.
  2. 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 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.
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

  1. Step 1: Put f(x) = y, where y B and x A.
  2. Step 2: Solve f(x) = y to obtainx in terms of y.
  3. Step 3: In the relation obtained in step 2, replace x by f-1 to obtain the inverse of f.
  4. Step 4: Let f(x) = x3 + 7 = y, then x3 = y - 7
  5. Step 5: x = (y-7)1/3 = >f-1(y) = (y - 7)1/3
Answer: Hence f-1 : R -> R : f-1(x) = (x-7)1/3for all x R.

Directions: Solve the following problems. Also write at least 5 examples of your own.
Q 1: If f:R -> R, write the inverse of f(x) = 3x + y.

Q 2: If f(x) = (2x+1)/3, for each x R, find f-1.

Q 3: f: R -> R be defined by f(x) = 10x - 7. if g = f-1 then g(x) is
1/(10x - 7)
1/(10x + 7)
(x + 7)/10

Q 4: Let f : A -> B, find f-1(B).
Inverse doesn't exist

Q 5: If f: R ->R is such that f(x) = log3x, f-1 is equal to
log x3

Q 6: If f: R -> R , defined by f(x) = x2 + 1, find the value f-1(17)

Q 7: If f: R=>R such that f(x) = e3x+2,find f-1e2

Q 8: Find the range of the function h(x) = x2 + 1
Range = Domain
Range doesn't exist
{(y: y is real,y>=1}

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Question 10: This question is available to subscribers only!

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