INVERSE FUNCTIONS USING TABLES

Example 1 :

Suppose f is the function whose domain is the four numbers {√2, 8, 17, 18}, with the values of f given in the table shown here in the margin.

(a) What is the range of f ?

(b) Explain why f is a one-to-one function.

(c) What is the table for the function f ⁻¹?

Solution :

(a) In the table above, x represents domain and f(x) represents range. Hence the range of the function is

{3, -5, 6, 1}

(b) From the table above, it is clear that every element of x is associated with an unique element of f(x). Hence it is one to one function.

(c)  From the table above, 

f(√2) = 3, f(8) = -5, f(17) = 6, f(18) = 1

f^-1(3) = √2, f^-1(-5) = 8, f^-1(6) = 17, f^-1(1) = 18

The table for the function f^-1 :

Example 2 :

For f and g are functions, each of whose domain consists of four numbers, with f and g defined by the tables below:

(a) What is the domain of f ?

(b) What is the range of f ?

(c) Sketch the graph of f.

(d) Give the table of values for f⁻¹.

(e) What is the domain of f⁻¹?

(f) What is the range of f⁻¹?

(g) Sketch the graph of f^-1.

(h) Give the table of values for f⁻¹ o f.

(i) Give the table of values for f o f^-1.

Solution :

(a) Domain of f = {1, 2, 3, 4}

(b) Range of f = {2, 3, 4, 5}

(c) The graph of f consists of all points of the form

(x, f(x))

as x varies over the domain of f . Thus the graph of f, shown below, consists of the four points :

(1, 4), (2, 5), (3, 2), (4, 3)

(d) The table for the inverse of a function is obtained by interchanging the two columns of the table for the function (after which one can, if desired, reorder the rows, as has been done below) :

(e) Domain of f^-1 = {2, 3, 4, 5}

(f) Range of f^-1 = {1, 2, 3, 4}

(g) The graph of f^-1 consists of all points of the form

(x, f^-1(x))

as x varies over the domain of f^-1. Thus the graph of f^-1, shown below, consists of the four points :

(4, 1), (5, 2), (2, 3), (3, 4)

(h)Table of values for f^-1 o f :

(f ⁻¹o f) (1) = f⁻¹[f(1)] = f⁻¹(4) = 1

(f ⁻¹o f) (2) = f⁻¹[f(2)] = f⁻¹(5) = 2

(f ⁻¹o f) (3) = f⁻¹[f(3)] = f⁻¹(2) = 3

(f ⁻¹o f) (4) = f⁻¹[f(4)] = f⁻¹(3) = 4

(i) Table of values for f o f^-1 :

(f o f ⁻¹) (2) = f[(f⁻¹(2)] = f(3) = 2

(f o f ⁻¹) (3) = f[f⁻¹(3)] = f(4) = 3

(f o f⁻¹) (4) = f[f ⁻¹(4)] = f(1) = 4

(f o f ⁻¹) (5) = f[f⁻¹(5)] = f(2) = 5

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