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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