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 ^{−1}o f) (1) = f^{−1}[f(1)] = f^{−1}(4) = 1
(f ^{−1}o f) (2) = f^{−1}[f(2)] = f^{−1}(5) = 2
(f ^{−1}o f) (3) = f^{−1}[f(3)] = f^{−1}(2) = 3
(f ^{−1}o 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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