Question #6d237

1 Answer
Nov 22, 2016

Answer:

#f(x) = 3x-1# is one to one, and has the inverse #f^(-1)(x) = (x+1)/3#

Explanation:

A function #f# is said to be one to one if for any #x_0, x_1# in the domain of #f#, #f(x_0) = f(x_1)# implies #x_0 = x_1#. In other words, there is only one element in the domain of #f# that maps to any given element in the range of #f#.

We will first show that #f(x) = 3x-1# has this property. Suppose #x_0# and #x_1# are real numbers such that #f(x_0) = f(x_1)#. Then

#3x_0-1 = 3x_1-1#

#=> 3x_0 = 3x_1#

#=> x_0 = x_1#

Thus #f(x) = 3x-1# is one to one.

As #f(x)# is one to one, it has an inverse function #f^(-1)(x)# where #f^(-1)(f(x)) = f(f^(-1)(x)) = x#.

One way of finding #f^(-1)# is to set #y=f(x) = 3x-1#, change all the #x#'s to #y#'s and vice versa, giving #x = f(y) = 3y-1#, and then solve for #y#. This results in an equation of the form #y = f^(-1)(x)#.

Set #x = f(y) = 3y-1#

#=> x+1 = 3y#

#=> y = (x+1)/3 = f^(-1)(x)#

Notice that given #f^(-1)(x) = (x+1)/3# we have #f^(-1)(f(x)) = f(f^(-1)(x))=x#, as desired.