Is the Function Invertible?

Date: 09/22/97 at 20:52:58
From: Lanka
Subject: Calculus

Hi Dr. Math, 

I have been working on a question and I have run into a problem. The 
question states: For the following functions f(x) decide if the 
function is invertible as a function from R to R. For those which are 
not invertible, find a subdomain such that the function becomes 
invertible when restricted to that subdomain. 

The one that I am inquiring about is f(x) = 3x^3 + 2 (three x cubed 
plus two). My approach was as follows: Since I know that all 
increasing and decreasing functions are invertible I attempted to 
prove that this function is one of these functions increasing or 
decreasing. I found f'(x) to be 9x^2. I then set 9x^2 > 0 to determine 
when the function was increasing (increasing f'n defined as a function 
in which f'(x) > 0) and on the next line I have x^2 > 0 (b/c I divided 
the nine out). 

However, for all values of x , x^2 will be greater than 0 except when 
x = 0. So, since f'(x) is not > 0 for all values of x would this still 
be considered an increasing f'n? Would it be valid to simply state 
that the f'n is increasing and since the slope is  only equal to zero 
in one pt the function is invertible.  

I don't like this situation because I like to prove things fully using 
mathematics rather than words. Is there any way around this?

Date: 09/23/97 at 08:14:59
From: Doctor Anthony
Subject: Re: Calculus

To be accurate one should say that 'strictly' increasing or decreasing 
functions are necessarily invertible. The word 'strictly' is necessary 
because a function that remained constant for an interval before 
increasing again would not be invertible. 

I like to describe matters in the following way. A relation becomes a 
function if there is a unique value of y for any given value of x.  
Thus y = ax^3 + bx^2 + cx + d is a function since for any value I 
ascribe to x there is one and only one value of y. However if I 
give y a value for the general cubic there is not necessarily only one 
value of x.  So for the general cubic f(x) is not invertible over its 
whole domain.  

You can visualize this by thinking of the shape of a cubic curve.  
Given any value of x, you find the corresponding y value by drawing a 
vertical line through that value of x parallel to the y axis.  This 
line can meet the curve in only ONE point. So this is a function in 
going from x to y. Now given a value of y, to find the corresponding 
x value you draw a horizontal line through the chosen y value. For 
some values of y, in the general cubic, this line can cut the curve in 
three places, thus giving 3 values of x for the one value of y, so 
this is NOT a function in going from y to x.  

For the function f(x) = 3x^3 + 2  since there are no maximum or 
minimum points, but only a point of inflection at x=0, it is easy to 
see that f(x) is invertible everywhere. Should you exclude x=0? No, 
because wherever x1<x2 we also have f(x1)<f(x2). So the function is 
strictly increasing. 

Another way to decide this is to show that ff^(-1)(x) = x for all x.

We had y = 3x^3 + 2  To get the inverse function, swap x and y and 
then make y the subject of the formula.

      x = 3y^3 + 2    on swapping x and y.  Now make y the subject. 
  
                           3y^3 = x - 2

                            y^3 = (x-2)/3

                      f^(-1)(x) = y = [(x-2)/3]^(1/3)


   ff^(-1)(x) = 3[[(x-2)/3]^(1/3)]^3 + 2

              = 3[(x-2)/3] + 2

              =  (x-2) + 2

              = x - 2 + 2

              = x   

And this is true for all x.

-Doctor Anthony,  The Math Forum
 Check out our web site!  http://mathforum.org/dr.math/