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/
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