Inverting FunctionsDate: 07/19/2002 at 02:25:51 From: Rich Subject: Inverse functions (this is long) I'm having some problems with inverse functions. I understand how to find the inverses of functions mathematically when possible. I also understand that the composition of two inverse functions is f(invf(x)) = x. I'm having trouble with graphs of these functions and the swithcing of the variables when applied to physics or engineering problems. This all started with the equation for the derivative of an inverse function. (I'm using invf as the notation for the inverse of f.) 1 invf'(x) = ------------- f'( invf(x) ) I found this formula in a text book and during its derivation the author used a graph, which I cannot draw here. During the analysis I became confused and after some hard thinking this is why I'm confused: Keeping in mind that Y is the vertical axis and X is the horizontal axis in a graph, if we have a function y = f(x) and if we reverse the x and y in the function the graph is a reflection about the y=x axis. But a graph reflected about the y = f(x) axis implies that it's an inverse function. So, if we say y = f(x) = x^3 it's inverse is y = invf(x) = x^(1/3) or the cube root of x. Now if we plot y = x^3 and y = x^(1/3) the graphs are reflections about the y=x axis, as is the composition f(invf(x)) = x for these two functions. No problem. But if we take y = f(x) = x^3 and switch the x and y, we get x = y^3 Now, if we plot this function its graph is exactly the same as the inverse function y = invf(x) = x^(1/3). So I thought ok, maybe they're the same function because they have the same graph and if you algebraically manipulate x = y^3 you will get y = x^(1/3). But the composition of f(y) = x = y^3 with y = f(x) = x^3 is f(f(y)) = (y^3)^3 = y^9 which is not x or y. This does not make sense from a graphical point of view since the graphs are reflection about y = x, but it does make sense if I just look at the functions. They are both cubing their independent variables so they cannot be inverses, even though the graphs suggest that. To complicate things more, the author writes: If we let y = invf(x) then if follows that f(y) = f(invf(x)) or f(y) = x. to suggest the following procedure: Step 1. Interchange x and y in the equation y = f(x) to produce the equation x = f(y). Step 2. Solve the equation x = f(y) for y as a function of x. Step 3. The resulting equation in step 2 will be y = invf(x), the right side of which is the formula for invf. Now if I use my functions above, y = f(x) = x^3 and interchange x and y we get f(y) = x = y^3 and solving for y we get y = invf(x) = x^(1/3) exactly as I showed above. Now if we do what he mentions above, that is, f(y) = f(invf(x)), then f(y) = f(x^(1/3)) = (x^(1/3))^3 = x which is correct. But this seems to imply that f(y) = y^3 = x is the inverse of y = invf(x) = x^(1/3). If so, then why are the graphs not reflected about the y=x axis as stated above? This is mind boggling to me. It has put me in a state of confusion where I don't know now whether interchanging x and y gives me the inverse or if I have to solve for y in terms of x; or how to choose between f(invf(x)) = x and f(invf(y) = y, since one of the equations above was in terms of y and it still worked. Then this gives rise to more confusion about the notation invf(x), invf(y), f(x), f(y). When I start playing with all these symbols and functions I keep going in circles. There was something unsettling for me when the author reached the above derivative equation. It took me a while to figure out but this was why. If you could help me understand this I would greatly appreciate it. Oh also, I tried to think of the variables in the functions f(x) or f(y) as dummy variables so that I could use (w),(e) .. or whatever variable and just look at the functions. But then when I tried to apply that hypothetically to functions such as F(a) = ma which is the force equation in physics, it was hard to keep track since I'm used to F = ma. This just made my head explode trying to figure out the meaning of this mathematically and physically. Thank you. Date: 07/19/2002 at 23:11:15 From: Doctor Peterson Subject: Re: Inverse functions (this is long) Hi, Rich. You seem to have all the basic facts, including the idea that the variables in a function definition are dummy variables (which is the key to your problem), yet you are getting tangled up. I'll try both to correct some misstatements you've made, and suggest better ways to approach the topic. You're right that the graphs of x = y^3 and y = x^(1/3) are identical. Switching x and y in the graph replaces y = f(x) with x = f(y), which is the same as y = invf(x). And two functions with the same graph ARE the same function. But your composition is inappropriate. You don't want to compose f with f, but with invf. This, I think, is the important point: although the EQUATIONS y = f(x) and x = invf(y) are equivalent (have the same graph), the FUNCTIONS are different. The composition you want in order to prove the functions are inverses is f(invf(x)) = f(x^(1/3)) = (x^(1/3))^3 = x not f(f(x)) = f(x^3) = (x^3)^3 = x^9 To put this a little differently, you talk about the functions f(x) = x^3 and f(y) = y^3 as if they were inverses; but they are the SAME function, just written using different dummy variables. It may be helpful to think of functions a "mappings" that take a value in one set (the domain) to a value in another set (the range). We can write your function f as f x -----> x^3 Its inverse is the mapping invf x <----- x^3 that undoes it. This inverse can also be written as invf x^(1/3) <----- x or you can use any variable, say y, in place of x, since the name of the dummy variable doesn't matter. The first form, invf(x^3) = x, is just a way of saying that this is the inverse, that invf(f(x)) = x for all x. The second form expresses it as a proper function definition, telling us for any x how to get invf(x), namely by taking the cube root. These functions are inverses because if you start in the left set, take f to the right set, and then take invf back to the left set, you will always end up where you started (and the same going in the other order). Seeing it this way, you would not even think of "composing" y = f(x) with x = f(y), because they just refer to the same function applied to different sets. Remember that a function is just something you can do to any variable. An equation does not define a function! It only relates two variables, in a way that can be viewed as two different functions (assuming it is invertible), depending on which variable you take as the dependent variable. I think this is the key to your confusion. Draw the graph of y=x^3. This can be seen as a graph of the function f(t) = t^3 (using a dummy variable as you suggest next), by thinking of it as graphing y = f(x); or as a graph of the function invf(t) = t^(1/3) by taking y as the independent variable, x = invf(y). The graph is the graph of the equation; it is only the graph of the function IF you take x as the independent variable as we conventionally do. The whole concept of functions is meant to avoid this confusion, but if you confuse functions with equations, nothing makes sense. You must consistently see the variables in a function definition as dummies, so that you see f(x) = x^3 and f(y) = y^3 as the same function, applied to different variables. When you deal with formulas where the variables stand for specific quantities, not just arbitrary variables, you can't use the "dummy variable" concept. The variable "a" represents the acceleration, and can't be used for the force without horrible confusion. In this case, you solve for the desired variable WITHOUT changing the names of the variables, because the names MEAN something. That is, the inverse of the function F(a) = ma is not "a = mF" as you might think, but simply a(F) = F/m which we obtain by solving F = ma for a. If you had managed to think of a as a force and F as an acceleration without your head exploding, I would be worried about you! If you have any further questions, feel free to write back. - Doctor Peterson, The Math Forum http://mathforum.org/dr.math/ Date: 07/20/2002 at 09:19:19 From: Rich Subject: Thank you (Inverse functions (this is long)) I would like to thank Dr. Peterson very very much, I greatly appreciate his timely response and analysis. The terminology and explanation has solidified my understanding. I believe my confusion was, as you pointed out, that "An equation does not define a function!" This is very important and I have never thought of it that way. It could be because I never read that somewhere or was never taught that or most importantly never had to think of them that way until now. Again thank you so much. Date: 07/20/2002 at 22:47:53 From: Doctor Peterson Subject: Re: Thank you (Inverse functions (this is long)) Hi, Rich. Thanks for writing back. This is definitely a subtle matter, and I would not be surprised if it is often taught without making all the distinctions clear. I wasn't sure if I had managed to do so; it can be very hard to communicate something that you just "see", when someone else doesn't see it that way. So it's helpful to know that what I said did so! - Doctor Peterson, The Math Forum http://mathforum.org/dr.math/ |
Search the Dr. Math Library: |
[Privacy Policy] [Terms of Use]
Ask Dr. Math^{TM}
© 1994-2015 The Math Forum
http://mathforum.org/dr.math/