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Inverting Functions

Date: 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 

  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 

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.)
   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 


  y = f(x) = x^3 


  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)), 


  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)


    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


    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

    x -----> x^3

Its inverse is the mapping

    x <----- x^3

that undoes it. This inverse can also be written as

    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 

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 

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 
Associated Topics:
College Calculus
High School Calculus
High School Functions

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