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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
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

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

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/
```
Associated Topics:
College Calculus
High School Calculus
High School Functions

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