Is there a function f(x) such that the derivative of f(x) is equal to the inverse of f(x)?

Prove it.

Is there a function f(x) such that the derivative of f(x) is equal to the inverse of f(x)?

Prove it.

this is kind of cheating but:

f(x)=0 where x=0

I assume you are asking this for a reason. Do you have a proof that no such function exists or something?

Interesting question.

—edit----

I was thinking he meant inverse function not 1/f(x)

Actually:

f(x)=0

f’(x)=0

f(x)^-1=1/0 which is a non-removable discontinuity. DNE

I dont think there is any, but then again I haven really taking time to look for a way.

Now you are gonna make me lose sleep on this.

Dave

The function you proposed (a constant) is not invertible, unfortunately.

I’m still giving this some thought, but I don’t think there’s such a function.

It is, indeed an interesting question. Does look like an assignment, though

More later…

[EDIT: David Guzman beat me to it…]

wait, when you say inverse, are you refering to 1/f(x) or the inverse function of f(x)

I was assuming he meant inverse function, but now that i think about it, he probably would have inserted the word “function” if that is what he meant. Please clarify this.

Judging by the question, phrontist seems to know his math. As such, if he meant 1/f(x), I believe he’d have used “the reciprocal” of f(x).

I might be wrong but:

f(x)=1

f’(x)=0

f(x)^-1=0/1=0

EDIT: OK, I am wrong

EDIT:

If you’re considering the reciprocal, then 1/1 is not equal to 0.

If you’re considering the inverse function, then I didn’t understand what you tried to do.

f(x)^-1 is a poor notation because it usually refers to the inverse of a function but can also be mistaken for the reciprocal of the function.

grr math things like this frusterate my simple mind. I am no math person and i don’t entirely understand why my solution is wrong. Maybie i am just stubborn. Lets define the inverse of a function as a being a function where f(x)=y and f^-1(y)=x for all x. By this logic if f(0)=0 then shouldn’t f^-1(0) be 0 as well and shouldn’t my solution work?

Rickertsen2,

For a function to be invertible, it has to be bijective - “one on one” and “onto”, that is, there’s only one *x* value associated with a given *y* value, and that holds for every x. As an example, y = x^2 is not invertible, because for x = ± 2, y = 4. To think of it in another way, if you were to invert that function (rotate its graph about the line y = x), you’d get a function that, for one value of x, is defined for two y values. That is not a function.

For a quick test, draw an horizontal line in this graph and sweep it up and down. If it ever touches two points on the curve, then the function is not invertible.

You can see that your constant function does not meet the requisites for an invertible function.

A Google search will provide thousands of good sources for information, this site is a nice one, with a simple explanation and pictures (who doesn’t like pictures? :p)

http://archives.math.utk.edu/visual.calculus/0/inverse.6/

I meant (and said) inverse, not reciprocal. This isn’t an assignment, just something that occured to me during my endless and tedious BC Caluclus (anyone else have Early Transcendentals?) homework.

I’ve been trying to approach the problem visually, the inverse of a function being that function “mirrored” about the line y=x. I think perhaps this is needlessly painful, I’m going to try some algerbraic manipulation of the definition of the derivative…

My guess is that no such function exists, but I wonder if there is an elegant proof of that.

You mean the Stewart book? Yep, had that. I’d say it’s not a bad book but there are some weird quirks every now and then where it won’t explain some things very well. Mine came with CDs but I’d say they were pretty much useless.

The following fact will lead you in the right direction.

Let f be a function that is differentialable on an interval I. Suppose that f has a defined inverse function, called g.

Using the defintion of an inverse, and the chain rule, it can be shown that

g’(x) = 1/f’(g(x)), where f’(g(x)) != 0

This should help with you algebraic manipulation.

You know what? I was tempted to try this problem, but my hatred of calculus overcame me while I was working some stuff out. Calculus tends to have more letters than numbers. This is math, not grammar… These letters also tend to not be in the English alphabet. Again, this is math, not a foreign language.

Today’s episode of Sesame Street was brought to you by… The number e …And, the letter mu.

There also tends to be odd quirks. Take Gabriel’s Horn. How can you possibly have something with an infinite surface area, infinite cross-section, but a finite volume!? And the volume isn’t just finite, it happens to be a very specific number/non-English letter: pi.

Yes, I realize the power of calculus. I also respect it. Seeing it work in physics still amazes me. But, I just don’t like it.

This one is seriously making my head hurt. I recollect hearing a similar question proposed but my brain is too fuddled to remember the question in its entirety.

I say if 3rd semester Calc doesn’t cover it, then it just isn’t worth the headache.

…not yet, at least

Ok well I cannot find a way to definitly prove it but I can prove that this is never the case in the following cases.

x^n where n is positive

x^n where n is negative

all linear functions,

All absolute value functions

x^n when x is not an integer

All trig functions

All step (integer) functions of x

Thats all I have time for now, Back to AP chem.

lets call f^-1(x) the inverse and (f(x))^-1 the recipricol

anyway

Surely if you have f(x) which has powers in, when you differentiate it the power will decrease by one (unless your using e^x)

reflecting it in y=x inverts the power

(eg f(x)=2x^2+3 f^-1(x) = (x-3)^(1/2) / 2

as such differentiating it only lowers the power by 1

so you can’t use integers as when you lower the power it won’t be afraction

if you want to try fractions, a/b - 1= b/a which the only thing that that has an equal distance from 1 on either side is 1/2 which would give 3/2 -1 (not equals) 2/3 so that can’t work

so not fractions

i’m wondering maybe something hyperbolic or complex?

C’mon, guys, I haven’t had calculus yet and I can tell you that there is a function like this. A function’s inverse is its reflection in the line y=x, right? A function is any line or relation where any input has one and only one output, right? So a straight diagonal line is a function, correct? Now, the only function that will equal its own inverse has to have a slope of -1. So, any function whose equation reads something like f(x)=-x+b with any value for b will be its own inverse. Was that so hard?

Well, you didn’t read the question with enough attention. Phrontist wants a function such that its **derivative** is equal to its inverse, not the function itself.

No, not at all. My previous example of y = x^2 would not be a function if you were correct. You described a bijective function, but there are also injective and, the ones that disproves you, surjective functions.

Yeah, that is correct if you disregard the derivative part of the problem.

I stand corrected.