Re: Calculus Query
 

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.

Re: Calculus Query
 

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?

Re: Calculus Query
 

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?

Re: Calculus Query
 

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.

Re: Calculus Query
 

I stand corrected.

Re: Calculus Query
 

but you can only actually call it a function if its 1-1 in theory...

Re: Calculus Query
 

Care to elaborate? Not sure if I understand what you wrote...

Re: Calculus Query
 

A 1-1 function (also written as one-to-one) is one that only goes down or up, not both, over its entire length. I don't understand why Denman says you can only call a function if it's 1-1 in theory, but he may know more than I do, which is quite likely. However, y=x^2 is a function, and it's not 1-1.

Re: Calculus Query
 

Well, that's not the definition of a one-to-one function; in general it is, indeed, a consequence of the function being one-to-one. However, it does not apply to every function. For example: y = x^3 would be one-to-one by your definition, but, if you consider its complex roots*, then it is not.

* - For the record, I never had much of a formal, theoretical education on complex numbers, just its practical applications. Therefore, I may be wrong on the above statement. ;)

Re: Calculus Query
 

sorry my bad i meant one-many, as long as its only one on the start so you cant get 2 values of f(x) for the same number.... however when you take the inverse (eg y=x^1/2) you need to limit your domain and so would need to limit the domain/range of the original function anyway

Re: Calculus Query
 

A one-to-one function is a function that has only one value of y (f(x)) for every value of x. You could also say that it passes the "horazantal line test" where if you were to draw horizantal lines, each would only pass though the graph once. One-to-one functions are important becuase they are functions that, when inversed, are still functions. y=x^2 is a function, but not one-to-one becuase f(1) and f(-1) have the same value. y=x^3 is one-to-one becuase no values of y repeat themselves. You can inverse y=x^3 and still have a function.
y=x^3 becomes x=y^3 when inversed.
x^(1/3)=y thus is also a function.
"One-to-one" comes from the fact that there is one value of x for every value y and one value y for every value x.
Nothing says that a functions has to be one-to-one. To inverse it and still have a function, yes. But a functions has to pass the virtical line test: if virtical lines were drawn, they would pass though the graph no more than once.
y^2=x passes the horizantal line test, so you can inverse it and have a function, but doesn't pass the virtical line test so it, in itself, is not a function.
The opposite is true of y=x^2. Is a function, but can't have the inverse be a function. Infact, it is the inverse of the above, but it isn't a function on the traditional x-y axis.

Things have been drifting, but I felt the need to clear the air about what is a function and a one-to-one function.

Re: Calculus Query
 

Well, turns out there is such a function.
;)

I took a more formal approach to the problem that didn't get me anywhere, so I talked to a buddy and here's what we found.
Explanation.
As you see, nothing formal. A mathematician will say it isn't a proof at all, but it does show that there's at least one said function.

PS - My statement on a previous post that y=x^3 was not bijective is wrong. :p

Re: Calculus Query
 

thats scary
do you know what 1/2(1+sqrt5) is!?
shudders
rep point for that

Re: Calculus Query
 

Amusing, isn't it? ;)
Of course, it isn't just a coincidence, it's a consequence of what we tried to do. We wanted an exponent n with the same characteristics as the Greeks wanted, just for different purposes. :D

Re: Calculus Query
 

Absolutely brilliant, Manoel! This is what I love about First. When phrontist in Virginia can put out a terrifically interesting question from Virginia, and have answers come in right on point from the UK and Brazil – well, we didn’t have anything like that going for us when I was in high school.

Manoel, however did you think of approaching the problem this way? You show extraordinary mathematical insight in choosing the right form for f(x). And phrontist – what made you think up this question? It’s absolutely first rate – you’ll do just fine at MIT when you get there.

For those who struggle with these concepts like I do, the inverse of a function exactly “undoes” what the original function did to the variable. If you perform a ‘function operation’ on a variable x, then follow up with the ‘inverse function operation’, you get the original x back.

So, if f(x) = 0.743 * x^1.618, and we take the first derivative,

then f’(x) = 0.743 * 1.618 * x^0.618 = 1.202 * x^0.618.

But, if f’(x) = Inv-f(x), we can substitute f(x) into the inverse function and get the original x back :

Inv-f(x) = 1.202 * ( f(x) )^0.618 =? x

Inv-f(x) = 1.202 * [ 0.743 * x^1.618] ^ 0.618 = 1.202 * 0.743^0.618 * (x^1.618)^0.618
= 1.202 * 0.832 * x = x, just what we started with.

So, the derivative of f(x) is the same as the inverse of f(x), just as Manoel showed us.

If we keep differentiating – which we can do as long as x >= 0, we can learn some interesting things about these functions, slopes, inflection points, concavity, etc.

f(x) = 0.743 * x^1.618
f’(x) = 1.202 * x^0.618
f’’(x) = 0.743 * x^-0.382
f’’’(x) = -0.284 * x^-1.382

From x=0 to infinity, the function f(x) increases continuously, as does the first derivative/inverse. But f(x) approaches infinity much faster than the derivative.

From f’(x) we can see that the slope of the function f(x) approaches infinity as x gets larger and larger. Eventually, f(x) is heading almost straight up.

From f ’’(x) we can see that the function is everywhere concave upwards, as f’’(x) is positive for all x>0. But f ’’(x) is the slope of the first derivative/inverse, so we can see that the inverse always has a positive slope. But f ’’(x) goes towards zero as x gets larger and larger, so the slope of the inverse goes to zero, i.e., the inverse function approaches a horizontal line. Not an asymptote, however – the inverse function can be made as large as desired by taking large enough values of x. There’s no upper limit for it, so no asymptote.

From f ’’’(x), which is always negative for all x>0, we can see that the inverse is always concave downwards.

(Manoel – you might want to re-check your graph of these functions. It’s always a good idea to double-check computer generated graphs.

Note that the function is numerically equal to its first derivative/inverse when x = 1.618.

A magical number, to be sure. More interesting properties than can be addressed here.

I haven't had this much fun since the last First competition.