Functions

[gene]

Ok. Let's say that there is a given function y=f(x)

And if your inverse of f(x) is exactly the same as f(x). What can you say about the symmetry of the graph of f(x)?

[Firedragon52]

Ok. Let's say that there is a given function y=f(x)

And if your inverse of f(x) is exactly the same as f(x). What can you say about the symmetry of the graph of f(x)?

When you say "inverse" do you mean y = 1/f(x) or (1/y) = f(x)?

[gene]

it is f^(-1) (x)

[gene]

the original equation is y=f(x)

 

but the inverse, which is supposedly suppose to reflect in the y=x line, happens to be y=f(x)

[Firedragon52]

it is f^(-1) (x)

So you're saying (1/f(x)) = f(x)? Sorry, I just want to make sure I understand you.

[gene]

Yup. the question appears so.

[Firedragon52]

Gee, thats strange. Unless that answer is always one, I don't see how (1/f(x)) = f(x). I think I'll leave this problem to a mathematician. Sorry...

[gene]

Thanks anyway. That qn came out for my exams.

[matt grime]

No, the inverse function is not 1/f.

 

If f is its own inverse (NOT RECIPROCAL) and you know the graph of f inverse is the graph of f reflected in the line y=x, then surely you can see that the graph of f must be unchanged afer reflection in that line.

[bloodhound]

the map f(x)->x from R to R is the only map which is self inverse

[gene]

No' date=' the inverse function is not 1/f.

 

If f is its own inverse (NOT RECIPROCAL) and you know the graph of f inverse is the graph of f reflected in the line y=x, then surely you can see that the graph of f must be unchanged afer reflection in that line.[/quote']

 

I'm not really sure of f inverse means 1/f. I don't really get you. Because apparently, the graph is not reflected that all.

 

Try this eqn.

 

y= (3x+11)/(x-3) , x is not 3.

 

if you find its inverse. You would get the exact same eqn.

[Primarygun]

Is the graph of the function inverted? Reflected in x or y axis or both?

[gene]

inverse functions reflect in the y=x .

 

No graph was given.

[matt grime]

If a function is self inverse then its graph is invariant under reflection in the line y=x. That IS the answer to your question.

 

And for whoeever said that f(x)=x is the only self inverse function, can I suggest f(x)=-x, or f(x)=1-x, or how about f(x)=2-x. Spotting a pattern? What about all the graphs of these? They're all invariant under reflection in the line y=x.

[gene]

Oh... it sounds like a great answer. Thanks.

[bloodhound]

 

And for whoeever said that f(x)=x is the only self inverse function' date=' can I suggest f(x)=-x, or f(x)=1-x, or how about f(x)=2-x. Spotting a pattern? What about all the graphs of these? They're all invariant under reflection in the line y=x.[/quote']oops sorry mate. i just remember doing group isomorphisms, and something about identity maps being self inverse, so i assumed (hastily) that it was the only one with that property

[matt grime]

The set of maps from R to R isn't a group under composition though, and even if it were there are still elements that square to the identity (if G is finite and |G| is even there are always non-identity elements that square to the identity).

[bloodhound]

ok , ill take your word for it. i am still a lilttle bit behind on my linear algebra module. probably the hardest module this term

[matt grime]

Well, why take my word for it? Let's prove it: if f(x) = 1 for all x, what is f^{-1}, the nominal inverse function? A function is invertible iff it is bijective. So obviously it fails to be a group.

 

If G is a finite group with even order, let S be a Sylow 2 subgroup. Let v be any non-identity element in S, then v has even order, say 2*r for some r. Then v^r has order 2.

[bloodhound]

lol. when i said ill take ur word for it i meant i am not up for that yet.

 

What is a sylow 2 subgroup by the way?

[matt grime]

Ah, that makes it a little harder then. Let G be any group, and suppose |G|=m.p^r, where p is a prime, and p doesn't divide m. Then G has a subgroup of order p^r called the Sylow p-subgroup. In fact it has potentially several of them, but the number of them satisfies a rule I can't recall precisely, they're all conjugate, and any subgroup of order p^s for some s is contained in some Sylow subgroup.

[bloodhound]

wow. thats a well amazing result.

[matt grime]

It is very strong, and very useful, though from what I recall it also doesn't have a proof that's at all worth memorizing, but it turns out to be very important in lots of areas.

[indignity]

the inverse is the opposite reciprocal... -1/f(x)

[matt grime]

erm, in what sense is the inverse of a functio the opposite of the reciprocal? The inverse of x^2 as a function from R+ to R+ is not -1/x^2 is it?