gene Posted November 5, 2004 Share Posted November 5, 2004 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)? Link to comment Share on other sites More sharing options...
Firedragon52 Posted November 5, 2004 Share Posted November 5, 2004 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)? Link to comment Share on other sites More sharing options...
gene Posted November 5, 2004 Author Share Posted November 5, 2004 it is f^(-1) (x) Link to comment Share on other sites More sharing options...
gene Posted November 5, 2004 Author Share Posted November 5, 2004 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) Link to comment Share on other sites More sharing options...
Firedragon52 Posted November 5, 2004 Share Posted November 5, 2004 it is f^(-1) (x)So you're saying (1/f(x)) = f(x)? Sorry, I just want to make sure I understand you. Link to comment Share on other sites More sharing options...
gene Posted November 5, 2004 Author Share Posted November 5, 2004 Yup. the question appears so. Link to comment Share on other sites More sharing options...
Firedragon52 Posted November 5, 2004 Share Posted November 5, 2004 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... Link to comment Share on other sites More sharing options...
gene Posted November 5, 2004 Author Share Posted November 5, 2004 Thanks anyway. That qn came out for my exams. Link to comment Share on other sites More sharing options...
matt grime Posted November 5, 2004 Share Posted November 5, 2004 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. Link to comment Share on other sites More sharing options...
bloodhound Posted November 5, 2004 Share Posted November 5, 2004 the map f(x)->x from R to R is the only map which is self inverse Link to comment Share on other sites More sharing options...
gene Posted November 6, 2004 Author Share Posted November 6, 2004 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. Link to comment Share on other sites More sharing options...
Primarygun Posted November 6, 2004 Share Posted November 6, 2004 Is the graph of the function inverted? Reflected in x or y axis or both? Link to comment Share on other sites More sharing options...
gene Posted November 6, 2004 Author Share Posted November 6, 2004 inverse functions reflect in the y=x . No graph was given. Link to comment Share on other sites More sharing options...
matt grime Posted November 6, 2004 Share Posted November 6, 2004 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. Link to comment Share on other sites More sharing options...
gene Posted November 6, 2004 Author Share Posted November 6, 2004 Oh... it sounds like a great answer. Thanks. Link to comment Share on other sites More sharing options...
bloodhound Posted November 6, 2004 Share Posted November 6, 2004 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 Link to comment Share on other sites More sharing options...
matt grime Posted November 6, 2004 Share Posted November 6, 2004 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). Link to comment Share on other sites More sharing options...
bloodhound Posted November 6, 2004 Share Posted November 6, 2004 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 Link to comment Share on other sites More sharing options...
matt grime Posted November 6, 2004 Share Posted November 6, 2004 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. Link to comment Share on other sites More sharing options...
bloodhound Posted November 6, 2004 Share Posted November 6, 2004 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? Link to comment Share on other sites More sharing options...
matt grime Posted November 6, 2004 Share Posted November 6, 2004 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. Link to comment Share on other sites More sharing options...
bloodhound Posted November 6, 2004 Share Posted November 6, 2004 wow. thats a well amazing result. Link to comment Share on other sites More sharing options...
matt grime Posted November 6, 2004 Share Posted November 6, 2004 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. Link to comment Share on other sites More sharing options...
indignity Posted November 6, 2004 Share Posted November 6, 2004 the inverse is the opposite reciprocal... -1/f(x) Link to comment Share on other sites More sharing options...
matt grime Posted November 6, 2004 Share Posted November 6, 2004 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? Link to comment Share on other sites More sharing options...
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