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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.

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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.

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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.

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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

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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.

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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.

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