Inverse functions?

There's a nice graphic-related proof.A function that is invertible on a domain is bijective,which means both injective & surjective.In terms of graphs,the graph of the function and its inverse (if it exists) are symmetrical wrt the graph of f(x)=x.If the two functions coincide,then they coincide with the identity function.

Daniel.

 

Do you mean to say, y=x is the only function where [tex]f(x)=f^{-1}(x)[/tex]?

Proof by counterexample:

[tex]y=\frac{1}{x}[/tex]

 

Any function which is symmetrical from the line y=x is the inverse of itself, in that sense there will be an infinite number of them.

Although perhaps very few simple functions, another easy one is [tex]y^2 + x^2 = r^2[/tex]

 

The identity function is invertible on all R...

Daniel.

 

how about y = -x, y = -1/x,....

 

If your question was simply "prove of disprove: y = x is the only function that is its own inverse", you can just give a counterexample and be done. Is this how the question you were given was worded?

 

precalculus question

Hello,

I have a similar question; I'm currently taking precalculus, and I have a question here that I'm stuck on:

"Let f(x)=(ax+b)/(cx-a), (cx-a)?0. Show that f(x) is it's own inverse."

I understand, graphically, that f(x) equals its inverse in this case, because they are reflections of each other over y=x. I've graphically solved this using technology and by-hand sketches. And I also understand that the restiction on the denominator come about because the equation would then be undefined (if it equalled 0). But I cannot figure out why this is, algebraically.

Can anyone help me, or give me a hint?

Thanks very much,
sam.grotz

 

Just compute f(f(x)).

 

Sam, just start with [tex]y = \frac{ax+b}{cx-a}[/tex] and then rearrage it to get "x" by itself on the LHS.

Whats happening. Are you just getting stuck on the algebra to do this rearrangement?

 

[tex]y^2 + x^2 = r^2[/tex] is not a function, because for any value of x < r, you can get two different values of y.

 

A silly conjecture... An infinite number of counterexamples.

 

Thanks!

Thanks to ZioX and uart for giving me the helpful direction. I think what I must have done before was just think too far ahead, and presume that it couldn't be simplified in that way. Thanks again! It turns out it actually ends up simplifying to just:

y=x

peace,
sam.grotz

 

what is the inverse function of sin x. I don't know if it is written correctly

 

arcsin, or sin^{-1}

and if you think that it is avoiding the issue, then you need to rethink your view of what a function is.