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According to a problem that I am working on a fixed point is defined as a point $x$ that satisfies $$f(x) = x.$$

A problem is asking me to find the fixed point of the inverse of $f(x)$, and the explanation tells me to find the inverse function first, and later use the definition.

Is this step necessary? I want to say no because the inverse of a function has an image that is reflected about the line $y=x$, so finding the fixed point of the original function should give you the fixed point of the inverse as well.

Is there an exception that I am not aware about?

dfeuer
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hyg17
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  • It's hard to say without context but 'I want to say no because the inverse of a function has an image that is reflected about the line y=x' is just geometric gibberish. I'm guessing they're looking for a proper proof. – Git Gud Sep 10 '13 at 17:40
  • @GitGud It's almost a complete proof. For $f:\mathbb R\to \mathbb R$ the set of fixed points is just a projection onto OX or OY of the intersection of the graph with the line $y=x$. Why attack perfect geometric intuitions of the OP and substitute them with algebraic abstract nonsense? – savick01 Sep 10 '13 at 19:37
  • @savick01 lol at algebraic abstract nonsense. – Git Gud Sep 10 '13 at 21:22
  • @hyg17: note that your argument - the one formalized a bit in my previous comment - works perfectly also in a general situation of a map $f:X\to X$. You just take the diagonal ${(x,x)\ |\ x\in X}\subseteq X\times X$ which is a generalization of the line $y=x$. – savick01 Sep 10 '13 at 21:42

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I believe the OP meant to say "if $f: R \rightarrow R$ is a function, then the graph of its inverse is simply the graph of $f$ reflected about the line $y = x$."

I'm going to assume that the original function is bijective (i.e., "one-to-one and onto"), so that it actually has an inverse function, rather than an inverse relation.

Let $g$ be the inverse of the function $f$. Suppose that $x_0$ is a fixed point of $g$. That means that

$g(x_0) = x_0$

Applying the function $f$ to both sides of this equation, you get

$f(g(x_0)) = f(x_0)$

Since $f$ and $g$ are inverses, we know that $f(g(x_0)) = x_0$, so this equation can be rewritten

$x_0 = f(x_0)$.

In other words, I've shown that any fixed point of $g$ is also a fixed point of $f$. (Essentially the same proof shows that fixed points of $f$ are fixed points of $g$ as well.) So if the problem asks you to find fixed points of the inverse of $f$, you might as well just look for fixed points of $f$, i.e., don't bother with trying to invert it.

-John

John Hughes
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