As a followup to this question. I'm trying to determine what must be true of the graph of $f$ in these cases. I've examined the two functions $f(x)= x$ and $f(x)= \frac{1}{x}$ and I'm not seeing any unifying graphic truth. Is there a graphic truth to be found for one-to-one function's where the inverse happens to be the same?
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I think the question asks what happens to the graph of $f$ if $f^{-1}(x) = f(x)$ – Darth Geek Aug 13 '14 at 15:33
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2Notice that $x \mapsto \frac{1}{x}$ is not the inverse of $x \mapsto x$. – Fly by Night Aug 13 '14 at 15:37
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2@Flybynight I don't think anyone is claiming that, rather $x$ and $1/x$ are their own inverses. – Jonas Dahlbæk Aug 13 '14 at 15:40
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HINT
The graphs $y=\mathrm{f}(x)$ and $y=\mathrm{f}^{-1}(x)$ are symmetric about the line $y=x$.
That means, to go from the graph $y=\mathrm{f}(x)$ to the graph $y=\mathrm{f}^{-1}(x)$, you reflect in the line $y=x$.
If the graphs $y=\mathrm{f}(x)$ and $y=\mathrm{f}^{-1}(x)$ are the same then what does that tell you?
Fly by Night
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What about function $f(x) = \frac{1}{x}$ which does not reflect on the line $y=x$? – James Aug 13 '14 at 16:15
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@James: $f(x)=\frac1x$ does reflect onto itself on the line $y=x$. Why do you write otherwise? – Rory Daulton Aug 13 '14 at 16:38
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Inversion of the graph corresponds to the operation $U(x,y)=(y,x)$. That is, the graph of the inverse of $f$ is $\{(f(x),x)\}$ (whereas the graph of $f$ is $\{(x,f(x))\}$). It follows that, in terms of graphs, inversion is reflection in the line $x=y$. To see this, consider the action of $U$ on the standard basis of $\mathbb R^2$ and compare to the action of reflection in the line $x=y$.
For the function $f(x)=1/x$, consider its graph. This is precisely its own reflection through the line $y=x$.
Jonas Dahlbæk
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Yes I see that, and this explains $f(x)=x$ nicely, but what about function $f(x) = \frac{1}{x}$ which does not reflect on the line $y=x$? – James Aug 13 '14 at 16:13
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