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Given $f(x)=(x-1)^2$, to make it injective and to obtain $f^{-1}(x)$, we need to restrict the domain, either from $(-\infty, 1]$ or $[1,\infty)$. Which is the larger domain?

I'm thinking that it's $(-\infty, 1]$ since $0$ is the midpoint of $(-\infty,\infty)$, it makes sense that it's the larger domain.

Is my reasoning sound or is it flawed?

Thanks.

Shaun
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Alfred
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    How "it makes sense"? It is correct, but for a much simpler reason: $;y=(x-1)^2;$ is an upwards parabola tangent at $;x=1;$ to the $;x$-axis and, thus, it is injective on any of its two branches: either the going-down one $;(-\infty,1];$ or the going-up one $;[1,\infty);$ – DonAntonio May 30 '14 at 14:00
  • @DonAntonio - It makes sense because the number of values from $(-\infty,1]$ is more than from $[1,-\infty)$. Am I correct in saying that the number of values for $(-\infty,0]$ is equal to $[0,\infty)$? – Alfred May 30 '14 at 14:04
  • Well @Alfred: no, the "number of values" in both cases is, in a very well defined and definite sense, exactly the same...and yes for your second question. – DonAntonio May 30 '14 at 14:08
  • @DonAntonio - If the second question is true, wouldn't that mean that range of values of $(-\infty,1]\neq [1,\infty)$? Since the values $[0,1]$ are "taken out" from $[0,\infty)$ and "put into" $(-\infty,0]$ to get $(-\infty,1]$? – Alfred May 30 '14 at 14:13
  • Oh, it may be that those ranges are different, @Alfred (though they are not, by the way), but that does not mean one has "more" values than the other. Within infinities realms things can go a little weird. – DonAntonio May 30 '14 at 14:15
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    @DonAntonio - I see, going to have to do more research, thanks for your help. – Alfred May 30 '14 at 14:24

1 Answers1

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Both domains are half-lines, so by any meaningful measurement they are the same size. Your argument is flawed because we want "size" to be reflection-invariant. Reflecting the half-line $[1,\infty)$ at the point $1$ gives the other half-line.

Also, we want "size" to be translation-invariant, so sliding $[1,\infty)$ to $[a,\infty)$ for any $a$ should give a set of the same "size". This seems paradoxical, since $[1,\infty)$ includes points that $[3,\infty)$ does not -- yet they are the same size. However this counterintuitive situation occurs all the time when dealing with infinite sets.

vadim123
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