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Assume that $f\colon\mathbb{E}\subset\mathbb{R}\rightarrow\mathbb{R}$, $f(x)=\max\lbrace2x-5,x-2\rbrace$.

Determine $\mathbb{E}$ for which $f$ is a bijection.

I was thinking it is $\mathbb{R}$, but I'm not sure. Can I get a confirmation and a way to prove it, please? Thank you very much!

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2 Answers 2

up vote 2 down vote accepted

I'd recommend graphing the two lines, then using the graph to determine the graph of $f$. It should be very clear that $\mathbb{R}$ is, in fact the answer.

It should also break $\mathbb{R}$ into four component intervals (two vertical, two horizontal). You'll have to show that the first horizontal interval maps bijectively to the first vertical interval, and the second likewise. Does that help?

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The function seems to be constant $x-2$ until $x=3$, when it starts being $2x-5$. However, it is a bijection whichever x might be, so I suppose the answer is $\mathbb{R}$. –  Grozav Alex Ioan May 26 '12 at 14:18
    
@GrozavAlexIoan: Exactly, and for proof, you can show (for instance) that $(-\infty,3]$ is mapped bijectively to $(-\infty,1]$ and $(3,\infty)$ is mapped bijectively to $(1,\infty)$. You could also have the first pair open and the second pair closed. Another approach would be to write $f$ as a piecewise function, then determine the piecewise function $g$ that is its inverse. The last may be the simplest way to proceed, actually, depending on what you're used to. Ultimately, it all amounts to the same thing. –  Cameron Buie May 26 '12 at 14:31

As Cameron noted, think of two lines below. It is clear that your function breaks into two rules at $x=3$. Isn't it a bijection?

enter image description here

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Nice graph! + 1 :-) –  amWhy Mar 5 '13 at 0:18

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