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Show that the function $f$ defined by $f(x):=\frac{x}{\sqrt{x^2+1}}\;,$ $x$ is an element of the reals, is a bijection of the reals onto $\{y:-1.

So we need to show that it is 1-1 and onto. I begin by trying to prove that it is 1-1:

$\frac{x}{\sqrt{x^2+1}} = \frac{y}{\sqrt{y^2+1}}$

solving, we get $x^2 = y^2$

How can we now prove that $x=y$? I think there must be two solutions but am not sure what to do.

Also, I am not entirely sure where I should begin to prove that it is onto.

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Once you know $x^2 = y^2$, you know that either $x = y$ or $x = -y$. Plug both into the original equation and you'll know that $x = -y$ is impossible.

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    Uh, I thought the question was all settled. I claim that $x = -x$ has only one solution which is $x = 0$. You can find various ways to "prove" it, depending on what axioms you accept.2012-08-24
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Alternately, you can look at the derivative:

$f'(x)=\frac{1}{\sqrt{x^2+1}} - \frac{x}{2}(x^2+1)^{-\frac{3}{2}}(2x) = \frac{x^2+1 - x^2}{(x^2+1)^{\frac{3}{2}}} = \frac{1}{(x^2+1)^{\frac{3}{2}}}>0$

Therefore $f(x)$ is strictly monotone increasing and is therefore one-to-one.

All that you have to do now is show that $f$ is onto: given some $y\in (-1,1)$, we need to find $x$ such that $f(x)=y$, or in other words, $y \sqrt{x^2+1} = x$

Just find all values of $x$ that satisfy this equation for a given $y$.

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    @mixedmath Seeing that yo$u$ are a moderator, it wo$u$ld have been more proper to omit the last sentence of your comment. (As an aside, I do not quite understand that you recommend to *remain civil at all times especially with first time users*... to a first time user.)2012-08-24
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In the proof that the function is 1-1 you are not using some important information. Suppose you know that $x^2=y^2$. What additional information does $x/\sqrt{x^2+1}=y/\sqrt{y^2+1}$ give you?

To show that the function is onto, unfortunately the supposed range of the function is not correctly displayed. But the range is $(-1,1)$, right? To see this, compute the limits of the functions as $x$ tends to $-\infty$ and to $\infty$ and think about what happens in between. Use the intermediate value theorem.

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    @SamuelGregory - No, it isn't.2012-08-24