Showing that a function is bijective 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 $\{y:-1<y<1\}$.
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. 
 A: 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.
A: 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 $y$.
A: 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.
