I am looking for help in regard to a practice question about functions. The question is
Show that a function $f$, defined by $f(x)=x/\sqrt{x^2+1}$ , $x \in \Bbb R$ is a bijection of $\Bbb R$ onto $\{ y: -1<y<1\}$.
So what I know that for it to be a bijection, it must be an injection and also a surjection.
So to proof this question, do I just need to prove both of those?
So for injection, when $x_1=x_2$ $f(x_1)=f(x_2)$
To do this I wrote $$\frac{x_1}{\sqrt{x_1^2+1}}=\frac{x_2}{\sqrt{x_2^2+1}},$$ squared both sides and expanded to solve that $x_1=x_2$.
Next for a surjection, must show that the range is contained,
so the bottom cannot be $0$ or negative because cannot square root a negative and cannot divide by $0$.
I believe in this situation you are supposed to write it as $y=x/\sqrt{x^2+1}$ and solve for $x$ in terms of $y$ but I have trouble doing that. Or can I jsut do it by solving an inequality such as $\sqrt{x^2+1}>0$,
Best Answer
Injective: suppose that $$\frac{x_1}{\sqrt{x_1^2+1}}=\frac{x_2}{\sqrt{x_2^2+1}}\ .\tag{$*$}$$ Squaring both sides and multiplying out denominators, $$x_1^2(x_2^2+1)=x_2^2(x_1^2+1)\quad\Rightarrow\quad x_1^2=x_2^2\ .$$ Now substituting back into the denominator on the RHS of $(*)$, $$\frac{x_1}{\sqrt{x_1^2+1}}=\frac{x_2}{\sqrt{x_1^2+1}}\quad\Rightarrow\quad x_1=x_2\ .$$ Doing it this way avoids having to (explicitly) consider whether or not $x_1$ and $x_2$ have the same sign.
Surjective: we want to solve $$y=\frac{x}{\sqrt{x^2+1}}\tag{$*\!*$}$$ for any $y\in(-1,1)$. Easy algebra gives $$x^2=\frac{y^2}{1-y^2}\ ,$$ and we now have to consider which square root to take in order to get the right $x$. It is not hard to see that $y$ should have the same sign as $x$, so we guess $$x=\frac{y}{\sqrt{1-y^2}}\ .$$ Note the word "guess": if you stop here, the solution is logically backwards and therefore incorrect. We need to actually substitute this expression for $x$ into the RHS of $(**)$ and confirm that it simplifies to $y$. This is easy, so I leave it to you.