Understanding the range of $f(x) = \frac{x}{\sqrt{1-x^2}}$

Question

Let's take the function $$f(x)=\frac{x}{\sqrt{1-x^{2}}}.$$

My question is, why is the range of the function is all real numbers?

Because doesn't the fact that the denominator must be $f(x)=\sqrt{1-x^{2}}$ and that the numerator must be $x$ limit the amount of values the function can produce? Because for every $x$, only one denominator value is possible. Doesn't this limit the output of this function, therefore preventing it from producing all real numbers? And furthermore, can you prove to me that the $x$ inputs needed to produce every single real number are ordered from smallest to largest? Basically, why is it the case here that the larger the $x$-value you put in, the larger the $y$-value?

Can someone please explain to me, as simply as possible and without calculus, why the range of the function is all real numbers? And furthermore, can you prove to me that the $x$ inputs needed to produce every single real number are ordered from smallest to largest? Basically, why is it the case here that the larger the $x$-value you put in, the larger the $y$-value?

Answer 1

We have that $f(x)\ge 0$ for $x\ge 0$ and $f(x)< 0$ for $x< 0$ and thus

$$y=\frac{x}{\sqrt{1-x^2}}\implies y^2(1-x^2)=x^2\implies x^2(1+y^2)=y^2\implies x=\frac{y}{\sqrt{1+y^2}}$$

which is defined for any $y\in \mathbb{R}$.

Answer 2

Shortly, let $x$ be a real numer, if you choose $t=\frac{x}{\sqrt{1+x^2}}$ (which is different from $1$), then your function's value on $t$ is $x$. This can be done to any real number $x$, therefore the function's range is the set of all real numbers.

Answer 3

Check that your function is defined and continuous on $]-1,1[$. Check where the function "starts" ($x\to -1$) and "ends" ($x\to 1$) and connect those "points" by a line, as you are allowed to do because your function is continuous.

Answer 4

The function is defined on $(-1,1)$ and is continuous.

Then in order to show that $\mathbb R$ is its range it is enough to prove that there are sequences $(x_n)_n$ and $(y_n)_n$ in $(-1,1)$ such that $f(x_n)\to+\infty$ and $f(y_n)\to-\infty$.

For this you can take $x_n=1-\frac1n$ and $y_n=-x_n$.

The intermediate value theorem then assures that for every $y\in\mathbb R$ we can find some $x\in(-1,1)$ that satisfies $y=f(x)$.