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

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?

• @drhab That's what I meant sorry, I'll correct it. Jun 30, 2018 at 13:47
• Please read this tutorial on how to typeset mathematics on this site. Jun 30, 2018 at 13:51
• Notice that the denominator gets very small as $x$ approaches either $1$ or $-1$, which allows the value of the function to become very large. Jun 30, 2018 at 13:52
• Yes, but what I'm asking is not if the function can produce very large numbers, but if the function can produce every single real number. And furthermore, can you prove to me that the x inputs needed to produce every single real number are ordered from smallest to largest? Why is it the case here that the larger the x-value you put in, the larger the y-value? Jun 30, 2018 at 13:59
• @EthanChan if $f(a)$ is very small and $f(b)$ then -because $f$ is continuous - the intermediate value theorem assures that $y=f(x)$ for some $x$ whenever $f(a)\leq y\leq f(b)$. Jun 30, 2018 at 14:02

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 $y\in \mathbb{R}$.

• Yes, but fortunately it is not needed to find an inverse for this. That can be quite difficult. Jun 30, 2018 at 14:04
• @dehab I also would use the surjectivity argument but the OP is asking for a solution without calculus concepts.
– user
Jun 30, 2018 at 14:07
• Yes. A recent comment of the OP learned me now that your answer is the best for him/her after all. Jun 30, 2018 at 14:09
• The implication that your arrows claim is not the one that is needed. Jun 30, 2018 at 16:18

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 $y=f(x)$.

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.

• and of course with some calculus you can see how the function cover all the real numbers by producing large numbers from the denominater. Jun 30, 2018 at 13:53

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.