# Thread: How to test whether a function is invertible or not without graphing it?

1. ## How to test whether a function is invertible or not without graphing it?

Let the real valued function $\displaystyle f : D \rightarrow \mathbb{R}$ be defined by:

$\displaystyle f(x) = \frac{x}{\sqrt{1-x^2}}$

If I was trying to determine if this function was invertible, I assume that to check if it was one to one/injective, I would take the first derivative:

$\displaystyle f'(x) = \frac{1}{(1-x^2)^{\frac{3}{2}}}$

So, $\displaystyle |x|$ has to be $\displaystyle < 1$ otherwise it is not defined. What does it mean that x cannot = all other values in the derivative? Either way, the function will always be positive which means it is always increasing and therefore one to one.

To check if the function is onto/surjective I would have to compare the codomain and the range. The codomain is $\displaystyle \mathbb{R}$ and the range would be a subset of that because x cannot = 1; the function is therefore not onto.

It is therefore not invertible. Is this reasoning sufficient?

Thanks.

2. ## Re: How to test whether a function is invertible or not without graphing it?

Originally Posted by terrorsquid
It is therefore not invertible. Is this reasoning sufficient?
Prove that $\displaystyle f'(x)>0$ for all $\displaystyle x\in (-1,1)$ that is, $\displaystyle f$ is strictly increasing and as a consequence injective. On the other hand $\displaystyle \lim_{x\to -1^+}f(x)=-\infty$ and $\displaystyle \lim_{x\to 1^-}f(x)=\infty$ . Hence by the intermediate value theorem for continuous functions, the range of $\displaystyle f$ is $\displaystyle \mathbb{R}$ . So, $\displaystyle f : (-1,1)\to\mathbb{R}$ is bijective.