Choosing the right sign for inverse functions?

Question

If I have to find an inverse function and through the algebra I get a $\pm$ sign how do I know which one to choose from if its in a given interval? For example a question asks:

The function $\left(\frac1{x^2}\right)$ is a one to one function on the interval $(0,\infty$) Find a formula for the inverse function on that domain. Sketch the graph of both functions from $0 < x < 4$

So I found the inverse function by doing this...

$$y = \left(\frac1{x^2}\right)$$ $$ x^2 = \left(\frac1y\right)$$ $$ x = \pm\left(\frac1{\sqrt{y}}\right)$$ $$ y = \pm\left(\frac1{\sqrt{x}}\right)$$

So now I have two inverse equations? How do I know which one to use?

Answer 1

When you want to find an inverse function, the inverse must exist, ie the function must be bijective.

Now since it is bijective on $(0, \infty)$, the inverse exists. Notice that $x>0$, so $x=\frac{1}{\sqrt{y}}$.

Answer 2

To summarize what has been said about this:

As Jonas Meyer noted in a comment, when you apply the inverse function the result has to be in the original function's domain. Since $-\frac{1}{\sqrt y} \not\in (0, \infty),$ that cannot be the correct answer, and there's only one possible answer remaining.

In order to prove that that the only possible remaining answer actually is an answer, you must show that the proposed inverse function is defined everywhere in its domain; in this case it must be defined on $(0,\infty).$ You can do this by showing that the original function is bijective (as explained in mich95's answer).

(Note that $y = \pm\left(\frac1{\sqrt{x}}\right)$ is not equivalent to any of the other equations and should not even be listed here.)