Finding the domain such that g(x) has an inverse

This is probably similar to a question I recently asked, but I can't quite make he connection.

**Question:**

$\displaystyle g(x) = \frac{3x}{5+x^2}$

Given that the domain of *g* is $\displaystyle x \geqslant a$, find the least value of *a* such that *g* has an inverse function.

Drawing a graph of the function, I notice that the first positive maximum is $\displaystyle \sqrt{5}$, which incidentally also happens to be the answer to the question. Why is this?

Thank you in advance.

Re: Finding the domain such that g(x) has an inverse

Just found this when searching for an explanation to this problem that was driving me crazy.

My problem now is that when you compute the inverse we have $\displaystyle g^-1(x) = \frac{3+sqrt(9-20x^2)}{2x}$.

This is confirmed in the markscheme.

Therefore when $\displaystyle x> \frac{3*sqrt(5)}{10}$ the inverse is complex. I confirmed this by graphing the answer to part c from the markscheme on my TI84. It doesn't graph as it's graphed in the markscheme. As x approaches zero from the right g-1(x) approaches zero as well, not infinity.

Am I missing something here?