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

1. ## 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:
$g(x) = \frac{3x}{5+x^2}$
Given that the domain of g is $x \geqslant a$, 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 $\sqrt{5}$, which incidentally also happens to be the answer to the question. Why is this?

2. there can only be an inverse if each x value corresponds to a single y value.

Your graph has a turning point a $x=\sqrt{5}$. starting from x=0:

as you increase x, y increases
after $x=\sqrt{5}$, y starts to decrease again

Can you see that this means there must be some y values that occur twice, once to the right of $x=\sqrt{5}$ and once to the left?

it might be clearer on this graph:
y&#61;3x&#47;&#40;5&#43;x&#94;2&#41; from -10,10 - Wolfram|Alpha

3. Originally Posted by mtpastille
This is probably similar to a question I recently asked, but I can't quite make he connection.
Question:
$g(x) = \frac{3x}{5+x^2}$ [Latex is misbehaving: g(x)=3x/(5+x^2)]
Use the $$...$$ tags
So that [TEX]g(x) = \frac{3x}{5+x^2}[/TEX]
becomes $g(x) = \frac{3x}{5+x^2}$.

4. Oh, so it's that simple! The horizontal line test springs to mind, that's what I'd forgotten. Thank you

5. ## 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 $g^-1(x) = \frac{3+sqrt(9-20x^2)}{2x}$.
This is confirmed in the markscheme.

Therefore when $x> \frac{3*sqrt(5)}{10}$ 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?