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

• May 1st 2011, 02:08 PM
mtpastille
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?

• May 1st 2011, 02:20 PM
SpringFan25
there can only be an inverse if each x value corresponds to a single y value.

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

as you increase x, y increases
after $\displaystyle x=\sqrt{5}$, 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 $\displaystyle x=\sqrt{5}$ 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
• May 1st 2011, 02:26 PM
Plato
Quote:

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

Use the $$...$$ tags
So that [TEX]g(x) = \frac{3x}{5+x^2}[/TEX]
becomes $\displaystyle g(x) = \frac{3x}{5+x^2}$.
• May 1st 2011, 02:55 PM
mtpastille
Oh, so it's that simple! The horizontal line test springs to mind, that's what I'd forgotten. Thank you :)
• Apr 2nd 2012, 09:17 AM
PhilMath
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?