# Inverse of a polynomial

1. Jan 10, 2012

### crybllrd

1. The problem statement, all variables and given/known data

The function

$y=x^{2}+4x-6$

has two inverses. What are they and which domains lead to these inverses?

2. Relevant equations

3. The attempt at a solution

$y=x^{2}+4x-6$

$x=y^{2}+4y-6$

$y(y+4)=x+6$

Not really sure where to go from here.

2. Jan 10, 2012

### Staff: Mentor

Write this equation as x2 + 4x - 6 - y = 0, and solve for x using the quadratic formula. That will give you x = f-1(y) (with some abuse of notation as f-1 is not a function).

This is no help at all.

3. Jan 10, 2012

### Mentallic

$$x^2+2x=3$$

and then taking the next step as follows, and getting stuck

$$x(x+2)=3$$

You need to factorize! Or if you can't, which will be the case if you have a number such as, say, 4 instead of the 3, then you need to use the quadratic formula. If the 3 is replaced with a constant or variable, such as

$$x^2+2x=k$$

then you definitely need to use the quadratic formula, applying all the same rules you know, but simply extending it to the realm outside of mere known constants.

4. Jan 10, 2012

### crybllrd

OK, I got it from here I believe. I'm on mobile, so I will work it out when I get home and post my answer.

5. Jan 10, 2012

### crybllrd

Alright, so I used the following for the quadratic formula:

$a=1, b=4, and, c=(-6-y)$

to get

$x=-2\pm2\sqrt{10+y}$

At this point, would I swap x and y to get the inverse of f(x)?

6. Jan 10, 2012

### Mentallic

Nearly, it should be

$$x=-2\pm\sqrt{10+y}$$

because when you factorize the 4 out of the surd, you need to take the root of that so what you would've had was

$$x=\frac{-4\pm\sqrt{4^2+4(6+y)}}{2}$$

$$x=\frac{-4\pm\sqrt{4(4+(6+y))}}{2}$$

$$x=\frac{-4\pm\sqrt{4}\sqrt{4+(6+y)}}{2}$$

$$x=\frac{-4\pm2\sqrt{4+(6+y)}}{2}$$

Yes but the inverse needs to be a function, and you can't possibly have that with a $\pm$ there. You need to restrict the domain of your original quadratic for there to be an inverse that's 1:1.

7. Jan 10, 2012

### crybllrd

OK, thank you.
I did have the quadratic right, I just typed it wrong.
I split it up into two functions from the +/-.
When I graph them, however, it does not look like it is mirrored over y=x.

EDIT: I figured it out, I didn't cancel out the 2 in the numerator.
All is well,
thanks again everyone~

Last edited: Jan 10, 2012