Inverse of a polynomial

[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.

[Mark44]

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
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).


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

y(y+4)=x+6
This is no help at all.
Not really sure where to go from here.

[Mentallic]

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.

That's similar to being asked to solve the quadratic

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.

[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.

[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}

That will give you x = f^{-1}(y)

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

[Mentallic]

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}
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}


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

[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~