Inverse of a Function: f(x) = 2x^2 +2x - 1

[geofox]

Hello, everyone!
I need help finding an inverse of a following function:

f(x) = 2x² +2x - 1

I was able to do the following:
x= 2y² +2y - 1
x-1 = 2y² +2y
(x-1)/2=y² +y

and I have no idea what to do next
I tried doing so:
(x-1)/2=y(y+1)
but that doesn't give me anything

Thank you!

 

[lookagain]

Hello, everyone!
I need help finding an inverse of a following function:

f(x) = 2x² +2x - 1

I was able to do the following:
x= 2y² +2y - 1
x-1 = 2y² +2y
(x-1)/2=y² +y

and I have no idea what to do next
I tried doing so:
(x-1)/2=y(y+1)
but that doesn't give me anything

Thank you!

geofox,

it doesn't have an inverse function. It is not one-to-one.

The graph fails the horizontal line test.

 

[stapel]

I need help finding an inverse of a following function:

f(x) = 2x² +2x - 1

As you know from the graphing you've done, this is a parabola, so it doesn't pass the Horizontal Line Test, so it does not have an inverse which is itself a function. But you can come up with a relation (a formula) for the inverse. For this sort of function, there's a trick:

You've got a quadratic:

. . . . .$\displaystyle y\, =\, 2x^2\, +\, 2x\, -\, 1$

You want to solve this for "x=", and then swap the variables. So let's start by getting everything over on one side:

. . . . .$\displaystyle 0\, =\, 2x^2\, +\, 2x\, -\, 1\, -\, y$

. . . . .$\displaystyle 0\, =\, 2x^2\, +\, 2x\, -\, (1\, +\, y)$

Now you can apply the Quadratic Formula, using:

. . . . .$\displaystyle a\, =\, 2,\, b\, =\, 2,\, c\, =\, -(1\, +\, y)$

Yes, the result will be lumpen, but it'll give you the relation that you need. Then swap the variables. You'll then have a formula with two halves, corresponding to the "plus" and the "minus" on the square root.

 

[pka]

I need help finding an inverse of a following function:
f(x) = 2x² +2x - 1

While this is not a one-to-one function, there are many times we can restrict the domain to achieve the result.
Consider the "variable swap" . $\displaystyle x=2y^2+2y-1\text{ or }2y^2+2y-(1+x)=0$.

Then $\displaystyle y=\dfrac{-2\pm\sqrt{4-4(2)[-(x+1)]}}{2(2)}$. So what restriction on that domain does one need to have an inverse?
Please show us what you find.