finding the inverse

[Craig is Lege]

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

f(x)= x^2 + 2x -1

2. Relevant equations



3. The attempt at a solution

x + 1 = y ( y + 2 )

how do i get y alone

[physicsnoob93]

Firstly, if you inverse this quadratic function straightaway, you wouldn't get a function anymore. This is because the definition of a function is that every element in the set of the domain will have a corresponding image in the range. (This simply means that every x value must have one corresponding y value.)

So when we consider just the quadratic function, it is a function right? Because every x value has only one y value to it.
So what if we inverse it? Then every x value would have 2 y values. This doesn't make it a function anymore. You would have to limit the domain of the quadratic function such that when you inverse it, no x value has 2 y values.
The next thing ur going to think about is in what situation the x value has 2 y values for an inverse. This is when the graph has a mirror image for any points along a horizontal line right? In other words, any vertical line would cut the graph more than once, and it is not a function.
This happens to quadratic functions at their minimum or maximum point.

After you've thought about all this, find the minimum or maximum point. Limit the domain to negative infinity to the minimum point or the maximum point to infinity.

Find the inverse.

x^2+2x-1= 0.
So your function is: y = x^2+2x-1
the first thing you'll do is swap all the x with the ys because when you find the inverse its like you're swapping the axes.

Then your function becomes: x = y^2+2y-1
For quadratic functions, its very easy to complete the square when u find the inverse.
Complete the square:
x = [(y-1)^2-1^2] - 1
= (y-1)^2 - 2

The last step is then to express y as the subject:
x + 2 = (y-1)^2
sqrt(x+2) + 1 = y
And lastly, don't forget the +/- in your square root and the domain of the function. The +/- of the square root depends on the domain that u find from finding the minimum point of the function before inversion.

[Craig is Lege]

thanks for the help...totally forgot about that trick.