Finding Inverse of f(x)= x^2 + 2x -1

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To find the inverse of the function f(x) = x^2 + 2x - 1, it is essential to recognize that the quadratic function is not one-to-one, as it produces two y-values for each x-value beyond its vertex. To create a valid inverse, the domain must be restricted to either the left or right of the vertex, ensuring that each x corresponds to a single y. The process involves swapping x and y, leading to the equation x = y^2 + 2y - 1, which can be solved by completing the square. After rearranging, the inverse can be expressed as y = ±√(x + 2) - 1, with the appropriate sign determined by the chosen domain. This method highlights the importance of domain restriction when dealing with quadratic functions in inverse calculations.
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Homework Statement



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

Homework Equations





The Attempt at a Solution



x + 1 = y ( y + 2 )

how do i get y alone
 
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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.
 
thanks for the help...totally forgot about that trick.
 
I tried to combine those 2 formulas but it didn't work. I tried using another case where there are 2 red balls and 2 blue balls only so when combining the formula I got ##\frac{(4-1)!}{2!2!}=\frac{3}{2}## which does not make sense. Is there any formula to calculate cyclic permutation of identical objects or I have to do it by listing all the possibilities? Thanks
Since ##px^9+q## is the factor, then ##x^9=\frac{-q}{p}## will be one of the roots. Let ##f(x)=27x^{18}+bx^9+70##, then: $$27\left(\frac{-q}{p}\right)^2+b\left(\frac{-q}{p}\right)+70=0$$ $$b=27 \frac{q}{p}+70 \frac{p}{q}$$ $$b=\frac{27q^2+70p^2}{pq}$$ From this expression, it looks like there is no greatest value of ##b## because increasing the value of ##p## and ##q## will also increase the value of ##b##. How to find the greatest value of ##b##? Thanks
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