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

  • Thread starter Craig is Lege
  • Start date
  • Tags
    Inverse
In summary, the conversation discusses the process of finding the inverse of a quadratic function. It emphasizes the importance of limiting the domain to ensure that the inverse is a function. The conversation also provides steps for finding the inverse and reminds to consider the +/- in the square root and the domain of the function.
  • #1
Craig is Lege
8
0

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
 
Physics news on Phys.org
  • #2
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.
 
  • #3
thanks for the help...totally forgot about that trick.
 

1. What is the inverse of the function f(x)= x^2 + 2x -1?

The inverse of a function is a new function that swaps the input and output values of the original function. In other words, if the original function maps x to y, the inverse function maps y to x.

2. How do you find the inverse of f(x)= x^2 + 2x -1?

To find the inverse of a function algebraically, you can follow the steps of:

  1. Replace f(x) with y.
  2. Switch the x and y variables.
  3. Solve for y.
  4. Replace y with f^-1(x) to represent the inverse function.

3. Can you use a graph to find the inverse of f(x)= x^2 + 2x -1?

Yes, you can also find the inverse of a function graphically by reflecting the points of the original function over the line y=x. The resulting graph will be the inverse function.

4. Is the inverse of f(x)= x^2 + 2x -1 a function?

Yes, the inverse of a function is also a function as long as the original function passes the horizontal line test, meaning that no horizontal line intersects the graph of the function more than once.

5. What is the domain and range of the inverse of f(x)= x^2 + 2x -1?

The domain and range of the inverse function will be the range and domain of the original function, respectively. In this case, the domain of the inverse function will be all real numbers (since the range of the original function is all real numbers), and the range of the inverse function will be all real numbers greater than or equal to -1/4 (since the domain of the original function is all real numbers greater than or equal to -1/4).

Similar threads

  • Precalculus Mathematics Homework Help
Replies
7
Views
396
  • Precalculus Mathematics Homework Help
Replies
3
Views
1K
  • Precalculus Mathematics Homework Help
Replies
15
Views
637
  • Precalculus Mathematics Homework Help
Replies
2
Views
517
  • Precalculus Mathematics Homework Help
Replies
11
Views
855
  • Precalculus Mathematics Homework Help
Replies
10
Views
578
  • Precalculus Mathematics Homework Help
Replies
10
Views
1K
  • Precalculus Mathematics Homework Help
Replies
4
Views
517
  • Precalculus Mathematics Homework Help
Replies
11
Views
2K
  • Precalculus Mathematics Homework Help
Replies
12
Views
484
Back
Top