SOLUTION: Please help me with this problem. From what I know, if a function f(x) is going to have an inverse then f(x) must 1-to-1. However the function below is not 1-to-1, but the problem

Algebra ->  Functions -> SOLUTION: Please help me with this problem. From what I know, if a function f(x) is going to have an inverse then f(x) must 1-to-1. However the function below is not 1-to-1, but the problem       Log On


   

Question 1196874: Please help me with this problem. From what I know, if a function f(x) is going to have an inverse then f(x) must 1-to-1. However the function below is not 1-to-1, but the problem is asking for it's inverse.
Would appreciate your guidance.
Find the inverse of a function:
f%28x%29+=+1%2Fx%5E2
Answer by math_tutor2020(3817) About Me  (Show Source):
You can put this solution on YOUR website!

If you were to graph f%28x%29+=+1%2F%28x%5E2%29, then you get this
graph%28400%2C400%2C-7%2C7%2C-7%2C7%2C0%2C1%2F%28x%5E2%29%29
Horizontal asymptote: x axis
Vertical asymptote: y axis
The green curve approaches each asymptote but never touches them. I like to think of it like an electric fence.

The graph fails the horizontal line test since it's possible to pass a single horizontal line through more than one point on the green function curve.

Therefore, f(x) is not one-to-one.

To get the inverse, the function must be one-to-one. So we must restrict the domain such that the function becomes one-to-one and hence passes the horizontal line test.

There are technically infinitely ways to do this. There isn't one single formula to follow. Usually we try to get the largest curve possible, i.e. avoid adding too much restriction. Also, many conventions tend to go positive domain values rather than negative domain values. It will depend on context.

With those two ideas in mind, one possible way to restrict the domain is to go with x > 0 which is the branch on the right hand side.

Let's now find the inverse
The idea is to swap x and y and solve for y
f%28x%29+=+1%2F%28x%5E2%29

y+=+1%2F%28x%5E2%29

x+=+1%2F%28y%5E2%29

xy%5E2+=+1

y%5E2+=+1%2Fx

y+=+sqrt%281%2Fx%29
Recall we made x > 0 the restricted domain of the original function.
The graph above shows the original function has a range of y > 0

Since the x and y values swap when finding the inverse, we'll swap the domain and range.
Domain of the inverse: x > 0
Range of the inverse: y > 0

Answer: The inverse is