# Inverse Function - algebra problem

1. Nov 30, 2011

### ZedCar

1. The problem statement, all variables and given/known data
I'm trying to find the inverse function of the following.

f(x) = [x^2 - 9]^0.5

x <= -3

3. The attempt at a solution

In the book I'm using it gives the inverse function as;

y = -(x^2 + 9)^0.5

[0, +∞)

Is this correct, as I'm getting an inverse function of;

y = [x^2 - 9]^0.5

[(18)^0.5, +∞)

2. Nov 30, 2011

### micromass

Staff Emeritus

3. Nov 30, 2011

### ZedCar

I've re-checked my work and I'm getting

y = (x^2 + 9)^0.5

Which is the same as the answer given in the book, but without the negative in front.

Someone was telling me that the square root returns the positive values of x but since the original range was negative, then the negative sign is required in front of the answer making the book answer correct.

What do you think?

4. Nov 30, 2011

### Staff: Mentor

What is the exact wording of the problem? In post #1 you have this:
The answer in the book would make sense if the original function happened to be f(x) = x2 - 9, x <= -3.

5. Nov 30, 2011

### ZedCar

The function given in the question is the square root of the function you have in your last post.

i.e. there's one radical covering the x^2 - 9

What do you think about the suggestion that was made to me about the square root returns the positive values of x but since the original range was negative, then the negative sign is required in front of the answer making the book answer correct?

6. Nov 30, 2011

### Staff: Mentor

On second thought, I agree with the book's answer (and have deleted an earlier response).

Starting with y = f(x) = (x2 - 9)1/2, x <= -3
The restricted domain here is (-∞, -3]. The range is [0, ∞)

If you solve the equation above for x, you'll have x = f-1(y). x will still be <= -3 and y will still be nonnegative. Once you switch variables, x will be nonnegative, and y will need to be <= -3.

Show us how you're getting from the equation above to your inverse.