Inverse Function - algebra problem

[ZedCar]

Homework Statement

I'm trying to find the inverse function of the following.

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

x <= -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, +∞)

 

[micromass]

Your answer is sadly incorrect. Can you show what you did to obtain your answer? Then we can tell you where you went wrong.

 

[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?

 

[Mark44]

What is the exact wording of the problem? In post #1 you have this:

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

x <= -3

The answer in the book would make sense if the original function happened to be f(x) = x² - 9, x <= -3.

 

[ZedCar]

The answer in the book would make sense if the original function happened to be f(x) = x² - 9, x <= -3.

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?

 

[Mark44]

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.