The given function f is one-to-one. Find f-1.
I can't seem to do the correct algebra to get the inverse. If someone could guide me a little, that would be wonderful. Thank you.
Thanks both of you. It was illogical of me to move the "y" over when it could have been isolated right away. At least I see that now.
I have a question in regards to range. You are saying it is because in the original equation, , we cannot have a negative under the square root thus it must be greater than zero. How do you know that this is your range and not domain? The domain would be , correct? Wait, did I just answer my own question? I just need clarification.