The given function f is one-to-one. Find f-1.
$\displaystyle f(x)=\sqrt{x-7}$
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, I understand that part. Afterwards, however, do I move the squareroot to the x? I tried...
$\displaystyle x^2=(y-7)$
$\displaystyle x=(y+7)^2$
So is that the inverse (incomplete answer)? I don't believe that's right. i'm just not getting something in the algebra. ugh..
You copied incorrectly.
$\displaystyle \begin{array}{rcl} x^2 & = & y - 7 \\ x^2 + 7 & = & y \qquad \text{Added 7 to both sides} \\ y & = & x^2 + 7 \end{array}$
However, note that the range of f(x) was $\displaystyle f(x) \geq 0$. Since we interchanged x's and y's, the domain of $\displaystyle f^{-1}(x)$ is $\displaystyle x \geq 0$
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 $\displaystyle f(x) \geq 0$ because in the original equation, $\displaystyle f(x)=\sqrt{x-7}$, 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 $\displaystyle f(x)= \geq7$, correct? Wait, did I just answer my own question? I just need clarification.