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.

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- Oct 6th 2008, 04:04 PMtiarFinding the inverse of f(x)
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. - Oct 6th 2008, 04:21 PMo_O
Let $\displaystyle y = f^{-1}(x)$ and change the x's and f(x)'s:

$\displaystyle x = \sqrt{y-7}$

Simply solve for y. - Oct 6th 2008, 04:42 PMtiar
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.. - Oct 6th 2008, 05:01 PMo_O
You're just solving for y in terms of x:

$\displaystyle \begin{array}{rcl} x & = & \sqrt{y - 7} \\ x^2 & = & y - 7 \\ & \vdots & \end{array}$

There's only one step left to do ... - Oct 6th 2008, 06:16 PMtiar
$\displaystyle y+x^2=-7$

$\displaystyle y=-x^2-7$ ??

Then turn it into...

$\displaystyle f^-1(x)=-x^2-7$??

Is that it? Things are always much simpler than they seem, lol. Thank you! - Oct 6th 2008, 06:25 PM11rdc11
- Oct 6th 2008, 06:25 PMo_O
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 $\displaystyle x \geq 0$ - Oct 6th 2008, 06:34 PMtiar
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.

- Oct 6th 2008, 06:45 PM11rdc11
When finding inverses just switch the domain and range.

So if f(x) domain is A and f(x) range is B then $\displaystyle f^{-1}(x)$ domain is B and $\displaystyle f^{-1}(x)$ range A

Does this make sense? - Oct 6th 2008, 07:06 PMtiar
Yes, that makes perfect sense, thanks.