Finding the inverse of f(x)

**tiar** · October 6th 2008, 04:04 PM

The given function f is one-to-one. Find f-1.
$f(x)=\sqrt{x-7}$

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.

**o_O** · October 6th 2008, 04:21 PM

Let $y = f^{-1}(x)$ and change the x's and f(x)'s:
$x = \sqrt{y-7}$

Simply solve for y.

**tiar** · October 6th 2008, 04:42 PM

Thanks, I understand that part. Afterwards, however, do I move the squareroot to the x? I tried...
$x^2=(y-7)$
$x=(y+7)^2$

So is that the inverse (incomplete answer)? I don't believe that's right. i'm just not getting something in the algebra. ugh..

**o_O** · October 6th 2008, 05:01 PM

You're just solving for y in terms of x:
$\begin{array}{rcl} x & = & \sqrt{y - 7} \\ x^2 & = & y - 7 \\ & \vdots & \end{array}$

There's only one step left to do ...

**tiar** · October 6th 2008, 06:16 PM

$y+x^2=-7$
$y=-x^2-7$ ??
Then turn it into...
$f^-1(x)=-x^2-7$ ??

Is that it? Things are always much simpler than they seem, lol. Thank you!

**11rdc11** · October 6th 2008, 06:25 PM

Originally Posted by tiar

$y+x^2=-7$
$y=-x^2-7$ ??
Then turn it into...
$f^-1(x)=-x^2-7$ ??

Is that it? Things are always much simpler than they seem, lol. Thank you!

Close it is

$x^2 + 7 = y$

**o_O** · October 6th 2008, 06:25 PM

You copied incorrectly.

$\begin{array}{rcl} x^2 & = & y - 7 \\ x^2 + 7 & = & y \qquad \text{Added 7 to both sides} \\ y & = & x^2 + 7 \end{array}$

However, note that the range of f(x) was $f(x) \geq 0$ . Since we interchanged x's and y's, the domain of $f^{-1}(x)$ is $x \geq 0$

**tiar** · October 6th 2008, 06:34 PM

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 $f(x) \geq 0$ because in the original equation, $f(x)=\sqrt{x-7}$ , 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 $f(x)= \geq7$ , correct? Wait, did I just answer my own question? I just need clarification.

**11rdc11** · October 6th 2008, 06:45 PM

When finding inverses just switch the domain and range.

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

Does this make sense?

**tiar** · October 6th 2008, 07:06 PM

Yes, that makes perfect sense, thanks.

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