# Finding the inverse of f(x)

• Oct 6th 2008, 04:04 PM
tiar
Finding 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 PM
o_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 PM
tiar
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 PM
o_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 PM
tiar
$\displaystyle y+x^2=-7$
$\displaystyle y=-x^2-7$ ??
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 PM
11rdc11
Quote:

Originally Posted by tiar
$\displaystyle y+x^2=-7$
$\displaystyle y=-x^2-7$ ??
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!

Close it is

$\displaystyle x^2 + 7 = y$
• Oct 6th 2008, 06:25 PM
o_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 PM
tiar
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 PM
11rdc11
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 PM
tiar
Yes, that makes perfect sense, thanks.