Graph and Find Inverse Function for f(x) = sqrt(x^2 - 2x)

[999iscool]

Homework Statement

Graph f(x) = sqrt(x^2 - 2x), and find an interval on which it is one-to-one. Find the inverse of the function restricted to that interval.

Homework Equations

The Attempt at a Solution

What I can't do is really finding the inverse function. It seems very simple, but somehow I got stuck in the process.

swap x and y in the original function
y = sqrt(x^2-2x)
x = sqrt(y^2-2y)
and solve for y

so i did
x^2 = y^2-2y, and i tried to factor out y
x^2 = y(y-2)
x^2/y-2 = y
now i am really stuck. how can i pull that y out?

Thank you for any kind of help!

---- edited

I was thinking about this formula: d/dy f-1(x) = 1/ f ' (y)
i guess i can then integrate the d/dy f-1(x) and get f-1(x)?

so i started working again
f ' = (1/2 (x^2-2x) ^-1/2) * 2x-2
so 1/f ' =[ 2 (x^2-2x)^1/2 ]/ 2x-2
which is d/dy f-1

but the integration doesn't work!

 

[Mark44]

Homework Statement

Graph f(x) = sqrt(x^2 - 2x), and find an interval on which it is one-to-one. Find the inverse of the function restricted to that interval.

Homework Equations

The Attempt at a Solution

What I can't do is really finding the inverse function. It seems very simple, but somehow I got stuck in the process.

Did you find the interval on which f is 1-to-1?

swap x and y in the original function
y = sqrt(x^2-2x)
x = sqrt(y^2-2y)
and solve for y

so i did
x^2 = y^2-2y, and i tried to factor out y

No, that's not what you need to do. Instead, complete the square on the right side.

x^2 = y(y-2)
x^2/y-2 = y
now i am really stuck. how can i pull that y out?

Thank you for any kind of help!

---- edited

I was thinking about this formula: d/dy f-1(x) = 1/ f ' (y)
i guess i can then integrate the d/dy f-1(x) and get f-1(x)?

so i started working again
f ' = (1/2 (x^2-2x) ^-1/2) * 2x-2
so 1/f ' =[ 2 (x^2-2x)^1/2 ]/ 2x-2
which is d/dy f-1

but the integration doesn't work!

This problem doesn't require calculus to solve.

 

[999iscool]

Hi. Mark. Yes, I did find the interval.
And thank you for your help. I got the answer with completing the square.