# Find Inverse of Radical Functions

• Oct 19th 2008, 06:29 AM
magentarita
Find the inverse of f(x) = sqrt{r^2 - x^2}, where x is greater than or = to 0 and x is less than or = to r.
• Oct 19th 2008, 06:53 AM
earboth
Quote:

Originally Posted by magentarita
Find the inverse of f(x) = sqrt{r^2 - x^2}, where x is greater than or = to 0 and x is less than or = to r.

The inverse function of f is f.

1. Your equation with it's constraints define a quarter circle with center at the origin placed in the 1st quadrant. If you reflect this quarter circle about the line y = x (first median of the coordinate system) you'll get the same quarter circle.

2. Change x and y in your equation and solve for y:

$y = \sqrt{r^2 - x^2}~\rightarrow~ x= \sqrt{r^2 - y^2}$

$x^2=r^2-y^2~\implies~y^2=r^2-x^2$

The constraints are the same.
• Oct 19th 2008, 09:12 PM
magentarita
Are you....
Quote:

Originally Posted by earboth
The inverse function of f is f.

1. Your equation with it's constraints define a quarter circle with center at the origin placed in the 1st quadrant. If you reflect this quarter circle about the line y = x (first median of the coordinate system) you'll get the same quarter circle.

2. Change x and y in your equation and solve for y:

$y = \sqrt{r^2 - x^2}~\rightarrow~ x= \sqrt{r^2 - y^2}$

$x^2=r^2-y^2~\implies~y^2=r^2-x^2$

The constraints are the same.

Are you saying that y^2 = r^2 - x^2 is the inverse?
Shouldn't we take the square root of both sides of the equation to isolate y?
• Nov 8th 2008, 10:05 PM
Jhevon
Quote:

Originally Posted by magentarita
Are you saying that y^2 = r^2 - x^2 is the inverse?
Shouldn't we take the square root of both sides of the equation to isolate y?

technically, as Earboth has it, it is fine. as long as you state the restrictions correctly.

if you wish you could take the square root of both sides to get $y = \pm \sqrt{r^2 - x^2}$

but take the positive one so that the restrictions match up
• Nov 9th 2008, 05:24 AM
magentarita
ok...
Quote:

Originally Posted by Jhevon
technically, as Earboth has it, it is fine. as long as you state the restrictions correctly.

if you wish you could take the square root of both sides to get $y = \pm \sqrt{r^2 - x^2}$

but take the positive one so that the restrictions match up

I got it now. Thanks.