# Thread: Find Inverse of Radical Functions

1. ## Find Inverse of Radical Functions

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.

2. 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:

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

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

The constraints are the same.

3. ## Are you....

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:

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

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

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?

4. 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 $\displaystyle y = \pm \sqrt{r^2 - x^2}$

but take the positive one so that the restrictions match up

5. ## ok...

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 $\displaystyle y = \pm \sqrt{r^2 - x^2}$

but take the positive one so that the restrictions match up
I got it now. Thanks.