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.

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- Oct 19th 2008, 06:29 AMmagentaritaFind 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.**

- Oct 19th 2008, 06:53 AMearboth
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. - Oct 19th 2008, 09:12 PMmagentaritaAre you....
- Nov 8th 2008, 10:05 PMJhevon
- Nov 9th 2008, 05:24 AMmagentaritaok...