## anonymous 5 years ago I can't find critical points for y=x(9-x^2)^1/2

1. anonymous

First take the derivative dy/dx=$(9-2x ^{2})/\sqrt{9-x ^{2}}$

2. anonymous

next we must find where dy/dx = 0 or where it is undefined

3. anonymous

dy/dx = 0 for $x = \pm \sqrt{9/2}$

4. anonymous

dy/dx is undefined for $x = \pm3$, when the denominator equals 0

5. anonymous

Are you looking for a specific type of critical point?

6. anonymous

$y = x \sqrt{9-x^2}$ the derivative of it is: $y'=\sqrt{9-x^2} -x^2/\sqrt{9-x^2}$ then y'=0 finely x =2.45

7. anonymous

that is not the only answer.

8. anonymous

and the derivative is not quite correct.

9. anonymous

The critical points should be $x = \pm \sqrt{9/2}$

10. anonymous

oops, sorry corec, the derivative is fine