Question

I can't find critical points for y=x(9-x^2)^1/2

Answer 1

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

Answer 2

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

Answer 3

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

Answer 4

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

Answer 5

Are you looking for a specific type of critical point?

Answer 6

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

Answer 7

that is not the only answer.

Answer 8

and the derivative is not quite correct.

Answer 9

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

Answer 10

oops, sorry corec, the derivative is fine