Question

Find Derivative, critical points, and determine the local extreme values

y=x^2 times spare root of (9-x)

Answer 1

\[y=x^2\sqrt{9-x}\] product rule for this one
use
\[(fg)'=f'g+g'f\] with
\[f(x)=x^2,f'(x)=2x,g(x)=\sqrt{9-x},g'(x)=-\frac{1}{2\sqrt{9-x}}\]

Answer 2

should get
\[y'=\frac{x(36-5x)}{2\sqrt{9-x}}\] after the algebra
now zeros are easy to find

Answer 3

How did you get x(36-5x) I got something different.

Answer 4

? I got x(36-3x)......