Question

Use the First Derivative test to determine the local extreme values of the function, and identify any absolute extrema.
y= x sqrt(8-x^2)

Answer 1

\[f(x)=x\sqrt{8-x^2}\] product rule for this one
\[f'(x)=\sqrt{8-x^2}-\frac{x^2}{\sqrt{8-x^2}}\]

Answer 2

subtract and get \[f'(x)=\frac{8-2x^2}{\sqrt{8-x^2}}\]

Answer 3

critical point where the numerator is 0, and also the denominator, but that is as the endpoints of the domain of the original function, you get \(f(\sqrt8)=f(-\sqrt8)=0\)

Answer 4

so set \[8-2x^2=0\] solve for \(x\)