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)

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)

anonymous · Answer

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

anonymous · Answer

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

anonymous · Answer

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 $f(\sqrt8)=f(-\sqrt8)=0$

anonymous · Answer

so set $$8-2x^2=0$$ solve for $x$