Answer
critical point is (1,3)
Work Step by Step
Given $y=(x)^2+\frac{2}{x}$
The critical point occurs where the first derivative is zero:
$\frac{dy}{dx}=\frac{d{((x)^2+\frac{2}{x})}}{dx}$
${\frac{dy}{dx}}=2x-2x^{-2}$
${\frac{dy}{dx}}=\frac{(2(x^3)-2)}{x^2}$
so the critical point is ${\frac{dy}{dx}}=0$
$f'(x)=\frac{(2(x^3)-2)}{x^2}=0$
$x=1,x=0,x=0$
the critical value of y can be obtained by putting in the value of x:
$y=(x)^2+\frac{2}{x}=(1)^2+\frac{2}{1}=3$
$y=(x)^2+\frac{2}{x}=(0)^2+\frac{2}{0}=undefined$
thus the final critical point is (1,3)