What is the extreme f(xy)=x^2+y^2+2/(xy) ?

The extreme f(xy)=x^2+y^2+2/(xy) is Minimum(1,1),Minimum(-1,-1)

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extreme f(xy)=x^2+y^2+2/(xy)

extreme points

f (x y) = x^{2} + y^{2} + \frac{2}{x y}

Minimum (1, 1), Minimum (- 1, - 1)

Solution steps

Find the critical points:

(1, 1), (- 1, - 1)

\frac{\partial ^{2} f}{\partial x ^{2}} = 2 + \frac{4}{x y ^{3}}

D (x, y) = 4 + \frac{8}{x y ^{3}} + \frac{8}{x ^{3} y} + \frac{12}{x ^{4} y ^{4}}

Check the sign of at each critical point

Check critical point

(1, 1) :

Minimum

Check critical point

(- 1, - 1) :

Minimum

Minimum (1, 1), Minimum (- 1, - 1)

f (x) = x^{3} - 12 x y + 8 y^{3}

f (x) = x 9 - x^{2}, - 1 \leq x \leq 3

x^{2} - 2 x - 8

f (x) = 2 x - y

f (x) = 2 x + y

What is the extreme f(xy)=x^2+y^2+2/(xy) ?
The extreme f(xy)=x^2+y^2+2/(xy) is Minimum(1,1),Minimum(-1,-1)