What is the extreme f(x,y)=x^4+y^4-2x^2-32y ?

The extreme f(x,y)=x^4+y^4-2x^2-32y is Saddle(0,2),Minimum(-1,2),Minimum(1,2)

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extreme f(x,y)=x^4+y^4-2x^2-32y

extreme points

f (x, y) = x^{4} + y^{4} - 2 x^{2} - 32 y

Saddle (0, 2), Minimum (- 1, 2), Minimum (1, 2)

Solution steps

Find the critical points:

(0, 2), (- 1, 2), (1, 2)

\frac{\partial ^{2} f}{\partial x ^{2}} = 12 y^{2}

D (x, y) = 12 y^{2} (12 x^{2} - 4)

Check the sign of at each critical point

Check critical point

(0, 2) :

Saddle

Check critical point

(- 1, 2) :

Minimum

Check critical point

(1, 2) :

Minimum

Saddle (0, 2), Minimum (- 1, 2), Minimum (1, 2)

y = x^{2} + 6 x

f (x) = 6 x - 12

f (x, y) = x^{2} - y^{2} 1 - x^{2} - y^{2}

f (x) = \frac{1}{x}

e^{- ∣ x ∣}

What is the extreme f(x,y)=x^4+y^4-2x^2-32y ?
The extreme f(x,y)=x^4+y^4-2x^2-32y is Saddle(0,2),Minimum(-1,2),Minimum(1,2)