What is the extreme f(x,y)=x^4+y^2-2x^2-5 ?

The extreme f(x,y)=x^4+y^2-2x^2-5 is Saddle(0,0),Minimum(-1,0),Minimum(1,0)

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extreme f(x,y)=x^4+y^2-2x^2-5

extreme points

f (x, y) = x^{4} + y^{2} - 2 x^{2} - 5

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

Solution steps

Find the critical points:

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

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

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

Check the sign of at each critical point

Check critical point

(0, 0) :

Saddle

Check critical point

(- 1, 0) :

Minimum

Check critical point

(1, 0) :

Minimum

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

f (x) = x + \frac{3}{x}

f (x) = \frac{x}{25 + x ^{2}}

f (x, y) = 2 x^{3} y - - x y

f (x) = x^{2} - 4 x + 9

f (x) = 3 x^{4} - 2 x^{3}

What is the extreme f(x,y)=x^4+y^2-2x^2-5 ?
The extreme f(x,y)=x^4+y^2-2x^2-5 is Saddle(0,0),Minimum(-1,0),Minimum(1,0)