What is the extreme f(x,y)=(1-x^2)^2-y^2 ?

The extreme f(x,y)=(1-x^2)^2-y^2 is Maximum(0,0),Saddle(-1,0),Saddle(1,0)

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extreme f(x,y)=(1-x^2)^2-y^2

extreme points

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

Maximum (0, 0), Saddle (- 1, 0), Saddle (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) = - 8 (3 x^{2} - 1)

Check the sign of at each critical point

Check critical point

(0, 0) :

Maximum

Check critical point

(- 1, 0) :

Saddle

Check critical point

(1, 0) :

Saddle

Maximum (0, 0), Saddle (- 1, 0), Saddle (1, 0)

f (x) = - x^{2} + 2 x + 8

f (x, y) = x y e^{x^{2}}

f (x) = x^{2} - 2 x - 5

f (x) = 1 - e^{- a x}

f (x, y) = - x^{3} + y^{3} + 6 x^{2} - 9 x - 3 y^{2} - 9 y

What is the extreme f(x,y)=(1-x^2)^2-y^2 ?
The extreme f(x,y)=(1-x^2)^2-y^2 is Maximum(0,0),Saddle(-1,0),Saddle(1,0)