What is the extreme f(x,y)=-x^4+x^2+y^2+3 ?

The extreme f(x,y)=-x^4+x^2+y^2+3 is Minimum(0,0),Saddle(-1/(sqrt(2)),0),Saddle(1/(sqrt(2)),0)

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extreme f(x,y)=-x^4+x^2+y^2+3

extreme points

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

Minimum (0, 0), Saddle (- \frac{1}{2}, 0), Saddle (\frac{1}{2}, 0)

Solution steps

Find the critical points:

(0, 0), (- \frac{1}{2}, 0), (\frac{1}{2}, 0)

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

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

Check the sign of at each critical point

Check critical point

(0, 0) :

Minimum

Check critical point

(- \frac{1}{2}, 0) :

Saddle

Check critical point

(\frac{1}{2}, 0) :

Saddle

Minimum (0, 0), Saddle (- \frac{1}{2}, 0), Saddle (\frac{1}{2}, 0)

\frac{6 x ^{2}}{x ^{2} - 9}

K (r, s) = 9 r - s

e^{x} + x^{2}

f (x) = x^{3} - 6 x^{2} - 15 x + 40

f (x) = 7 x - 5 x ln (x)

What is the extreme f(x,y)=-x^4+x^2+y^2+3 ?
The extreme f(x,y)=-x^4+x^2+y^2+3 is Minimum(0,0),Saddle(-1/(sqrt(2)),0),Saddle(1/(sqrt(2)),0)