What is the extreme f(x,y)=(x^2+y^2)e^{x^2-y^2} ?

The extreme f(x,y)=(x^2+y^2)e^{x^2-y^2} is Minimum(0,0),Saddle(0,1),Saddle(0,-1)

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

extreme points

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

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

Solution steps

Find the critical points:

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

\frac{\partial ^{2} f}{\partial x ^{2}} = 2 (2 e^{(x + y) (x - y)} y^{4} + 2 e^{(x + y) (x - y)} x^{2} y^{2} - 5 e^{(x + y) (x - y)} y^{2} - e^{(x + y) (x - y)} x^{2} + e^{(x + y) (x - y)})

D (x, y) = 4 (2 e^{(x + y) (x - y)} x^{4} + 2 e^{(x + y) (x - y)} x^{2} y^{2} + 5 e^{(x + y) (x - y)} x^{2} + e^{(x + y) (x - y)} y^{2} + e^{(x + y) (x - y)}) (2 e^{(x + y) (x - y)} y^{4} + 2 e^{(x + y) (x - y)} x^{2} y^{2} - 5 e^{(x + y) (x - y)} y^{2} - e^{(x + y) (x - y)} x^{2} + e^{(x + y) (x - y)}) - 4 x^{2} (- 2 e^{(x + y) (x - y)} y^{3} - 2 e^{(x + y) (x - y)} x^{2} y)^{2}

Check the sign of at each critical point

Check critical point

(0, 0) :

Minimum

Check critical point

(0, 1) :

Saddle

Check critical point

(0, - 1) :

Saddle

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

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

x^{3} e^{- x}

f (x, y) = x ln (x^{2} + y^{2})

f (x) = x^{\frac{12}{5}} - 6 x^{\frac{2}{5}} + 2

10 x - 5 x ln (x)

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