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

The extreme f(x,y)=e^{(x-y)}(x^2-2y^2) is Saddle(0,0),Maximum(-4,-2),Saddle(0,0)

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

extreme points

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

Saddle (0, 0), Maximum (- 4, - 2), Saddle (0, 0)

Solution steps

Find the critical points:

(0, 0), (- 4, - 2), (0, 0)

\frac{\partial ^{2} f}{\partial x ^{2}} = e^{- y + x} x^{2} - 2 e^{- y + x} y^{2} + 8 e^{- y + x} y - 4 e^{- y + x}

D (x, y) = e^{- y + x} (x^{2} + 4 x - 2 y^{2} + 2) (e^{- y + x} x^{2} - 2 e^{- y + x} y^{2} + 8 e^{- y + x} y - 4 e^{- y + x}) - e^{2 (- y + x)} (x^{2} - 2 y^{2} + 2 x + 4 y)^{2}

Check the sign of at each critical point

Check critical point

(0, 0) :

Saddle

Check critical point

(- 4, - 2) :

Maximum

Check critical point

(0, 0) :

Saddle

Saddle (0, 0), Maximum (- 4, - 2), Saddle (0, 0)

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

\frac{x ^{3}}{3} - \frac{x ^{2}}{2} - 2 x + \frac{1}{3}

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

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

f (x) = 4 cos^{2} (x)

What is the extreme f(x,y)=e^{(x-y)}(x^2-2y^2) ?
The extreme f(x,y)=e^{(x-y)}(x^2-2y^2) is Saddle(0,0),Maximum(-4,-2),Saddle(0,0)