What is the extreme ye^{x^2-2y^2} ?

The extreme ye^{x^2-2y^2} is Saddle(-1/2 ,0),Saddle(1/2 ,0),Saddle(-1/2 ,0),Saddle(1/2 ,0)

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extreme ye^{x^2-2y^2}

extreme points

y e^{x^{2} - 2 y^{2}}

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

Solution steps

Find the critical points:

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

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

D (y, x) = 2 y (16 e^{(x + 2 y) (x - 2 y)} y^{3} - 12 e^{(x + 2 y) (x - 2 y)} y) (2 e^{x^{2} - 2 y^{2}} x^{2} + e^{x^{2} - 2 y^{2}}) - 4 e^{2 (x + 2 y) (x - 2 y)} x^{2} (2 y + 1)^{2} (2 y - 1)^{2}

Check the sign of at each critical point

Check critical point

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

Saddle

Check critical point

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

Saddle

Check critical point

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

Saddle

Check critical point

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

Saddle

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

f (x) = 2 - 2 cos (x), (0, 2 π)

\frac{d}{d o} (2 x + 1 o n [0.2])

f (x) = \frac{( x ^{2} - 5 ) ^{3}}{125}

4 cos (x)

f (x) = 2 x^{3} + 3 x^{2} - 36 x - 5

What is the extreme ye^{x^2-2y^2} ?
The extreme ye^{x^2-2y^2} is Saddle(-1/2 ,0),Saddle(1/2 ,0),Saddle(-1/2 ,0),Saddle(1/2 ,0)