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

The extreme f(y,x)=ye^{x^2-y^2} is

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

extreme points

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

D (y, x) = 2 y (4 e^{(y + x) (- y + x)} y^{3} - 6 e^{(y + x) (- y + x)} y) (2 e^{x^{2} - y^{2}} x^{2} + e^{x^{2} - y^{2}}) - 4 e^{2 (y + x) (- y + x)} 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, y) = \frac{2 x ^{2} - 3 y ^{2}}{y ^{2} + 1}

x^{3} + x, - 1 \leq x \leq 2

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

f (x) = 5 - 3 x^{2}, - 5 \leq x \leq 1

f (x) = 16 x + 21 x^{\frac{16}{21}}