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

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

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extreme f(x,y)=e^{2y}(2x^2-2xy+1)

extreme points

f (x, y) = e^{2 y} (2 x^{2} - 2 x y + 1)

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

Solution steps

Find the critical points:

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

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

D (x, y) = - 32 e^{4 y} x^{2} + 32 e^{4 y} x y + 12 e^{4 y} - 16 e^{4 y} y^{2} - 16 e^{4 y} y

Check the sign of at each critical point

Check critical point

(- 1, - 2) :

Minimum

Check critical point

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

Saddle

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

f (x) = 2 x^{2} - 4 x + y^{2} - 6 y + 1

f (x) = x^{3} - 12 x^{2} - 27 x + 2, - 2 \leq x \leq 0

f (x) = - 4 x^{3} + 12 x^{2}

f (x) = \frac{8 x}{x ^{2} + 16}

\frac{( x ^{2} - 1 )}{( x - 3 ) ^{2}}

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