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

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

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

extreme points

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

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

Solution steps

Find the critical points:

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

\frac{\partial ^{2} f}{\partial x ^{2}} = 2 e^{2 x}

D (x, y) = - 8 e^{4 x} y^{2} - 8 e^{2 x} y - 1

Check the sign of at each critical point

Check critical point

(0, 0) :

Saddle

Check critical point

(1, - \frac{1}{2 e ^{2}}) :

Minimum

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

3 x^{5} - 5 x^{3}

\frac{2}{x ^{12}} - \frac{1}{x ^{6}}

f (x) = \frac{x ^{2} - 4}{x - 1}

f (x, y) = 2 x - 5 y + 3

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

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