What is the extreme f(x,y)=e^{2x}(x+y^2+2y)+10 ?

The extreme f(x,y)=e^{2x}(x+y^2+2y)+10 is Minimum(1/2 ,-1)

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extreme f(x,y)=e^{2x}(x+y^2+2y)+10

extreme points

f (x, y) = e^{2 x} (x + y^{2} + 2 y) + 10

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

Solution steps

Find the critical points:

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

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

D (x, y) = - 8 e^{4 x} y^{2} - 16 e^{4 x} y - 8 e^{4 x} + 8 e^{4 x} x

Check the sign of at each critical point

Check critical point

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

Minimum

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

P (x, y) = x^{2} - y^{2}

f (x) = x \cdot ln^{2} (x)

f (x) = f (x, y) = x^{3} + y^{3} - 3 x^{2} - 9 y^{2} - 9 x

\frac{1}{x ^{3}}

What is the extreme f(x,y)=e^{2x}(x+y^2+2y)+10 ?
The extreme f(x,y)=e^{2x}(x+y^2+2y)+10 is Minimum(1/2 ,-1)