What is the extreme f(x)=14x^2-2x^32y^2+4xy ?

The extreme f(x)=14x^2-2x^32y^2+4xy is Saddle(1/(\sqrt[3]{28)},2^{1/3}*7^{2/3})

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extreme f(x)=14x^2-2x^32y^2+4xy

extreme points

f (x) = 14 x^{2} - 2 x^{3} 2 y^{2} + 4 x y

Solution steps

Find the critical points:

\frac{\partial ^{2} f}{\partial x ^{2}} = - 8 x^{3}

D (x, y) = - 384 x^{4} y^{2} - 224 x^{3} + 192 x^{2} y - 16

Check the sign of at each critical point

Check critical point

Saddle

2 x^{2} + y^{2} + x y - 3 x

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

f (x) = x^{4} - 72 x^{2}

f (x) = - 3 x^{4} ln (3 x), (0, 4)

What is the extreme f(x)=14x^2-2x^32y^2+4xy ?
The extreme f(x)=14x^2-2x^32y^2+4xy is Saddle(1/(\sqrt[3]{28)},2^{1/3}*7^{2/3})