What is the extreme f(x,y)=x^3+12xy+y^4 ?

The extreme f(x,y)=x^3+12xy+y^4 is Saddle(0,0),Minimum(2\sqrt[5]{6},-6^{2/5})

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extreme f(x,y)=x^3+12xy+y^4

extreme points

f (x, y) = x^{3} + 12 x y + y^{4}

Solution steps

Find the critical points:

\frac{\partial ^{2} f}{\partial x ^{2}} = 12 y^{2}

D (x, y) = 72 x y^{2} - 144

Check the sign of at each critical point

Check critical point

(0, 0) :

Saddle

Check critical point

Minimum

f (x) = x^{3} e^{2 x} + 1

f (x) = sin (x) + 4

- x^{2} + 3 x + 5, - 3 \leq x \leq 3

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

What is the extreme f(x,y)=x^3+12xy+y^4 ?
The extreme f(x,y)=x^3+12xy+y^4 is Saddle(0,0),Minimum(2\sqrt[5]{6},-6^{2/5})