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

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

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

extreme points

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

Saddle (0, 0), Minimum (256, - 6^{\frac{2}{5}})

Solution steps

Find the critical points:

(0, 0), (256, - 6^{\frac{2}{5}})

\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

(256, - 6^{\frac{2}{5}}) :

Minimum

Saddle (0, 0), Minimum (256, - 6^{\frac{2}{5}})

f (x) = 3 x - 5 ln (4 x)

f (x, y) = - \frac{4}{x ^{2}}, 0.5 \leq x \leq 4

x e^{- 5 x}

f (x) = 4 x y + 50 y - 9 y^{2} - \frac{1}{10} x^{2} y - 130

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

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