What is the extreme f(x,y)=4x^3-11xy^2+15y+12 ?

The extreme f(x,y)=4x^3-11xy^2+15y+12 is

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extreme f(x,y)=4x^3-11xy^2+15y+12

extreme points

f (x, y) = 4 x^{3} - 11 x y^{2} + 15 y + 12

Saddle (\frac{3 5 4 1331}{22 4 27}, 4 \frac{675}{1331}), Saddle (- \frac{3 5 4 1331}{22 4 27}, - 4 \frac{675}{1331})

Solution steps

Find the critical points:

(\frac{3 5 4 1331}{22 4 27}, 4 \frac{675}{1331}), (- \frac{3 5 4 1331}{22 4 27}, - 4 \frac{675}{1331})

\frac{\partial ^{2} f}{\partial x ^{2}} = - 22 x

D (x, y) = - 528 x^{2} - 484 y^{2}

Check the sign of at each critical point

Check critical point

(\frac{3 5 4 1331}{22 4 27}, 4 \frac{675}{1331}) :

Saddle

Check critical point

(- \frac{3 5 4 1331}{22 4 27}, - 4 \frac{675}{1331}) :

Saddle

Saddle (\frac{3 5 4 1331}{22 4 27}, 4 \frac{675}{1331}), Saddle (- \frac{3 5 4 1331}{22 4 27}, - 4 \frac{675}{1331})

f (x) = sin (x) + cos^{2} (x)

f (x) = cos (5 x)

\frac{x}{( x - 1 ) ^{2}}

f (x) = cos (4 x)

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

What is the extreme f(x,y)=4x^3-11xy^2+15y+12 ?
The extreme f(x,y)=4x^3-11xy^2+15y+12 is