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

The extreme f(x,y)=y^3+2x^2+3xy-y^2-3x-y+3 is Saddle(7/16 , 5/12),Minimum(0,1)

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extreme f(x,y)=y^3+2x^2+3xy-y^2-3x-y+3

extreme points

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

Saddle (\frac{7}{16}, \frac{5}{12}), Minimum (0, 1)

Solution steps

Find the critical points:

(\frac{7}{16}, \frac{5}{12}), (0, 1)

\frac{\partial ^{2} f}{\partial x ^{2}} = 6 y - 2

D (x, y) = 24 y - 17

Check the sign of at each critical point

Check critical point

(\frac{7}{16}, \frac{5}{12}) :

Saddle

Check critical point

(0, 1) :

Minimum

Saddle (\frac{7}{16}, \frac{5}{12}), Minimum (0, 1)

f (x) = e^{x} (x - 3)

f (x) = 4 x^{3} + 12 x^{2} - 36 x

f (x, y) = 8 x ln (x) y

f (x) = \frac{1}{4} x^{4} - 2 x^{3} + 2

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

What is the extreme f(x,y)=y^3+2x^2+3xy-y^2-3x-y+3 ?
The extreme f(x,y)=y^3+2x^2+3xy-y^2-3x-y+3 is Saddle(7/16 , 5/12),Minimum(0,1)