What is the extreme f(x,y)=x^2+y^3+xy-3x-4y+5 ?

The extreme f(x,y)=x^2+y^3+xy-3x-4y+5 is Saddle(23/12 ,-5/6),Minimum(1,1)

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extreme f(x,y)=x^2+y^3+xy-3x-4y+5

extreme points

f (x, y) = x^{2} + y^{3} + x y - 3 x - 4 y + 5

Saddle (\frac{23}{12}, - \frac{5}{6}), Minimum (1, 1)

Solution steps

Find the critical points:

(\frac{23}{12}, - \frac{5}{6}), (1, 1)

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

D (x, y) = 12 y - 1

Check the sign of at each critical point

Check critical point

(\frac{23}{12}, - \frac{5}{6}) :

Saddle

Check critical point

(1, 1) :

Minimum

Saddle (\frac{23}{12}, - \frac{5}{6}), Minimum (1, 1)

\frac{1 - x ^{3}}{x ^{2} - 4}

y = 4 x^{3} - x

(\frac{x}{4})^{4} - \frac{4}{3} x^{3} + \frac{1}{2} x^{2} + 6 x + 2

\frac{1}{3} x^{3} - 4 x^{2} + 12 x + 200

f (x, y) = \frac{1}{2} (x - 2)^{2} + (y - 1)^{2}

What is the extreme f(x,y)=x^2+y^3+xy-3x-4y+5 ?
The extreme f(x,y)=x^2+y^3+xy-3x-4y+5 is Saddle(23/12 ,-5/6),Minimum(1,1)