What is the extreme f(x,y)=x^3+y^2-2xy+7x-8y+2 ?

The extreme f(x,y)=x^3+y^2-2xy+7x-8y+2 is Minimum(1,5),Saddle(-1/3 , 11/3)

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extreme f(x,y)=x^3+y^2-2xy+7x-8y+2

extreme points

f (x, y) = x^{3} + y^{2} - 2 x y + 7 x - 8 y + 2

Minimum (1, 5), Saddle (- \frac{1}{3}, \frac{11}{3})

Solution steps

Find the critical points:

(1, 5), (- \frac{1}{3}, \frac{11}{3})

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

D (x, y) = 12 x - 4

Check the sign of at each critical point

Check critical point

(1, 5) :

Minimum

Check critical point

(- \frac{1}{3}, \frac{11}{3}) :

Saddle

Minimum (1, 5), Saddle (- \frac{1}{3}, \frac{11}{3})

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

F (x) = x^{3} - 6 x^{2} + 9 x + 15

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

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

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

What is the extreme f(x,y)=x^3+y^2-2xy+7x-8y+2 ?
The extreme f(x,y)=x^3+y^2-2xy+7x-8y+2 is Minimum(1,5),Saddle(-1/3 , 11/3)