What is the extreme f(x,y)=x^3+y^2-2xy+7x-8y+4 ?

The extreme f(x,y)=x^3+y^2-2xy+7x-8y+4 is Minimum(1,5),Saddle(-1/3 , 11/3)

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extreme f(x,y)=x^3+y^2-2xy+7x-8y+4

extreme points

f (x, y) = x^{3} + y^{2} - 2 x y + 7 x - 8 y + 4

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) = x^{2} y + y^{3} - 12 y

\frac{e ^{x}}{x + 1}

f (x) = x^{2} - 2 x - 3, 0 \leq x \leq 2

f (x) = x (14 - 2 x) (15 - x)

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

What is the extreme f(x,y)=x^3+y^2-2xy+7x-8y+4 ?
The extreme f(x,y)=x^3+y^2-2xy+7x-8y+4 is Minimum(1,5),Saddle(-1/3 , 11/3)