What is the extreme f(x,y)= 1/3 x^3+1/3 y^3-x-y+10 ?

The extreme f(x,y)= 1/3 x^3+1/3 y^3-x-y+10 is Saddle(-1,1),Maximum(-1,-1),Minimum(1,1),Saddle(1,-1)

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extreme f(x,y)= 1/3 x^3+1/3 y^3-x-y+10

extreme points

f (x, y) = \frac{1}{3} x^{3} + \frac{1}{3} y^{3} - x - y + 10

Saddle (- 1, 1), Maximum (- 1, - 1), Minimum (1, 1), Saddle (1, - 1)

Solution steps

Find the critical points:

(- 1, 1), (- 1, - 1), (1, 1), (1, - 1)

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

D (x, y) = 4 x y

Check the sign of at each critical point

Check critical point

(- 1, 1) :

Saddle

Check critical point

(- 1, - 1) :

Maximum

Check critical point

(1, 1) :

Minimum

Check critical point

(1, - 1) :

Saddle

Saddle (- 1, 1), Maximum (- 1, - 1), Minimum (1, 1), Saddle (1, - 1)

f (x) = \frac{x ^{4}}{4} - 8 x

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

f (x) = x - 5 x^{\frac{1}{5}}

f (x) = 4 x^{3} - 16 x

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

What is the extreme f(x,y)= 1/3 x^3+1/3 y^3-x-y+10 ?
The extreme f(x,y)= 1/3 x^3+1/3 y^3-x-y+10 is Saddle(-1,1),Maximum(-1,-1),Minimum(1,1),Saddle(1,-1)