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

The extreme f(x,y)=x^5+y^5-5x-5y is Saddle(-1,1),Maximum(-1,-1),Minimum(1,1),Saddle(1,-1)

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extreme f(x,y)=x^5+y^5-5x-5y

extreme points

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

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}} = 20 y^{3}

D (x, y) = 400 x^{3} y^{3}

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, y) = y^{2} + x y + 3 x + 2 y + 5

y = x e^{3 - \frac{x}{4}}

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

\frac{x ^{2} - 9}{2 x - 4}

f (x) = 2 x^{3} - 15 x^{2} - 36

What is the extreme f(x,y)=x^5+y^5-5x-5y ?
The extreme f(x,y)=x^5+y^5-5x-5y is Saddle(-1,1),Maximum(-1,-1),Minimum(1,1),Saddle(1,-1)