What is the extreme f(x,y)=2^3-y^2+2xy+1 ?

The extreme f(x,y)=2^3-y^2+2xy+1 is Saddle(0,0)

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extreme f(x,y)=2^3-y^2+2xy+1

extreme points

f (x, y) = 2^{3} - y^{2} + 2 x y + 1

Saddle (0, 0)

Solution steps

Find the critical points:

(0, 0)

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

D (x, y) = - 4

Check the sign of at each critical point

Check critical point

(0, 0) :

Saddle

Saddle (0, 0)

y = x^{2} + 2

f (x) = 2^{(x + 0.528)} - 10 (x + 0.528)

y = x^{2} - 5

- x^{2} - 4 x + 1

f (x) = 3 x^{3} - 24 x^{2} + 48 x - 6

What is the extreme f(x,y)=2^3-y^2+2xy+1 ?
The extreme f(x,y)=2^3-y^2+2xy+1 is Saddle(0,0)