What is the extreme f(x)=-x^2-y^2-2y+5 ?

The extreme f(x)=-x^2-y^2-2y+5 is Maximum(0,-1)

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extreme f(x)=-x^2-y^2-2y+5

extreme points

f (x) = - x^{2} - y^{2} - 2 y + 5

Maximum (0, - 1)

Solution steps

Find the critical points:

(0, - 1)

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

D (x, y) = 4

Check the sign of at each critical point

Check critical point

(0, - 1) :

Maximum

Maximum (0, - 1)

f (x) = - 2 x^{4} + 16 x^{3} - 26

f (x) = (x - 5)^{2} + 3

f (x, y) = 1 - x + y

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

f (x y) = 3 x^{4} + 8 x^{3} - 18 x^{2} + 6 y^{2} + 12 y - 4

What is the extreme f(x)=-x^2-y^2-2y+5 ?
The extreme f(x)=-x^2-y^2-2y+5 is Maximum(0,-1)