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

The extreme f(x,y)=-x^2-y^3+4x+3y is Maximum(2,1),Saddle(2,-1)

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

extreme points

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

Maximum (2, 1), Saddle (2, - 1)

Solution steps

Find the critical points:

(2, 1), (2, - 1)

\frac{\partial ^{2} f}{\partial x ^{2}} = - 6 y

D (x, y) = 12 y

Check the sign of at each critical point

Check critical point

(2, 1) :

Maximum

Check critical point

(2, - 1) :

Saddle

Maximum (2, 1), Saddle (2, - 1)

x^{2} + x y + 3 x + 2 y + 5

f (x) = x^{3} + 2 x^{2} - 5 x - 6

f (x) = \frac{1}{x ^{2} - 9}

y = 2 x^{3} - 3 x + 5

What is the extreme f(x,y)=-x^2-y^3+4x+3y ?
The extreme f(x,y)=-x^2-y^3+4x+3y is Maximum(2,1),Saddle(2,-1)