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

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

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extreme f(x,y)=-x^3+y^2+3x-4y-1

extreme points

f (x, y) = - x^{3} + y^{2} + 3 x - 4 y - 1

Minimum (- 1, 2), Saddle (1, 2)

Solution steps

Find the critical points:

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

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

D (x, y) = - 12 x

Check the sign of at each critical point

Check critical point

(- 1, 2) :

Minimum

Check critical point

(1, 2) :

Saddle

Minimum (- 1, 2), Saddle (1, 2)

f (x, y) = x^{3} + y^{3} - 9 x y

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

x^{3} - 6 x^{2} + 5

f (x) = 3 - x, - 1 \leq x \leq 2

f (x) = x + 2 cos (x)

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