What is the extreme f(x,y)=-x^2-y^3+3y^2-2x+45y+1 ?

The extreme f(x,y)=-x^2-y^3+3y^2-2x+45y+1 is Saddle(-1,-3),Maximum(-1,5)

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extreme f(x,y)=-x^2-y^3+3y^2-2x+45y+1

extreme points

f (x, y) = - x^{2} - y^{3} + 3 y^{2} - 2 x + 45 y + 1

Saddle (- 1, - 3), Maximum (- 1, 5)

Solution steps

Find the critical points:

(- 1, - 3), (- 1, 5)

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

D (x, y) = - 2 (- 6 y + 6)

Check the sign of at each critical point

Check critical point

(- 1, - 3) :

Saddle

Check critical point

(- 1, 5) :

Maximum

Saddle (- 1, - 3), Maximum (- 1, 5)

y = - 12 x^{5} - 45 x^{4} + 20 x^{3} - 1

f (x) = x^{3} - 2 x - 4 x + 4 (0.3)

x^{2} + x^{- 2}

(x - 3)^{2} (x - 2)

f (x) = 6 x^{2} - 6 x + 9

What is the extreme f(x,y)=-x^2-y^3+3y^2-2x+45y+1 ?
The extreme f(x,y)=-x^2-y^3+3y^2-2x+45y+1 is Saddle(-1,-3),Maximum(-1,5)