What is the extreme f(x,y)=3x^3-(y+1)^3-9xy+3y+1 ?

The extreme f(x,y)=3x^3-(y+1)^3-9xy+3y+1 is Saddle(0,0),Maximum(-1,1)

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extreme f(x,y)=3x^3-(y+1)^3-9xy+3y+1

extreme points

f (x, y) = 3 x^{3} - (y + 1)^{3} - 9 x y + 3 y + 1

Saddle (0, 0), Maximum (- 1, 1)

Solution steps

Find the critical points:

(0, 0), (- 1, 1)

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

D (x, y) = 18 x (- 6 y - 6) - 81

Check the sign of at each critical point

Check critical point

(0, 0) :

Saddle

Check critical point

(- 1, 1) :

Maximum

Saddle (0, 0), Maximum (- 1, 1)

f (x) = \frac{x ^{4}}{4} - x^{3} + 3

f (x) = - 3 (x + 2) e^{4 x}, (- 5, 0)

sin (5 x), - \frac{π}{5} \leq x \leq \frac{π}{4}

f (x) = \frac{x ^{4} - 4 x ^{3}}{2}

f (x) = 0.5 x^{2} + 15 x + 5000

What is the extreme f(x,y)=3x^3-(y+1)^3-9xy+3y+1 ?
The extreme f(x,y)=3x^3-(y+1)^3-9xy+3y+1 is Saddle(0,0),Maximum(-1,1)