What is the extreme f(x,y)=x^3-3x+y^2+6y+8 ?

The extreme f(x,y)=x^3-3x+y^2+6y+8 is Saddle(-1,-3),Minimum(1,-3)

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extreme f(x,y)=x^3-3x+y^2+6y+8

extreme points

f (x, y) = x^{3} - 3 x + y^{2} + 6 y + 8

Saddle (- 1, - 3), Minimum (1, - 3)

Solution steps

Find the critical points:

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

\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, - 3) :

Saddle

Check critical point

(1, - 3) :

Minimum

Saddle (- 1, - 3), Minimum (1, - 3)

f (x, y) = 8 + 76 x y + 38 x^{2} + 240 y + \frac{y ^{4}}{4}

y = \frac{x ^{2}}{ln ( x )}

f (x) = x^{3} - 6 x^{2} + 9 + 9, - 1 \leq x \leq 9

f (x, y) = 16 - x^{2} - 16 y^{2}

f (x) = x^{3} + y^{3} + 9 x^{2} - 3 y^{2} - 8

What is the extreme f(x,y)=x^3-3x+y^2+6y+8 ?
The extreme f(x,y)=x^3-3x+y^2+6y+8 is Saddle(-1,-3),Minimum(1,-3)