What is the extreme f(x,y)=x^3+xy^2-8x+4y-1 ?

The extreme f(x,y)=x^3+xy^2-8x+4y-1 is

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extreme f(x,y)=x^3+xy^2-8x+4y-1

extreme points

f (x, y) = x^{3} + x y^{2} - 8 x + 4 y - 1

Saddle (- \frac{2}{3}, 6), Saddle (\frac{2}{3}, - 6), Maximum (- 2, 2), Minimum (2, - 2)

Solution steps

Find the critical points:

(- \frac{2}{3}, 6), (\frac{2}{3}, - 6), (- 2, 2), (2, - 2)

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

D (x, y) = 12 x^{2} - 4 y^{2}

Check the sign of at each critical point

Check critical point

(- \frac{2}{3}, 6) :

Saddle

Check critical point

(\frac{2}{3}, - 6) :

Saddle

Check critical point

(- 2, 2) :

Maximum

Check critical point

(2, - 2) :

Minimum

Saddle (- \frac{2}{3}, 6), Saddle (\frac{2}{3}, - 6), Maximum (- 2, 2), Minimum (2, - 2)

f (x) = ln (8 + 8 x^{3})

f (x) = sin (x) + x

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

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

f (x) = 4 x^{3} - 9 x

What is the extreme f(x,y)=x^3+xy^2-8x+4y-1 ?
The extreme f(x,y)=x^3+xy^2-8x+4y-1 is