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

The extreme f(x)=y^3-x^2+2y^2-4x-3 is Saddle(-2,0),Maximum(-2,-4/3)

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

extreme points

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

Saddle (- 2, 0), Maximum (- 2, - \frac{4}{3})

Solution steps

Find the critical points:

(- 2, 0), (- 2, - \frac{4}{3})

\frac{\partial ^{2} f}{\partial x ^{2}} = 6 y + 4

D (x, y) = - 2 (6 y + 4)

Check the sign of at each critical point

Check critical point

(- 2, 0) :

Saddle

Check critical point

(- 2, - \frac{4}{3}) :

Maximum

Saddle (- 2, 0), Maximum (- 2, - \frac{4}{3})

f (x, y) = 2 x^{3} + x y^{2} + 5 x^{2} + y^{2} + 3

\frac{x ^{2} - 48}{x - 7}

4 x^{3} - x^{4}

f (x, y) = 2 x^{2} + y^{2} + 2 x y + 2 x + 2 y

f (x, y) = 3 x + 4 y

What is the extreme f(x)=y^3-x^2+2y^2-4x-3 ?
The extreme f(x)=y^3-x^2+2y^2-4x-3 is Saddle(-2,0),Maximum(-2,-4/3)