What is the extreme f(x)=2x^3+y^3+3x^2-3y-12x-4 ?

The extreme f(x)=2x^3+y^3+3x^2-3y-12x-4 is Minimum(1,1),Saddle(1,-1),Saddle(-2,1),Maximum(-2,-1)

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extreme f(x)=2x^3+y^3+3x^2-3y-12x-4

extreme points

f (x) = 2 x^{3} + y^{3} + 3 x^{2} - 3 y - 12 x - 4

Minimum (1, 1), Saddle (1, - 1), Saddle (- 2, 1), Maximum (- 2, - 1)

Solution steps

Find the critical points:

(1, 1), (1, - 1), (- 2, 1), (- 2, - 1)

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

D (x, y) = 6 y (12 x + 6)

Check the sign of at each critical point

Check critical point

(1, 1) :

Minimum

Check critical point

(1, - 1) :

Saddle

Check critical point

(- 2, 1) :

Saddle

Check critical point

(- 2, - 1) :

Maximum

Minimum (1, 1), Saddle (1, - 1), Saddle (- 2, 1), Maximum (- 2, - 1)

f (x) = \frac{2 x}{x ^{2} + 4}, - 4 \leq x \leq 4

\frac{5 x}{x ^{2} - 9}

3 x^{2} + 24 x + 45

f (x) = \frac{e ^{x}}{x ^{7}}

f (x) = 7 x^{4} - 2 x^{2} - 8 x + 8

What is the extreme f(x)=2x^3+y^3+3x^2-3y-12x-4 ?
The extreme f(x)=2x^3+y^3+3x^2-3y-12x-4 is Minimum(1,1),Saddle(1,-1),Saddle(-2,1),Maximum(-2,-1)