What is the extreme f(x,y)=1-3x^4-6x^2y+2y^3 ?

The extreme f(x,y)=1-3x^4-6x^2y+2y^3 is Maximum(1,-1),Maximum(-1,-1)

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extreme f(x,y)=1-3x^4-6x^2y+2y^3

extreme points

f (x, y) = 1 - 3 x^{4} - 6 x^{2} y + 2 y^{3}

Maximum (1, - 1), Maximum (- 1, - 1)

Solution steps

Find the critical points:

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

\frac{\partial ^{2} f}{\partial x ^{2}} = 12 y

D (x, y) = 12 y (- 36 x^{2} - 12 y) - 144 x^{2}

Check the sign of at each critical point

Check critical point

(1, - 1) :

Maximum

Check critical point

(- 1, - 1) :

Maximum

Maximum (1, - 1), Maximum (- 1, - 1)

f (x) = 8 x^{3} + 13 x

f (x, y) = 18 y^{2} + x^{2} - x^{2} y

2 (\frac{1}{2})^{x}

f (x) = \frac{( x - 4 ) ^{2}}{x + 10}

f (x) = 4 x - 1

What is the extreme f(x,y)=1-3x^4-6x^2y+2y^3 ?
The extreme f(x,y)=1-3x^4-6x^2y+2y^3 is Maximum(1,-1),Maximum(-1,-1)