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

The extreme f(x,y)=2x^3+y^3+3x^2-3y+12x-4 is None

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

extreme points

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

None

Solution steps

Find the critical points:

None

2 x^{3} + y^{3} + 3 x^{2} - 3 y + 12 x - 4

has no critical points, therefore there are no extreme points

None

f (x) = \frac{x ^{2} + 3 x - 40}{x + 1}

f (x) = (x + 8)^{\frac{2}{3}}

f (x) = \frac{( x + 3 )}{x ^{2} - 5 x + 6}

f (x) = x^{4} - \frac{3 x ^{2}}{2} - x + \frac{29}{16}

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

What is the extreme f(x,y)=2x^3+y^3+3x^2-3y+12x-4 ?
The extreme f(x,y)=2x^3+y^3+3x^2-3y+12x-4 is None