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

The extreme f(x,y)=x^3ln(y^2+1)-2y is None

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

extreme points

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

None

Solution steps

Find the critical points:

None

x^{3} ln (y^{2} + 1) - 2 y

has no critical points, therefore there are no extreme points

None

f (x) = (\frac{x}{5})^{1} \cdot (\frac{y}{5})^{x - 1}

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

f (x) = x^{3} - 3 x^{2} + 10

f (x) = (x^{2} - 3) e^{x}

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

What is the extreme f(x,y)=x^3ln(y^2+1)-2y ?
The extreme f(x,y)=x^3ln(y^2+1)-2y is None