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

The extreme f(x,y)=x^3+y^2+xy is Saddle(0,0),Minimum(1/6 ,-1/12)

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extreme f(x,y)=x^3+y^2+xy

extreme points

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

Saddle (0, 0), Minimum (\frac{1}{6}, - \frac{1}{12})

Solution steps

Find the critical points:

(0, 0), (\frac{1}{6}, - \frac{1}{12})

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

D (x, y) = 12 x - 1

Check the sign of at each critical point

Check critical point

(0, 0) :

Saddle

Check critical point

(\frac{1}{6}, - \frac{1}{12}) :

Minimum

Saddle (0, 0), Minimum (\frac{1}{6}, - \frac{1}{12})

f (x) = 54 x^{4} - 8 x

\frac{x ^{3}}{x ^{2} + x - 2}

f (x) = x^{4} + 18 x^{2} + 5, - 6 \leq x \leq 6

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

f (x) = (x^{2} - 4 y^{2}) e^{1 - x^{2} - y^{2}}

What is the extreme f(x,y)=x^3+y^2+xy ?
The extreme f(x,y)=x^3+y^2+xy is Saddle(0,0),Minimum(1/6 ,-1/12)