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

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

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extreme f(x,y)=x^3+6xy+2y^2+6x

extreme points

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

Saddle (1, - \frac{3}{2}), Minimum (2, - 3)

Solution steps

Find the critical points:

(1, - \frac{3}{2}), (2, - 3)

\frac{\partial ^{2} f}{\partial x ^{2}} = 4

D (x, y) = 24 x - 36

Check the sign of at each critical point

Check critical point

(1, - \frac{3}{2}) :

Saddle

Check critical point

(2, - 3) :

Minimum

Saddle (1, - \frac{3}{2}), Minimum (2, - 3)

f (x, y) = 5 x^{2} + 3 x y + y^{2}

f (x, y) = x + y - x y

\frac{- 3 x + 48}{( x - 120 ) ^{9}}

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

e^{x}

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