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

The extreme f(x,y)=2y^3-2y^2-2xy+x^2-12x is Minimum(8,2),Saddle(5,-1)

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

extreme points

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

Minimum (8, 2), Saddle (5, - 1)

Solution steps

Find the critical points:

(8, 2), (5, - 1)

\frac{\partial ^{2} f}{\partial x ^{2}} = 12 y - 4

D (x, y) = 24 y - 12

Check the sign of at each critical point

Check critical point

(8, 2) :

Minimum

Check critical point

(5, - 1) :

Saddle

Minimum (8, 2), Saddle (5, - 1)

f (x) = (1 - x) e^{x}

\frac{x ^{2} + 4 x - 5}{x - 1}

f (x) = 2 x^{2} + 4 x + 5

f (x) = 9 - x^{2}

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

What is the extreme f(x,y)=2y^3-2y^2-2xy+x^2-12x ?
The extreme f(x,y)=2y^3-2y^2-2xy+x^2-12x is Minimum(8,2),Saddle(5,-1)