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

The extreme f(x,y)=8x^3+2xy-3x^2+y^2+1 is Minimum(1/3 ,-1/3),Saddle(0,0)

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extreme f(x,y)=8x^3+2xy-3x^2+y^2+1

extreme points

f (x, y) = 8 x^{3} + 2 x y - 3 x^{2} + y^{2} + 1

Minimum (\frac{1}{3}, - \frac{1}{3}), Saddle (0, 0)

Solution steps

Find the critical points:

(\frac{1}{3}, - \frac{1}{3}), (0, 0)

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

D (x, y) = 96 x - 16

Check the sign of at each critical point

Check critical point

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

Minimum

Check critical point

(0, 0) :

Saddle

Minimum (\frac{1}{3}, - \frac{1}{3}), Saddle (0, 0)

f (x) = 3 x + 4 y

\frac{1 - x ^{2}}{x}

f (x) = 3 x^{2} - 18 x + 24

f (x, y) = x^{3} + y^{3} - 24 x y

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

What is the extreme f(x,y)=8x^3+2xy-3x^2+y^2+1 ?
The extreme f(x,y)=8x^3+2xy-3x^2+y^2+1 is Minimum(1/3 ,-1/3),Saddle(0,0)