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

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

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

extreme points

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

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

Solution steps

Find the critical points:

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

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

D (x, y) = - 24 x - 4

Check the sign of at each critical point

Check critical point

(0, 0) :

Saddle

Check critical point

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

Maximum

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

f (x, y) = 1 - x y

f (x) = \frac{1}{7} x^{7} - x

f (x, y) = \frac{1}{x} + x y - \frac{8}{y}

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

y = 3 x^{2} - 12 x + 13

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