What is the extreme f(x,y)=-x^3+4xy-2y^2+4x+1 ?

The extreme f(x,y)=-x^3+4xy-2y^2+4x+1 is Saddle(-2/3 ,-2/3),Maximum(2,2)

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extreme f(x,y)=-x^3+4xy-2y^2+4x+1

extreme points

f (x, y) = - x^{3} + 4 x y - 2 y^{2} + 4 x + 1

Saddle (- \frac{2}{3}, - \frac{2}{3}), Maximum (2, 2)

Solution steps

Find the critical points:

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

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

D (x, y) = 24 x - 16

Check the sign of at each critical point

Check critical point

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

Saddle

Check critical point

(2, 2) :

Maximum

Saddle (- \frac{2}{3}, - \frac{2}{3}), Maximum (2, 2)

f (x, y) = x^{3} + y^{3} + 9 x^{2} - 3 y^{2} + 15 x - 9 y

f (x) = 2 x + sin (4 x)

y = \frac{x ^{2} - 7}{x - 4}

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

f (x) = - 2 x^{3} + 39 x^{2} - 216 x + 9

What is the extreme f(x,y)=-x^3+4xy-2y^2+4x+1 ?
The extreme f(x,y)=-x^3+4xy-2y^2+4x+1 is Saddle(-2/3 ,-2/3),Maximum(2,2)