What is the extreme f(x)=-2x^3-2y^3+6xy+10 ?

The extreme f(x)=-2x^3-2y^3+6xy+10 is Saddle(0,0),Maximum(1,1)

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extreme f(x)=-2x^3-2y^3+6xy+10

extreme points

f (x) = - 2 x^{3} - 2 y^{3} + 6 x y + 10

Saddle (0, 0), Maximum (1, 1)

Solution steps

Find the critical points:

(0, 0), (1, 1)

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

D (x, y) = 144 x y - 36

Check the sign of at each critical point

Check critical point

(0, 0) :

Saddle

Check critical point

(1, 1) :

Maximum

Saddle (0, 0), Maximum (1, 1)

8 x^{3} + 2 x y - 3 x^{2} + y^{2} + 1

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

P (a, b) = 2 a + b

y = x^{2} - 2 ln (x)

f (x) = - e^{\frac{2}{7} x}

What is the extreme f(x)=-2x^3-2y^3+6xy+10 ?
The extreme f(x)=-2x^3-2y^3+6xy+10 is Saddle(0,0),Maximum(1,1)