What is the extreme f(x)=-27xy+x^3+y^3 ?

The extreme f(x)=-27xy+x^3+y^3 is Saddle(0,0),Minimum(9,9)

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extreme f(x)=-27xy+x^3+y^3

extreme points

f (x) = - 27 x y + x^{3} + y^{3}

Saddle (0, 0), Minimum (9, 9)

Solution steps

Find the critical points:

(0, 0), (9, 9)

\frac{\partial ^{2} f}{\partial x ^{2}} = 6 y

D (x, y) = 36 x y - 729

Check the sign of at each critical point

Check critical point

(0, 0) :

Saddle

Check critical point

(9, 9) :

Minimum

Saddle (0, 0), Minimum (9, 9)

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

y = - 11 sin (22 x + 33) + 44

f (t) = 2 e^{l n (x)} (t)

f (x) = \frac{2 ( x + 6 ) ^{2}}{x + 8}

f (x) = \frac{x ^{3} + x - 2}{x ^{2} - 4 x + 3}

What is the extreme f(x)=-27xy+x^3+y^3 ?
The extreme f(x)=-27xy+x^3+y^3 is Saddle(0,0),Minimum(9,9)