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

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

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extreme f(x,y)=-x^3+6xy-3y^2+1

extreme points

f (x, y) = - x^{3} + 6 x y - 3 y^{2} + 1

Saddle (0, 0), Maximum (2, 2)

Solution steps

Find the critical points:

(0, 0), (2, 2)

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

D (x, y) = 36 x - 36

Check the sign of at each critical point

Check critical point

(0, 0) :

Saddle

Check critical point

(2, 2) :

Maximum

Saddle (0, 0), Maximum (2, 2)

f (x) = \frac{- 8}{x ^{2} - 1}

f (x) = 2 - x^{2}

x^{4} - 32 x^{2}

f (x, y) = 16 + (x - 2)^{2} - 4 (y + 1)^{2}

f (x) = 3 x^{4} - 4 x^{3} - 8, - 1 \leq x \leq 2

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