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

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

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

extreme points

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

Maximum (- 1, - \frac{3}{2})

Solution steps

Find the critical points:

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

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

D (x, y) = 8

Check the sign of at each critical point

Check critical point

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

Maximum

Maximum (- 1, - \frac{3}{2})

f (x, y) = x^{3} + 8 y^{3} - 12 x y

f (x) = 4 x e^{- 4 x + 1}

f (x) = \frac{( x ^{2} - 9 x + 50 )}{x - 7}

f (x) = x^{4} - 32 x + 9

f (x) = - 6 x^{2} - 12 x - 9, - 3 \leq x \leq 2

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