What is the extreme f(x)=-3x^2+3xy-2x+y^2+6y+12 ?

The extreme f(x)=-3x^2+3xy-2x+y^2+6y+12 is Saddle(-22/21 ,-10/7)

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extreme f(x)=-3x^2+3xy-2x+y^2+6y+12

extreme points

f (x) = - 3 x^{2} + 3 x y - 2 x + y^{2} + 6 y + 12

Saddle (- \frac{22}{21}, - \frac{10}{7})

Solution steps

Find the critical points:

(- \frac{22}{21}, - \frac{10}{7})

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

D (x, y) = - 21

Check the sign of at each critical point

Check critical point

(- \frac{22}{21}, - \frac{10}{7}) :

Saddle

Saddle (- \frac{22}{21}, - \frac{10}{7})

f (x) = \frac{x}{16 + x ^{2}}

f (x) = 4 - x

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

\frac{2}{1 + x ^{2}}

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

What is the extreme f(x)=-3x^2+3xy-2x+y^2+6y+12 ?
The extreme f(x)=-3x^2+3xy-2x+y^2+6y+12 is Saddle(-22/21 ,-10/7)