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

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

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

extreme points

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

Saddle (0, 0)

Solution steps

Find the critical points:

(0, 0)

\frac{\partial ^{2} f}{\partial x ^{2}} = 0

D (x, y) = - 9

Check the sign of at each critical point

Check critical point

(0, 0) :

Saddle

Saddle (0, 0)

f (x) = - x^{3} - 3 x^{2} + 24 x - 6

f (x) = 1 - 2 x, x \geq - 1

4 x^{3} e^{- x}

f (x) = 9 sin (x) + 9 cos (x), 0 \leq x \leq 2 π

f (x) = ln (8 - ln (x))

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