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

The extreme f(x,y)=-y^2+xy-x^3+y+3 is Saddle(-1/3 , 1/3),Maximum(1/2 , 3/4)

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

extreme points

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

Saddle (- \frac{1}{3}, \frac{1}{3}), Maximum (\frac{1}{2}, \frac{3}{4})

Solution steps

Find the critical points:

(- \frac{1}{3}, \frac{1}{3}), (\frac{1}{2}, \frac{3}{4})

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

D (x, y) = 12 x - 1

Check the sign of at each critical point

Check critical point

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

Saddle

Check critical point

(\frac{1}{2}, \frac{3}{4}) :

Maximum

Saddle (- \frac{1}{3}, \frac{1}{3}), Maximum (\frac{1}{2}, \frac{3}{4})

2 - cos (x)

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

f (x) = x^{3} - 6 x^{2} + 9 x + 6

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

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

What is the extreme f(x,y)=-y^2+xy-x^3+y+3 ?
The extreme f(x,y)=-y^2+xy-x^3+y+3 is Saddle(-1/3 , 1/3),Maximum(1/2 , 3/4)