What is the extreme f(xy)=x^3+y^2-6xy+9x+5y+2 ?

The extreme f(xy)=x^3+y^2-6xy+9x+5y+2 is Minimum(4, 19/2),Saddle(2, 7/2)

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extreme f(xy)=x^3+y^2-6xy+9x+5y+2

extreme points

f (x y) = x^{3} + y^{2} - 6 x y + 9 x + 5 y + 2

Minimum (4, \frac{19}{2}), Saddle (2, \frac{7}{2})

Solution steps

Find the critical points:

(4, \frac{19}{2}), (2, \frac{7}{2})

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

D (x, y) = 12 x - 36

Check the sign of at each critical point

Check critical point

(4, \frac{19}{2}) :

Minimum

Check critical point

(2, \frac{7}{2}) :

Saddle

Minimum (4, \frac{19}{2}), Saddle (2, \frac{7}{2})

f (x, y) = e^{- x - y^{2}} \cdot (3 x - y^{2})

x^{2} + y^{2} + 2 x - 4 y + 8

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

f (x, y) = 3 x^{3} - y^{3} - 9 x + 12 y + 3

f (x) = \frac{x}{5} + \frac{4 + ( 6 - x ) ^{2}}{3}

What is the extreme f(xy)=x^3+y^2-6xy+9x+5y+2 ?
The extreme f(xy)=x^3+y^2-6xy+9x+5y+2 is Minimum(4, 19/2),Saddle(2, 7/2)