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

The extreme f(x)=x^3+y^2-6xy+6x+3y-2 is Minimum(5, 27/2),Saddle(1, 3/2)

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

extreme points

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

Minimum (5, \frac{27}{2}), Saddle (1, \frac{3}{2})

Solution steps

Find the critical points:

(5, \frac{27}{2}), (1, \frac{3}{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

(5, \frac{27}{2}) :

Minimum

Check critical point

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

Saddle

Minimum (5, \frac{27}{2}), Saddle (1, \frac{3}{2})

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

f (x) = 2 x^{4} - 7 x^{3} + 4 x^{2} + 2 x + 3

f (x) = 5 t^{2} + 24 t + 136

4 x^{3} - 3 x^{2} - 36 x + 17

f (x) = \frac{ln ( x )}{x ^{\frac{5}{2}}}

What is the extreme f(x)=x^3+y^2-6xy+6x+3y-2 ?
The extreme f(x)=x^3+y^2-6xy+6x+3y-2 is Minimum(5, 27/2),Saddle(1, 3/2)