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

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

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

extreme points

f (x) = 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) = 6 x + 4 x^{- 1}

f (x) = - 3 x^{3} + 81 x + 9

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

f (r, s) = \frac{2 r}{r ^{4} + 7 s}

f (x) = 1 + \frac{7}{x} - \frac{2}{x ^{2}}

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