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

The extreme f(x,y)=3x^2y^3-y^2-x^2+6 is

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

extreme points

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

Maximum (0, 0), Saddle (\frac{2}{6 3 ^{5}}, \frac{1}{3 3}), Saddle (- \frac{2}{6 3 ^{5}}, \frac{1}{3 3})

Solution steps

Find the critical points:

(0, 0), (\frac{2}{6 3 ^{5}}, \frac{1}{3 3}), (- \frac{2}{6 3 ^{5}}, \frac{1}{3 3})

\frac{\partial ^{2} f}{\partial x ^{2}} = 18 x^{2} y - 2

D (x, y) = - 216 x^{2} y^{4} - 12 y^{3} - 36 x^{2} y + 4

Check the sign of at each critical point

Check critical point

(0, 0) :

Maximum

Check critical point

(\frac{2}{6 3 ^{5}}, \frac{1}{3 3}) :

Saddle

Check critical point

(- \frac{2}{6 3 ^{5}}, \frac{1}{3 3}) :

Saddle

Maximum (0, 0), Saddle (\frac{2}{6 3 ^{5}}, \frac{1}{3 3}), Saddle (- \frac{2}{6 3 ^{5}}, \frac{1}{3 3})

f (x) = \frac{x ^{2} + x + 6}{x - 1}

y = x^{\frac{4}{7}} (\frac{11}{4} + x)

f (x) = ln (x^{2} + 9)

\frac{1}{3} x^{3} + 2 x^{2} + 4 x + 1

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

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