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

The extreme f(x,y)=3y^2+x^3-6xy-9x is Minimum(3,3),Saddle(-1,-1)

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

extreme points

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

Minimum (3, 3), Saddle (- 1, - 1)

Solution steps

Find the critical points:

(3, 3), (- 1, - 1)

\frac{\partial ^{2} f}{\partial x ^{2}} = 6

D (x, y) = 36 x - 36

Check the sign of at each critical point

Check critical point

(3, 3) :

Minimum

Check critical point

(- 1, - 1) :

Saddle

Minimum (3, 3), Saddle (- 1, - 1)

f (x) = 5 sin (x) cos (x)

f (x, y) = x^{4} + y^{2} - 2 x^{2} - 5

f (x) = x + \frac{3}{x}

f (x) = \frac{x}{25 + x ^{2}}

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

What is the extreme f(x,y)=3y^2+x^3-6xy-9x ?
The extreme f(x,y)=3y^2+x^3-6xy-9x is Minimum(3,3),Saddle(-1,-1)