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

The extreme f(x)=x^3+y^2-3x^2-3y^2-9x is Maximum(-1,0),Saddle(3,0)

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

extreme points

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

Maximum (- 1, 0), Saddle (3, 0)

Solution steps

Find the critical points:

(- 1, 0), (3, 0)

\frac{\partial ^{2} f}{\partial x ^{2}} = - 4

D (x, y) = - 4 (6 x - 6)

Check the sign of at each critical point

Check critical point

(- 1, 0) :

Maximum

Check critical point

(3, 0) :

Saddle

Maximum (- 1, 0), Saddle (3, 0)

f (x, y) = 5 + 4 x - 2 x^{2} + 3 y - y^{2}

f (x, y) = x^{3} + y^{3} - 243 x - 27 y - 2

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

\frac{( x - 1 )}{e ^{x + 2}}

14 + 14 x^{- \frac{3}{17}}

What is the extreme f(x)=x^3+y^2-3x^2-3y^2-9x ?
The extreme f(x)=x^3+y^2-3x^2-3y^2-9x is Maximum(-1,0),Saddle(3,0)