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

The extreme f(x,y)=9-3y^2+6x^2-2y-3x^3 is Saddle(0,-1/3),Maximum(4/3 ,-1/3)

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

extreme points

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

Saddle (0, - \frac{1}{3}), Maximum (\frac{4}{3}, - \frac{1}{3})

Solution steps

Find the critical points:

(0, - \frac{1}{3}), (\frac{4}{3}, - \frac{1}{3})

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

D (x, y) = - 6 (- 18 x + 12)

Check the sign of at each critical point

Check critical point

(0, - \frac{1}{3}) :

Saddle

Check critical point

(\frac{4}{3}, - \frac{1}{3}) :

Maximum

Saddle (0, - \frac{1}{3}), Maximum (\frac{4}{3}, - \frac{1}{3})

f (x) = \frac{1}{x - 5}, (1, 3)

f (x, y, λ) = 9 - x^{2} - y^{2} + λ (4 - x - y)

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

g (t) = u^{1} (t) + 2 u^{3} (t) - 6 u^{4} (t)

f (x) = ∣ x ∣ + x^{2} - 5 x + 6, (- 1, 4)

What is the extreme f(x,y)=9-3y^2+6x^2-2y-3x^3 ?
The extreme f(x,y)=9-3y^2+6x^2-2y-3x^3 is Saddle(0,-1/3),Maximum(4/3 ,-1/3)