What is the extreme f(x)=x^3+2xy^2-5x-4y+20 ?

The extreme f(x)=x^3+2xy^2-5x-4y+20 is

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extreme f(x)=x^3+2xy^2-5x-4y+20

extreme points

f (x) = x^{3} + 2 x y^{2} - 5 x - 4 y + 20

Saddle (\frac{2}{3}, \frac{3}{2}), Saddle (- \frac{2}{3}, - \frac{3}{2}), Minimum (1, 1), Maximum (- 1, - 1)

Solution steps

Find the critical points:

(\frac{2}{3}, \frac{3}{2}), (- \frac{2}{3}, - \frac{3}{2}), (1, 1), (- 1, - 1)

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

D (x, y) = 24 x^{2} - 16 y^{2}

Check the sign of at each critical point

Check critical point

(\frac{2}{3}, \frac{3}{2}) :

Saddle

Check critical point

(- \frac{2}{3}, - \frac{3}{2}) :

Saddle

Check critical point

(1, 1) :

Minimum

Check critical point

(- 1, - 1) :

Maximum

Saddle (\frac{2}{3}, \frac{3}{2}), Saddle (- \frac{2}{3}, - \frac{3}{2}), Minimum (1, 1), Maximum (- 1, - 1)

f (t) = 5 e^{a t}

f (x) = I x - 4 I

f (x) = x ((\frac{1}{3}) x - 2)^{3} + 10

f (x) = \frac{x}{x ^{2} - x + 25}, 0 \leq x \leq 27

A - \frac{2}{X - 2} + \frac{1}{( X - 2 ) ^{2}}

What is the extreme f(x)=x^3+2xy^2-5x-4y+20 ?
The extreme f(x)=x^3+2xy^2-5x-4y+20 is