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

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

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extreme f(x,y)=4-y^3-x^2-3xy

extreme points

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

Saddle (0, 0), Maximum (- \frac{9}{4}, \frac{3}{2})

Solution steps

Find the critical points:

(0, 0), (- \frac{9}{4}, \frac{3}{2})

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

D (x, y) = 12 y - 9

Check the sign of at each critical point

Check critical point

(0, 0) :

Saddle

Check critical point

(- \frac{9}{4}, \frac{3}{2}) :

Maximum

Saddle (0, 0), Maximum (- \frac{9}{4}, \frac{3}{2})

- 74 + 30 x - 3 x^{2}

f (x, y) = 50 - 7 x^{2} - 2 y^{2}

4^{x} + 1

f (x) = x + \frac{2 \cdot 1800}{x}

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

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