What is the extreme f(x,y)=2y^3-3xy+x^2-1 ?

The extreme f(x,y)=2y^3-3xy+x^2-1 is Minimum(9/8 , 3/4),Saddle(0,0)

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extreme f(x,y)=2y^3-3xy+x^2-1

extreme points

f (x, y) = 2 y^{3} - 3 x y + x^{2} - 1

Minimum (\frac{9}{8}, \frac{3}{4}), Saddle (0, 0)

Solution steps

Find the critical points:

(\frac{9}{8}, \frac{3}{4}), (0, 0)

\frac{\partial ^{2} f}{\partial x ^{2}} = 12 y

D (x, y) = 24 y - 9

Check the sign of at each critical point

Check critical point

(\frac{9}{8}, \frac{3}{4}) :

Minimum

Check critical point

(0, 0) :

Saddle

Minimum (\frac{9}{8}, \frac{3}{4}), Saddle (0, 0)

- \frac{cos ( 2 x )}{2} - 2 sin (x)

f (x) = 2 x^{3} - x^{2} - 4 x - 1

x 1 - x^{2}

f (x) = 2 x^{4} - 16 x^{2} + 3

f (x) = \frac{2}{5} x^{5} - 7 x^{4} + 30 x^{3} - 27

What is the extreme f(x,y)=2y^3-3xy+x^2-1 ?
The extreme f(x,y)=2y^3-3xy+x^2-1 is Minimum(9/8 , 3/4),Saddle(0,0)