What is the extreme f(x,y)=2x^3+y^2-6xy-23 ?

The extreme f(x,y)=2x^3+y^2-6xy-23 is Minimum(3,9),Saddle(0,0)

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extreme f(x,y)=2x^3+y^2-6xy-23

extreme points

f (x, y) = 2 x^{3} + y^{2} - 6 x y - 23

Minimum (3, 9), Saddle (0, 0)

Solution steps

Find the critical points:

(3, 9), (0, 0)

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

D (x, y) = 24 x - 36

Check the sign of at each critical point

Check critical point

(3, 9) :

Minimum

Check critical point

(0, 0) :

Saddle

Minimum (3, 9), Saddle (0, 0)

f (x) = \frac{x}{x - 4}, 6 \leq x \leq 10

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

f (x) = \frac{3 x ^{2}}{x ^{2} - 16}

f (x) = x^{4} + 32 x

p (x) = 3 x^{3} - 4 x^{2} + a x - 50

What is the extreme f(x,y)=2x^3+y^2-6xy-23 ?
The extreme f(x,y)=2x^3+y^2-6xy-23 is Minimum(3,9),Saddle(0,0)