What is the extreme f(x,y)=8x^4+y^2-24xy ?

The extreme f(x,y)=8x^4+y^2-24xy is Saddle(0,0),Minimum(-3,-36),Minimum(3,36)

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extreme f(x,y)=8x^4+y^2-24xy

extreme points

f (x, y) = 8 x^{4} + y^{2} - 24 x y

Saddle (0, 0), Minimum (- 3, - 36), Minimum (3, 36)

Solution steps

Find the critical points:

(0, 0), (- 3, - 36), (3, 36)

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

D (x, y) = 192 x^{2} - 576

Check the sign of at each critical point

Check critical point

(0, 0) :

Saddle

Check critical point

(- 3, - 36) :

Minimum

Check critical point

(3, 36) :

Minimum

Saddle (0, 0), Minimum (- 3, - 36), Minimum (3, 36)

f (x) = x^{3} - 3 x^{2} - 9 x - 5

ln (x + 3) - 1

f (x) = 6 sin (x)

f (x) = x + 1 + 1 - x

f (x) = x^{3} - 3 x^{2} + 5 x - 6

What is the extreme f(x,y)=8x^4+y^2-24xy ?
The extreme f(x,y)=8x^4+y^2-24xy is Saddle(0,0),Minimum(-3,-36),Minimum(3,36)