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

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

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

extreme points

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

Saddle (0, 0), Minimum (4, - 2), Minimum (- 4, 2)

Solution steps

Find the critical points:

(0, 0), (4, - 2), (- 4, 2)

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

D (x, y) = 48 y^{2} - 64

Check the sign of at each critical point

Check critical point

(0, 0) :

Saddle

Check critical point

(4, - 2) :

Minimum

Check critical point

(- 4, 2) :

Minimum

Saddle (0, 0), Minimum (4, - 2), Minimum (- 4, 2)

f (x) = 2 x^{3} + 9 x^{2} + 1

F (m, a) = m \cdot a

f (x) = x - x^{2}

f (x, y) = - 2 x^{3} - 2 y^{3} + 6 x y + 10

ln (x) + ln (2 - x)

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