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

The extreme f(x)=4xy-x^4-y^4 is Saddle(0,0),Maximum(-1,-1),Maximum(1,1)

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extreme f(x)=4xy-x^4-y^4

extreme points

f (x) = 4 x y - x^{4} - y^{4}

Saddle (0, 0), Maximum (- 1, - 1), Maximum (1, 1)

Solution steps

Find the critical points:

(0, 0), (- 1, - 1), (1, 1)

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

D (x, y) = 144 x^{2} y^{2} - 16

Check the sign of at each critical point

Check critical point

(0, 0) :

Saddle

Check critical point

(- 1, - 1) :

Maximum

Check critical point

(1, 1) :

Maximum

Saddle (0, 0), Maximum (- 1, - 1), Maximum (1, 1)

f (x, y) = 6 - 3 x - 2 y

f (x) = cos (x) + sin (x)

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

f (x) = 2 x^{3} + 3 x^{2} - 120 x, 0 \leq x \leq 5

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

What is the extreme f(x)=4xy-x^4-y^4 ?
The extreme f(x)=4xy-x^4-y^4 is Saddle(0,0),Maximum(-1,-1),Maximum(1,1)