What is the extreme y^4-y^2-2xy+x^2 ?

The extreme y^4-y^2-2xy+x^2 is Saddle(0,0),Minimum(-1,-1),Minimum(1,1)

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extreme y^4-y^2-2xy+x^2

extreme points

y^{4} - y^{2} - 2 x y + x^{2}

Saddle (0, 0), Minimum (- 1, - 1), Minimum (1, 1)

Solution steps

Find the critical points:

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

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

D (y, x) = 24 y^{2} - 8

Check the sign of at each critical point

Check critical point

(0, 0) :

Saddle

Check critical point

(- 1, - 1) :

Minimum

Check critical point

(1, 1) :

Minimum

Saddle (0, 0), Minimum (- 1, - 1), Minimum (1, 1)

f (x, y) = lo g_{10} (\frac{( x ^{2} + 2 \cdot y )}{( x + y ) ^{2}})

f (x) = y = tan (\frac{π x}{8}), 0 \leq x \leq 2

x^{4} - y^{3} - 2 x^{2} + 3 y + 1

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

f (x, y) = \frac{1}{x - y - 4}

What is the extreme y^4-y^2-2xy+x^2 ?
The extreme y^4-y^2-2xy+x^2 is Saddle(0,0),Minimum(-1,-1),Minimum(1,1)