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

The extreme f(x)=8y^2+x^2-x^2y is Minimum(0,0),Saddle(4,1),Saddle(-4,1)

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extreme f(x)=8y^2+x^2-x^2y

extreme points

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

Minimum (0, 0), Saddle (4, 1), Saddle (- 4, 1)

Solution steps

Find the critical points:

(0, 0), (4, 1), (- 4, 1)

\frac{\partial ^{2} f}{\partial x ^{2}} = 16

D (x, y) = 16 (2 - 2 y) - 4 x^{2}

Check the sign of at each critical point

Check critical point

(0, 0) :

Minimum

Check critical point

(4, 1) :

Saddle

Check critical point

(- 4, 1) :

Saddle

Minimum (0, 0), Saddle (4, 1), Saddle (- 4, 1)

f (x) = \frac{4 x - 3}{x}

f (x) = \frac{x ^{2} - 1}{- 5 x + \frac{1}{2}}

f (x) = - 2 x^{3} - 33 x^{2} - 108 x - 1

3 x + 1

3 x + 5

What is the extreme f(x)=8y^2+x^2-x^2y ?
The extreme f(x)=8y^2+x^2-x^2y is Minimum(0,0),Saddle(4,1),Saddle(-4,1)