What is the extreme f(x,y)=5x^4-x^2+6y^2 ?

The extreme f(x,y)=5x^4-x^2+6y^2 is Saddle(0,0),Minimum(-1/(sqrt(10)),0),Minimum(1/(sqrt(10)),0)

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extreme f(x,y)=5x^4-x^2+6y^2

extreme points

f (x, y) = 5 x^{4} - x^{2} + 6 y^{2}

Saddle (0, 0), Minimum (- \frac{1}{10}, 0), Minimum (\frac{1}{10}, 0)

Solution steps

Find the critical points:

(0, 0), (- \frac{1}{10}, 0), (\frac{1}{10}, 0)

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

D (x, y) = 12 (60 x^{2} - 2)

Check the sign of at each critical point

Check critical point

(0, 0) :

Saddle

Check critical point

(- \frac{1}{10}, 0) :

Minimum

Check critical point

(\frac{1}{10}, 0) :

Minimum

Saddle (0, 0), Minimum (- \frac{1}{10}, 0), Minimum (\frac{1}{10}, 0)

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

f (x) = \frac{2}{x + 7}

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

f (x) = - 16 - 7 x - x^{2}

f (x, y) = - 5 x^{2} - 8 y^{2} - 2 x y + 42 x + 102 y

What is the extreme f(x,y)=5x^4-x^2+6y^2 ?
The extreme f(x,y)=5x^4-x^2+6y^2 is Saddle(0,0),Minimum(-1/(sqrt(10)),0),Minimum(1/(sqrt(10)),0)