What is the extreme f(x,y)=3x^2-y^2+4 ?

The extreme f(x,y)=3x^2-y^2+4 is Saddle(0,0)

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extreme f(x,y)=3x^2-y^2+4

extreme points

f (x, y) = 3 x^{2} - y^{2} + 4

Saddle (0, 0)

Solution steps

Find the critical points:

(0, 0)

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

D (x, y) = - 12

Check the sign of at each critical point

Check critical point

(0, 0) :

Saddle

Saddle (0, 0)

f (x) = 12 x^{2} - 48 x + 20

f (x) = 4 x^{4} - 32 x^{3}

y = 3 x^{\frac{2}{3}} - 2 x [- 1.1]

f (x) = \frac{7 + x}{8 - x}

f (x) = x^{7} - x^{5}

What is the extreme f(x,y)=3x^2-y^2+4 ?
The extreme f(x,y)=3x^2-y^2+4 is Saddle(0,0)