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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

extreme f(x)=x^2-4y^2-4

extreme points

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

Saddle (0, 0)

Solution steps

Find the critical points:

(0, 0)

\frac{\partial ^{2} f}{\partial x ^{2}} = - 8

D (x, y) = - 16

Check the sign of at each critical point

Check critical point

(0, 0) :

Saddle

Saddle (0, 0)

f (x, y) = \frac{5 - x ^{2} + 3 y}{y}

2 cos (x) + sin (2 x), 0 \leq x \leq \frac{π}{2}

f (x) = x ln (x) + 1

f (x) = \frac{4 x ^{3}}{3} - 7 x + 6

f (x) = x^{4} + x^{3} + x^{2} + 1

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