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

The extreme f(x,y)=x^2-y^2sqrt(4+y) is Saddle(0,0),Minimum(0,-16/5)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

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

extreme points

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

Saddle (0, 0), Minimum (0, - \frac{16}{5})

Solution steps

Find the critical points:

(0, 0), (0, - \frac{16}{5})

\frac{\partial ^{2} f}{\partial x ^{2}} = - \frac{15 y ^{2} + 96 y + 128}{4 ( y + 4 ) y + 4}

D (x, y) = - \frac{( 15 y ^{2} + 96 y + 128 ) y + 4}{2 ( y + 4 ) ^{2}}

Check the sign of at each critical point

Check critical point

(0, 0) :

Saddle

Check critical point

(0, - \frac{16}{5}) :

Minimum

Saddle (0, 0), Minimum (0, - \frac{16}{5})

What is the extreme f(x,y)=x^2-y^2sqrt(4+y) ?
The extreme f(x,y)=x^2-y^2sqrt(4+y) is Saddle(0,0),Minimum(0,-16/5)