What is the extreme f(x)=x^2+3xy+y^2+x+3 ?

The extreme f(x)=x^2+3xy+y^2+x+3 is Saddle(2/5 ,-3/5)

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+3xy+y^2+x+3

extreme points

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

Saddle (\frac{2}{5}, - \frac{3}{5})

Solution steps

Find the critical points:

(\frac{2}{5}, - \frac{3}{5})

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

D (x, y) = - 5

Check the sign of at each critical point

Check critical point

(\frac{2}{5}, - \frac{3}{5}) :

Saddle

Saddle (\frac{2}{5}, - \frac{3}{5})

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

f (x, y) = x^{2} + y^{2} + x y - 8 x - 7 y + 400

f (x) = \frac{( x ^{2} - 12 )}{x - 4}

f (t) = t e^{\frac{w}{t}}

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

What is the extreme f(x)=x^2+3xy+y^2+x+3 ?
The extreme f(x)=x^2+3xy+y^2+x+3 is Saddle(2/5 ,-3/5)