What is the extreme f(x)=(1/9 x^2+y^2)e^{(-x)/3+y} ?

The extreme f(x)=(1/9 x^2+y^2)e^{(-x)/3+y} is Saddle(3,-1),Minimum(0,0),Saddle(3,-1)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

extreme f(x)=(1/9 x^2+y^2)e^{(-x)/3+y}

extreme points

f (x) = (\frac{1}{9} x^{2} + y^{2}) e^{\frac{- x}{3} + y}

Saddle (3, - 1), Minimum (0, 0), Saddle (3, - 1)

Solution steps

Find the critical points:

(3, - 1), (0, 0), (3, - 1)

\frac{\partial ^{2} f}{\partial x ^{2}} = e^{y - \frac{x}{3}} (y^{2} + 4 y + \frac{x ^{2}}{9} + 2)

D (x, y) = (\frac{e ^{\frac{- x + 3 y}{3}} x ^{2}}{81} + \frac{e ^{\frac{- x + 3 y}{3}} y ^{2}}{9} - \frac{4 e ^{\frac{- x + 3 y}{3}} x}{27} + \frac{2}{9} e^{\frac{- x + 3 y}{3}}) e^{y - \frac{x}{3}} (y^{2} + 4 y + \frac{x ^{2}}{9} + 2) - ((- \frac{1}{3} e^{y - \frac{x}{3}} (y^{2} + \frac{x ^{2}}{9} + 2 y) + \frac{2 e ^{\frac{- x + 3 y}{3}} x}{9}))^{2}

Check the sign of at each critical point

Check critical point

(3, - 1) :

Saddle

Check critical point

(0, 0) :

Minimum

Check critical point

(3, - 1) :

Saddle

Saddle (3, - 1), Minimum (0, 0), Saddle (3, - 1)

f (x) = 16 x^{3} (x + 1)^{2}

2 x^{\frac{3}{2}} - 9 x + 6

f (x) = 5 x^{2} - 8

f (x) = 5 x^{2} + 1

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

What is the extreme f(x)=(1/9 x^2+y^2)e^{(-x)/3+y} ?
The extreme f(x)=(1/9 x^2+y^2)e^{(-x)/3+y} is Saddle(3,-1),Minimum(0,0),Saddle(3,-1)