What is the extreme f(y)=14x^2-2x^3+2y^2+4xy ?

The extreme f(y)=14x^2-2x^3+2y^2+4xy is Saddle(-4,4),Minimum(0,0)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

extreme f(y)=14x^2-2x^3+2y^2+4xy

extreme points

f (y) = 14 x^{2} - 2 x^{3} + 2 y^{2} + 4 x y

Saddle (- 4, 4), Minimum (0, 0)

Solution steps

Find the critical points:

(- 4, 4), (0, 0)

\frac{\partial ^{2} f}{\partial y ^{2}} = - 12 x + 28

D (y, x) = - 48 x + 96

Check the sign of at each critical point

Check critical point

(- 4, 4) :

Saddle

Check critical point

(0, 0) :

Minimum

Saddle (- 4, 4), Minimum (0, 0)

f (x) = x (152 - x)

f (x, y) = 7 x^{2} y + 9 x y^{2}

f (x) = x^{2} + x y + y^{2} + 2 x - 2 y + 6

f (x) = 165000 x - 100 x^{2}

\frac{x}{x ^{2} + 81}

What is the extreme f(y)=14x^2-2x^3+2y^2+4xy ?
The extreme f(y)=14x^2-2x^3+2y^2+4xy is Saddle(-4,4),Minimum(0,0)