What is the extreme f(x,y)=2y^3-3y^2-12y+2x^2-6x+3 ?

The extreme f(x,y)=2y^3-3y^2-12y+2x^2-6x+3 is Minimum(3/2 ,2),Saddle(3/2 ,-1)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

extreme f(x,y)=2y^3-3y^2-12y+2x^2-6x+3

extreme points

f (x, y) = 2 y^{3} - 3 y^{2} - 12 y + 2 x^{2} - 6 x + 3

Minimum (\frac{3}{2}, 2), Saddle (\frac{3}{2}, - 1)

Solution steps

Find the critical points:

(\frac{3}{2}, 2), (\frac{3}{2}, - 1)

\frac{\partial ^{2} f}{\partial x ^{2}} = 12 y - 6

D (x, y) = 4 (12 y - 6)

Check the sign of at each critical point

Check critical point

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

Minimum

Check critical point

(\frac{3}{2}, - 1) :

Saddle

Minimum (\frac{3}{2}, 2), Saddle (\frac{3}{2}, - 1)

f (x) = 2 x^{4} - 4 x^{2} + 6

f (x) = x e^{7 x^{2}}

f (x, y) = 5 x + 3 y + 2

f (x) = ln (5 - 3 x^{2})

f (x) = \frac{x ^{3} + 3}{2 x ^{2} - 6 x - 4}

What is the extreme f(x,y)=2y^3-3y^2-12y+2x^2-6x+3 ?
The extreme f(x,y)=2y^3-3y^2-12y+2x^2-6x+3 is Minimum(3/2 ,2),Saddle(3/2 ,-1)