What is the extreme f(x)=3x^2-4y^3-6x+48y-1 ?

The extreme f(x)=3x^2-4y^3-6x+48y-1 is Saddle(1,2),Minimum(1,-2)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

extreme f(x)=3x^2-4y^3-6x+48y-1

extreme points

f (x) = 3 x^{2} - 4 y^{3} - 6 x + 48 y - 1

Saddle (1, 2), Minimum (1, - 2)

Solution steps

Find the critical points:

(1, 2), (1, - 2)

\frac{\partial ^{2} f}{\partial x ^{2}} = - 24 y

D (x, y) = - 144 y

Check the sign of at each critical point

Check critical point

(1, 2) :

Saddle

Check critical point

(1, - 2) :

Minimum

Saddle (1, 2), Minimum (1, - 2)

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

f (x) = 12 x^{3} - 9 x^{2} - 108 x + 4

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

f (x) = 3 x - 3

f (x) = - x^{3} - x^{2} + 5 x + 1

What is the extreme f(x)=3x^2-4y^3-6x+48y-1 ?
The extreme f(x)=3x^2-4y^3-6x+48y-1 is Saddle(1,2),Minimum(1,-2)