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

The extreme f(x)=(2^2+3x)^2-(2x^3-4y^2) is Minimum(-1,0),Saddle(4,0)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

extreme points

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

Minimum (- 1, 0), Saddle (4, 0)

Solution steps

Find the critical points:

(- 1, 0), (4, 0)

\frac{\partial ^{2} f}{\partial x ^{2}} = 8

D (x, y) = 8 (- 12 x + 18)

Check the sign of at each critical point

Check critical point

(- 1, 0) :

Minimum

Check critical point

(4, 0) :

Saddle

Minimum (- 1, 0), Saddle (4, 0)

x^{2} - 10 x

f (x, y) = - x^{2} - y^{2} + 16

f (x) = 3 x^{4} - 294 x^{2} + 5

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

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

What is the extreme f(x)=(2^2+3x)^2-(2x^3-4y^2) ?
The extreme f(x)=(2^2+3x)^2-(2x^3-4y^2) is Minimum(-1,0),Saddle(4,0)