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

The extreme f(x)=(x^2-4)^2-4 is Minimum(-2,-4),Maximum(0,12),Minimum(2,-4)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

extreme f(x)=(x^2-4)^2-4

extreme points

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

Minimum (- 2, - 4), Maximum (0, 12), Minimum (2, - 4)

Solution steps

f^{'} (x) = 4 x^{3} - 16 x

Find intervals:

Decreasing

: - \infty < x < - 2,

Increasing

: - 2 < x < 0,

Decreasing

: 0 < x < 2,

Increasing

: 2 < x < \infty

Plug

x = - 2

into

(x^{2} - 4)^{2} - 4 : - 4

Minimum (- 2, - 4)

Plug

x = 0

into

(x^{2} - 4)^{2} - 4 : 12

Maximum (0, 12)

Plug

x = 2

into

(x^{2} - 4)^{2} - 4 : - 4

Minimum (2, - 4)

Minimum (- 2, - 4), Maximum (0, 12), Minimum (2, - 4)

What is the extreme f(x)=(x^2-4)^2-4 ?
The extreme f(x)=(x^2-4)^2-4 is Minimum(-2,-4),Maximum(0,12),Minimum(2,-4)