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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

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

extreme

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

Solution steps

f^{'} (x) = 4 x^{3} - 8 x

Find intervals:

Decreasing

: - \infty < x < - 2,

Increasing

: - 2 < x < 0,

Decreasing

: 0 < x < 2,

Increasing

: 2 < x < \infty

Plug

x = - 2

into

x^{4} - 4 x^{2} + 2 : - 2

Minimum (- 2, - 2)

Plug

x = 0

into

x^{4} - 4 x^{2} + 2 : 2

Maximum (0, 2)

Plug

x = 2

into

x^{4} - 4 x^{2} + 2 : - 2

Minimum (2, - 2)

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

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