What is the extreme f(x)=x^4-30x^2+29 ?

The extreme f(x)=x^4-30x^2+29 is Minimum(-sqrt(15),-196),Maximum(0,29),Minimum(sqrt(15),-196)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

extreme f(x)=x^4-30x^2+29

extreme

Minimum (- 15, - 196), Maximum (0, 29), Minimum (15, - 196)

Solution steps

f^{'} (x) = 4 x^{3} - 60 x

Find intervals:

Decreasing

: - \infty < x < - 15,

Increasing

: - 15 < x < 0,

Decreasing

: 0 < x < 15,

Increasing

: 15 < x < \infty

Plug

x = - 15

into

x^{4} - 30 x^{2} + 29 : - 196

Minimum (- 15, - 196)

Plug

x = 0

into

x^{4} - 30 x^{2} + 29 : 29

Maximum (0, 29)

Plug

x = 15

into

x^{4} - 30 x^{2} + 29 : - 196

Minimum (15, - 196)

Minimum (- 15, - 196), Maximum (0, 29), Minimum (15, - 196)

What is the extreme f(x)=x^4-30x^2+29 ?
The extreme f(x)=x^4-30x^2+29 is Minimum(-sqrt(15),-196),Maximum(0,29),Minimum(sqrt(15),-196)