What is the extreme f(x)=17x^4-102x^2 ?

The extreme f(x)=17x^4-102x^2 is Minimum(-sqrt(3),-153),Maximum(0,0),Minimum(sqrt(3),-153)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

extreme f(x)=17x^4-102x^2

extreme

Minimum (- 3, - 153), Maximum (0, 0), Minimum (3, - 153)

Solution steps

f^{'} (x) = 68 x^{3} - 204 x

Find intervals:

Decreasing

: - \infty < x < - 3,

Increasing

: - 3 < x < 0,

Decreasing

: 0 < x < 3,

Increasing

: 3 < x < \infty

Plug

x = - 3

into

17 x^{4} - 102 x^{2} : - 153

Minimum (- 3, - 153)

Plug

x = 0

into

17 x^{4} - 102 x^{2} : 0

Maximum (0, 0)

Plug

x = 3

into

17 x^{4} - 102 x^{2} : - 153

Minimum (3, - 153)

Minimum (- 3, - 153), Maximum (0, 0), Minimum (3, - 153)

What is the extreme f(x)=17x^4-102x^2 ?
The extreme f(x)=17x^4-102x^2 is Minimum(-sqrt(3),-153),Maximum(0,0),Minimum(sqrt(3),-153)