What is the extreme f(x)=3x^4-30x^2+27 ?

The extreme f(x)=3x^4-30x^2+27 is Minimum(-sqrt(5),-48),Maximum(0,27),Minimum(sqrt(5),-48)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

extreme f(x)=3x^4-30x^2+27

extreme points

f (x) = 3 x^{4} - 30 x^{2} + 27

Minimum (- 5, - 48), Maximum (0, 27), Minimum (5, - 48)

Solution steps

f^{'} (x) = 12 x^{3} - 60 x

Find intervals:

Decreasing

: - \infty < x < - 5,

Increasing

: - 5 < x < 0,

Decreasing

: 0 < x < 5,

Increasing

: 5 < x < \infty

Plug

x = - 5

into

3 x^{4} - 30 x^{2} + 27 : - 48

Minimum (- 5, - 48)

Plug

x = 0

into

3 x^{4} - 30 x^{2} + 27 : 27

Maximum (0, 27)

Plug

x = 5

into

3 x^{4} - 30 x^{2} + 27 : - 48

Minimum (5, - 48)

Minimum (- 5, - 48), Maximum (0, 27), Minimum (5, - 48)

f (x) = 3 x + 10

\frac{2 x}{x - 4}

- 3 x^{2} + 3 x - 2

(- 2, 2), (1, 2)

What is the extreme f(x)=3x^4-30x^2+27 ?
The extreme f(x)=3x^4-30x^2+27 is Minimum(-sqrt(5),-48),Maximum(0,27),Minimum(sqrt(5),-48)