What is the extreme 8x^4-48x^2 ?

The extreme 8x^4-48x^2 is Minimum(-sqrt(3),-72),Maximum(0,0),Minimum(sqrt(3),-72)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

extreme 8x^4-48x^2

extreme points

8 x^{4} - 48 x^{2}

Minimum (- 3, - 72), Maximum (0, 0), Minimum (3, - 72)

Solution steps

f^{'} (x) = 32 x^{3} - 96 x

Find intervals:

Decreasing

: - \infty < x < - 3,

Increasing

: - 3 < x < 0,

Decreasing

: 0 < x < 3,

Increasing

: 3 < x < \infty

Plug

x = - 3

into

8 x^{4} - 48 x^{2} : - 72

Minimum (- 3, - 72)

Plug

x = 0

into

8 x^{4} - 48 x^{2} : 0

Maximum (0, 0)

Plug

x = 3

into

8 x^{4} - 48 x^{2} : - 72

Minimum (3, - 72)

Minimum (- 3, - 72), Maximum (0, 0), Minimum (3, - 72)

c

f (x) = 5 + (8 + 7 x)^{\frac{2}{7}}

F

y = \frac{6 x}{x ^{2} + 4}

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

What is the extreme 8x^4-48x^2 ?
The extreme 8x^4-48x^2 is Minimum(-sqrt(3),-72),Maximum(0,0),Minimum(sqrt(3),-72)