What is the extreme f(x)=3\sqrt[3]{x}-4x ?

The extreme f(x)=3\sqrt[3]{x}-4x is Minimum(-1/8 ,-1),Minimum(1/8 ,1)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

extreme f(x)=3\sqrt[3]{x}-4x

extreme points

Minimum (- \frac{1}{8}, - 1), Minimum (\frac{1}{8}, 1)

Solution steps

f^{'} (x) = \frac{1}{x ^{\frac{2}{3}}} - 4

Find intervals:

Increasing

: - \infty < x < - 8,

Decreasing

: - 8 < x < 8,

Increasing

: 8 < x < \infty

Plug

x = - \frac{1}{8}

into

Minimum (- \frac{1}{8}, - 1)

Plug

x = \frac{1}{8}

into

Minimum (\frac{1}{8}, 1)

Minimum (- \frac{1}{8}, - 1), Minimum (\frac{1}{8}, 1)

y = \frac{( ln ( x ) ) ^{2}}{x}

f (x, y) = - x^{3} + 2 y^{3} + 27 x - 24 y + 3

f (x) = \frac{x ^{2}}{e ^{x}}

F (I, J) = (- 3 I + 2 J)

f (x) = x_{1} x

What is the extreme f(x)=3\sqrt[3]{x}-4x ?
The extreme f(x)=3\sqrt[3]{x}-4x is Minimum(-1/8 ,-1),Minimum(1/8 ,1)