What is the extreme f(x)=\sqrt[3]{x}(8-x) ?

The extreme f(x)=\sqrt[3]{x}(8-x) is Maximum(2,6\sqrt[3]{2})

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

extreme f(x)=\sqrt[3]{x}(8-x)

extreme points

Solution steps

Find the critical points:

x = 0, x = 2

Domain of

Find intervals:

Increasing

: - \infty < x < 0,

Increasing

: 0 < x < 2,

Decreasing

: 2 < x < \infty

Plug the extreme point

x = 2

into

x^{2} \cdot e^{- x}

y = \frac{1}{x - 2} - 3

f (x) = x^{3} + x^{2} + x

f (x, y) = 8 y^{2} - 5 x^{2} - 10 y + 2 x y

f (x) = \frac{x}{x ^{2} + 9 x + 20}

What is the extreme f(x)=\sqrt[3]{x}(8-x) ?
The extreme f(x)=\sqrt[3]{x}(8-x) is Maximum(2,6\sqrt[3]{2})