What is the extreme f(x)=x^{1/3}(8-x) ?

The extreme f(x)=x^{1/3}(8-x) is Maximum(2,6*2^{1/3})

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

extreme f(x)=x^{1/3}(8-x)

extreme points

f (x) = x^{\frac{1}{3}} (8 - x)

Maximum (2, 6 \cdot 2^{\frac{1}{3}})

Solution steps

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

Find intervals:

Increasing

: - \infty < x < 0,

Increasing

: 0 < x < 2,

Decreasing

: 2 < x < \infty

Plug

x = 2

into

x^{\frac{1}{3}} (8 - x) : 6 \cdot 2^{\frac{1}{3}}

Maximum (2, 6 \cdot 2^{\frac{1}{3}})

f (x, y) = x^{3} + x y^{2} - 8 x + 4 y - 1

f (x) = ln (8 + 8 x^{3})

f (x) = sin (x) + x

f (x) = x^{2} + 2 x + 4

f (x, y) = ln (4 - x^{2} - y^{2})

What is the extreme f(x)=x^{1/3}(8-x) ?
The extreme f(x)=x^{1/3}(8-x) is Maximum(2,6*2^{1/3})