What is the extreme f(x)=\sqrt[3]{x}(8-x),0<= x<= 8 ?

The extreme f(x)=\sqrt[3]{x}(8-x),0<= x<= 8 is Minimum(0,0),Maximum(2,6\sqrt[3]{2}),Minimum(8,0)

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),0<= x<= 8

extreme points

Solution steps

Find the critical points:

x = 0, x = 2, x = 8

Domain of

Find intervals:

Increasing

: 0 < x < 2,

Decreasing

: 2 < x < 8

Plug the extreme point

x = 0

into

Minimum (0, 0)

Plug the extreme point

x = 2

into

Plug the extreme point

x = 8

into

Minimum (8, 0)

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

What is the extreme f(x)=\sqrt[3]{x}(8-x),0<= x<= 8 ?
The extreme f(x)=\sqrt[3]{x}(8-x),0<= x<= 8 is Minimum(0,0),Maximum(2,6\sqrt[3]{2}),Minimum(8,0)