What is the extreme f(x)=3^2\sqrt[3]{x^2}-2x ?

The extreme f(x)=3^2\sqrt[3]{x^2}-2x is Minimum(0,0),Maximum(27,27)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

extreme f(x)=3^2\sqrt[3]{x^2}-2x

extreme

Minimum (0, 0), Maximum (27, 27)

Solution steps

Find the critical points:

x = 0, x = 27

Domain of

3^{2} 3 x^{2} - 2 x : - \infty < x < \infty

Find intervals:

Decreasing

: - \infty < x < 0,

Increasing

: 0 < x < 27,

Decreasing

: 27 < x < \infty

Plug the extreme point

x = 0

into

3^{2} 3 x^{2} - 2 x \Rightarrow y = 0

Minimum (0, 0)

Plug the extreme point

x = 27

into

3^{2} 3 x^{2} - 2 x \Rightarrow y = 27

Maximum (27, 27)

Minimum (0, 0), Maximum (27, 27)

What is the extreme f(x)=3^2\sqrt[3]{x^2}-2x ?
The extreme f(x)=3^2\sqrt[3]{x^2}-2x is Minimum(0,0),Maximum(27,27)