What is the extreme f(x)=x^{4/3}+4^{1/3} ?

The extreme f(x)=x^{4/3}+4^{1/3} is Minimum(0,4^{1/3})

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

extreme f(x)=x^{4/3}+4^{1/3}

extreme

Minimum (0, 4^{\frac{1}{3}})

Solution steps

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

Find intervals:

Decreasing

: - \infty < x < 0,

Increasing

: 0 < x < \infty

Plug

x = 0

into

x^{\frac{4}{3}} + 4^{\frac{1}{3}} : 4^{\frac{1}{3}}

Minimum (0, 4^{\frac{1}{3}})

What is the extreme f(x)=x^{4/3}+4^{1/3} ?
The extreme f(x)=x^{4/3}+4^{1/3} is Minimum(0,4^{1/3})