What is the extreme f(x)=2x^2+(28)/x ?

The extreme f(x)=2x^2+(28)/x is Minimum(\sqrt[3]{7},6*7^{2/3})

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

extreme f(x)=2x^2+(28)/x

extreme

Minimum (37, 6 \cdot 7^{\frac{2}{3}})

Solution steps

f^{'} (x) = 4 x - \frac{28}{x ^{2}}

Find intervals:

Decreasing

: - \infty < x < 0,

Decreasing

: 0 < x < 37,

Increasing

: 37 < x < \infty

Plug

x = 37

into

2 x^{2} + \frac{28}{x} : 6 \cdot 7^{\frac{2}{3}}

Minimum (37, 6 \cdot 7^{\frac{2}{3}})

What is the extreme f(x)=2x^2+(28)/x ?
The extreme f(x)=2x^2+(28)/x is Minimum(\sqrt[3]{7},6*7^{2/3})