What is the extreme f(x)=x^2+(520)/x ?

The extreme f(x)=x^2+(520)/x is Minimum(\sqrt[3]{260},260^{2/3}+(8*65^{2/3})/(\sqrt[3]{4)})

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

extreme f(x)=x^2+(520)/x

extreme

Minimum (3260, 26 0^{\frac{2}{3}} + \frac{8 \cdot 6 5 ^{\frac{2}{3}}}{3 4})

Solution steps

f^{'} (x) = 2 x - \frac{520}{x ^{2}}

Find intervals:

Decreasing

: - \infty < x < 0,

Decreasing

: 0 < x < 3260,

Increasing

: 3260 < x < \infty

Plug

x = 3260

into

x^{2} + \frac{520}{x} : 26 0^{\frac{2}{3}} + \frac{8 \cdot 6 5 ^{\frac{2}{3}}}{3 4}

Minimum (3260, 26 0^{\frac{2}{3}} + \frac{8 \cdot 6 5 ^{\frac{2}{3}}}{3 4})

What is the extreme f(x)=x^2+(520)/x ?
The extreme f(x)=x^2+(520)/x is Minimum(\sqrt[3]{260},260^{2/3}+(8*65^{2/3})/(\sqrt[3]{4)})