What is the extreme f(x)=x^2+(800)/x ?

The extreme f(x)=x^2+(800)/x is Minimum(2\sqrt[3]{50},4*50^{2/3}+(200*2^{2/3})/(\sqrt[3]{25)})

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

extreme f(x)=x^2+(800)/x

extreme

Minimum (2350, 4 \cdot 5 0^{\frac{2}{3}} + \frac{200 \cdot 2 ^{\frac{2}{3}}}{3 25})

Solution steps

f^{'} (x) = 2 x - \frac{800}{x ^{2}}

Find intervals:

Decreasing

: - \infty < x < 0,

Decreasing

: 0 < x < 2350,

Increasing

: 2350 < x < \infty

Plug

x = 2350

into

x^{2} + \frac{800}{x} : 4 \cdot 5 0^{\frac{2}{3}} + \frac{200 \cdot 2 ^{\frac{2}{3}}}{3 25}

Minimum (2350, 4 \cdot 5 0^{\frac{2}{3}} + \frac{200 \cdot 2 ^{\frac{2}{3}}}{3 25})

What is the extreme f(x)=x^2+(800)/x ?
The extreme f(x)=x^2+(800)/x is Minimum(2\sqrt[3]{50},4*50^{2/3}+(200*2^{2/3})/(\sqrt[3]{25)})