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

Popular >

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

extreme points

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

Solution steps

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

Find intervals:

Decreasing

: - \infty < x < 0,

Decreasing

Increasing

Plug

into

f (x) = \frac{1}{3} x^{3} + 5 x^{2} + 24

f (x) = x ln (x) - 2 x, 1 \leq x \leq 4

f (x) = - 2 x^{2} - 1, - 4 \leq x \leq 3

f (x) = \frac{x}{ln ( 2 x )}

f (x, y) = 4 - x^{2} - y^{2} + x y + x

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)})