What is the extreme f(x)=x^2+(200)/x ?

The extreme f(x)=x^2+(200)/x is Minimum(\sqrt[3]{100},100^{2/3}+(200)/(\sqrt[3]{100)})

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extreme f(x)=x^2+(200)/x

extreme points

f (x) = x^{2} + \frac{200}{x}

Minimum (3100, 10 0^{\frac{2}{3}} + \frac{200}{3 100})

Solution steps

f^{'} (x) = 2 x - \frac{200}{x ^{2}}

Find intervals:

Decreasing

: - \infty < x < 0,

Decreasing

: 0 < x < 3100,

Increasing

: 3100 < x < \infty

Plug

x = 3100

into

x^{2} + \frac{200}{x} : 10 0^{\frac{2}{3}} + \frac{200}{3 100}

Minimum (3100, 10 0^{\frac{2}{3}} + \frac{200}{3 100})

- 0.1 t^{2} + 0.8 t + 98.8, 0 \leq t \geq 8

x^{8} ln (x)

f (x) = x^{6} + 4 x^{5} + 10

F (x, y) = 2 x^{5} + x^{3} y^{2} + 6 x y^{4}

f (x) = \frac{9 x}{x ^{2} + 9}

What is the extreme f(x)=x^2+(200)/x ?
The extreme f(x)=x^2+(200)/x is Minimum(\sqrt[3]{100},100^{2/3}+(200)/(\sqrt[3]{100)})