What is the extreme f(x)=(x^3+8)/x ?

The extreme f(x)=(x^3+8)/x is Minimum(\sqrt[3]{4},(12)/(\sqrt[3]{4)})

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extreme f(x)=(x^3+8)/x

extreme

Minimum (34, \frac{12}{3 4})

Solution steps

f^{'} (x) = \frac{2 x ^{3} - 8}{x ^{2}}

Find intervals:

Decreasing

: - \infty < x < 0,

Decreasing

: 0 < x < 34,

Increasing

: 34 < x < \infty

Plug

x = 34

into

\frac{x ^{3} + 8}{x} : \frac{12}{3 4}

Minimum (34, \frac{12}{3 4})

What is the extreme f(x)=(x^3+8)/x ?
The extreme f(x)=(x^3+8)/x is Minimum(\sqrt[3]{4},(12)/(\sqrt[3]{4)})