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 points

f (x) = \frac{x ^{3} + 8}{x}

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})

P (x, y) = - 3 (20 x - 400)^{2} - \frac{6}{7} (2 y - 44)^{2} + 8

f (x) = x^{3} - \frac{9}{2} x^{2} + 1

y = 2 x^{2} + z^{2}

f (x, y) = 1 - 3 x^{4} - 6 x^{2} y + 2 y^{3}

f (x) = 8 x^{3} + 13 x

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)})