What is the extreme y=(2x^3+2)/(x^2) ?

The extreme y=(2x^3+2)/(x^2) is Minimum(\sqrt[3]{2},3*2^{1/3})

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extreme y=(2x^3+2)/(x^2)

extreme points

y = \frac{2 x ^{3} + 2}{x ^{2}}

Minimum (32, 3 \cdot 2^{\frac{1}{3}})

Solution steps

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

Find intervals:

Increasing

: - \infty < x < 0,

Decreasing

: 0 < x < 32,

Increasing

: 32 < x < \infty

Plug

x = 32

into

\frac{2 x ^{3} + 2}{x ^{2}} : 3 \cdot 2^{\frac{1}{3}}

Minimum (32, 3 \cdot 2^{\frac{1}{3}})

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

f (x) = (x^{3} + 8)^{5}

\frac{x ^{3}}{x ^{2} + 1}

y = x^{2} + 4 x + 4

f (x) = x

What is the extreme y=(2x^3+2)/(x^2) ?
The extreme y=(2x^3+2)/(x^2) is Minimum(\sqrt[3]{2},3*2^{1/3})