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}}

Solution steps

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

Find intervals:

Increasing

: - \infty < x < 0,

Decreasing

Increasing

Plug

into

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

f (x) = 5 x - 4

y = - 2 (x - 3)^{2} + 1

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})