What is the extreme f(x)=(x^3+1)/(x^2) ?

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

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extreme f(x)=(x^3+1)/(x^2)

extreme points

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

Solution steps

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

Find intervals:

Increasing

: - \infty < x < 0,

Decreasing

Increasing

Plug

into

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

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

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

f (x) = 5 x - 4

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

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

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