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

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

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

extreme points

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

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

Solution steps

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

Find intervals:

Increasing

: - \infty < x < 0,

Decreasing

: 0 < x < 32,

Increasing

: 32 < x < \infty

Plug

x = 32

into

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

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

x^{3} - 9 x^{2} + 24 x

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

p (x) = 2 x^{3} - 11 x^{2} + a x - 40

\frac{x ^{4} - 98 x ^{2}}{8}

2 x e^{3 x}

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