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

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

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

extreme points

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

Solution steps

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

Find intervals:

Decreasing

: - \infty < x < 0,

Decreasing

Increasing

Plug

into

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

f (x) = - 2 x^{2} + 12 x

cos (x) - sin (x)

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

2 t^{2} + 7 t + 9

f (x) = 6 x 16 - x^{2}

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