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

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

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

extreme points

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

Solution steps

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

Find intervals:

Increasing

Decreasing

Increasing

: 0 < x < \infty

Plug

into

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

2 x - 3 y = 24

f (x) = - 2 x^{5} + 11 x^{3}

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

\frac{- 2 x ^{2} - 2 x + 4}{x ^{2} + 5 x + 6}

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