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}}

Maximum (- 32, \frac{- 3 - 2 ^{\frac{2}{3}}}{2 ^{\frac{2}{3}}})

Solution steps

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

Find intervals:

Increasing

: - \infty < x < - 32,

Decreasing

: - 32 < x < 0,

Increasing

: 0 < x < \infty

Plug

x = - 32

into

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

Maximum (- 32, \frac{- 3 - 2 ^{\frac{2}{3}}}{2 ^{\frac{2}{3}}})

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

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

f (x) = - 9 y^{2}

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

f (x) = x^{2} - 10 x + 24

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)})