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

The extreme f(x)=(x^2-1)/(x^3) is Minimum(-sqrt(3),-2/(3sqrt(3))),Maximum(sqrt(3), 2/(3sqrt(3)))

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

extreme points

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

Minimum (- 3, - \frac{2}{3 3}), Maximum (3, \frac{2}{3 3})

Solution steps

f^{'} (x) = \frac{- x ^{2} + 3}{x ^{4}}

Find intervals:

Decreasing

: - \infty < x < - 3,

Increasing

: - 3 < x < 0,

Increasing

: 0 < x < 3,

Decreasing

: 3 < x < \infty

Plug

x = - 3

into

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

Minimum (- 3, - \frac{2}{3 3})

Plug

x = 3

into

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

Maximum (3, \frac{2}{3 3})

Minimum (- 3, - \frac{2}{3 3}), Maximum (3, \frac{2}{3 3})

f (x, y) = e^{- y} \cdot (x^{2} + y^{2})

f (x, y) = x^{3} - 3 x

f (x) = 2 x^{2} + 3, - 1 \leq x \leq 2

f (x) = sin (π x) (- \frac{1}{3}, \frac{1}{3})

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

What is the extreme f(x)=(x^2-1)/(x^3) ?
The extreme f(x)=(x^2-1)/(x^3) is Minimum(-sqrt(3),-2/(3sqrt(3))),Maximum(sqrt(3), 2/(3sqrt(3)))