What is the extreme f(x)=(((x^2-1)^{0.5})/(x^3)) ?

The extreme f(x)=(((x^2-1)^{0.5})/(x^3)) is

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extreme f(x)=(((x^2-1)^{0.5})/(x^3))

extreme points

f (x) = (\frac{( x ^{2} - 1 ) ^{0.5}}{x ^{3}})

Minimum (- \frac{3}{2}, - \frac{2}{3 \cdot 3 ^{\frac{1}{2}}}), Maximum (- 1, 0), Minimum (1, 0), Maximum (\frac{3}{2}, \frac{2}{3 \cdot 3 ^{\frac{1}{2}}})

Solution steps

f^{'} (x) = \frac{- 2 x ^{2} + 3}{x ^{4} ( x ^{2} - 1 ) ^{\frac{1}{2}}}

Find intervals:

Decreasing

: - \infty < x < - \frac{3}{2},

Increasing

: - \frac{3}{2} < x < - 1,

Increasing

: 1 < x < \frac{3}{2},

Decreasing

: \frac{3}{2} < x < \infty

Plug

x = - \frac{3}{2}

into

\frac{( x ^{2} - 1 ) ^{0.5}}{x ^{3}} : - \frac{2}{3 \cdot 3 ^{\frac{1}{2}}}

Minimum (- \frac{3}{2}, - \frac{2}{3 \cdot 3 ^{\frac{1}{2}}})

Plug

x = - 1

into

\frac{( x ^{2} - 1 ) ^{0.5}}{x ^{3}} : 0

Maximum (- 1, 0)

Plug

x = 1

into

\frac{( x ^{2} - 1 ) ^{0.5}}{x ^{3}} : 0

Minimum (1, 0)

Plug

x = \frac{3}{2}

into

\frac{( x ^{2} - 1 ) ^{0.5}}{x ^{3}} : \frac{2}{3 \cdot 3 ^{\frac{1}{2}}}

Maximum (\frac{3}{2}, \frac{2}{3 \cdot 3 ^{\frac{1}{2}}})

Minimum (- \frac{3}{2}, - \frac{2}{3 \cdot 3 ^{\frac{1}{2}}}), Maximum (- 1, 0), Minimum (1, 0), Maximum (\frac{3}{2}, \frac{2}{3 \cdot 3 ^{\frac{1}{2}}})

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

f (x, y) = 6 x^{2} + 9 x y + 3 y^{2} + 8 x y^{2} + 5 x^{2} y

\frac{x ^{2} - 11}{x - 6}

f (x, y) = \frac{2 x y}{( 2 x + 3 y - 4 )}

f (x) = - 2 x + 3 ln (4 x), (1, 5)

What is the extreme f(x)=(((x^2-1)^{0.5})/(x^3)) ?
The extreme f(x)=(((x^2-1)^{0.5})/(x^3)) is