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

The extreme f(x)=(2(-x+1))/(x^3) is Minimum(3/2 ,-8/27)

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

extreme points

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

Minimum (\frac{3}{2}, - \frac{8}{27})

Solution steps

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

Find intervals:

Decreasing

: - \infty < x < 0,

Decreasing

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

Increasing

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

Plug

x = \frac{3}{2}

into

\frac{2 ( - x + 1 )}{x ^{3}} : - \frac{8}{27}

Minimum (\frac{3}{2}, - \frac{8}{27})

x^{4} + 2 x^{3} + 6 x^{2} - 9 x + 12

f (x) = ∣ 2 x - 4 ∣

y (t, x) = - 2 t + x

f (x) = 2 x^{2} + 3 x - 5, - 4 \leq x \leq 4

g (x) = y (x - 3) (- x + 1)

What is the extreme f(x)=(2(-x+1))/(x^3) ?
The extreme f(x)=(2(-x+1))/(x^3) is Minimum(3/2 ,-8/27)