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

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

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

extreme points

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

Minimum (- 3, - \frac{1}{8})

Solution steps

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

Find intervals:

Decreasing

: - \infty < x < - 3,

Increasing

: - 3 < x < 1,

Decreasing

: 1 < x < \infty

Plug

x = - 3

into

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

Minimum (- 3, - \frac{1}{8})

y = - 4 x - 5

f (x) = ln (2 - x)

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

I^{22}

(1.2) (- 3.5)

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