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

The extreme f(x)=((x+1)^3)/((x-1)^2) is Saddle(-1,0),Minimum(5, 27/2)

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

extreme points

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

Saddle (- 1, 0), Minimum (5, \frac{27}{2})

Solution steps

f^{'} (x) = \frac{( x + 1 ) ^{2} ( x - 5 )}{( x - 1 ) ^{3}}

Find intervals:

Increasing

: - \infty < x < - 1,

Increasing

: - 1 < x < 1,

Decreasing

: 1 < x < 5,

Increasing

: 5 < x < \infty

Plug

x = - 1

into

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

Saddle (- 1, 0)

Plug

x = 5

into

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

Minimum (5, \frac{27}{2})

Saddle (- 1, 0), Minimum (5, \frac{27}{2})

What is the extreme f(x)=((x+1)^3)/((x-1)^2) ?
The extreme f(x)=((x+1)^3)/((x-1)^2) is Saddle(-1,0),Minimum(5, 27/2)