What is the critical ((x+1)^3)/((x-1)^2) ?

The critical ((x+1)^3)/((x-1)^2) is x=-1,x=5

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critical ((x+1)^3)/((x-1)^2)

critical points

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

x = - 1, x = 5

Solution steps

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

Find where

f^{'} (x)

is equal to zero or undefined

x = - 1, x = 5, x = 1

Identify critical points not in

f (x)

domain

Domain of

\frac{( x + 1 ) ^{3}}{( x - 1 ) ^{2}} : x < 1 or x > 1

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

is not defined at

x = 1

, therefore:

x = - 1, x = 5

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

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

f (x) = \frac{e ^{4 x}}{x + 1}

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

f (x) = (x - 4)^{2} + 10

What is the critical ((x+1)^3)/((x-1)^2) ?
The critical ((x+1)^3)/((x-1)^2) is x=-1,x=5