What is the critical (x^4-3x^2)/((x^2-1)^2) ?

The critical (x^4-3x^2)/((x^2-1)^2) is x=0

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critical (x^4-3x^2)/((x^2-1)^2)

critical points

\frac{x ^{4} - 3 x ^{2}}{( x ^{2} - 1 ) ^{2}}

x = 0

Solution steps

\frac{x ^{4} - 3 x ^{2}}{( x ^{2} - 1 ) ^{2}}

Find where

f^{'} (x)

is equal to zero or undefined

x = 0, x = - 1, x = 1

Identify critical points not in

f (x)

domain

Domain of

\frac{x ^{4} - 3 x ^{2}}{( x ^{2} - 1 ) ^{2}} : x < - 1 or - 1 < x < 1 or x > 1

\frac{x ^{4} - 3 x ^{2}}{( x ^{2} - 1 ) ^{2}}

is not defined at

x = - 1, x = 1

, therefore:

x = 0

f (x) = x^{3} - \frac{15}{2} x^{2} + 12 x

f (x) = 4 x^{3} - 24 x^{2}

f (x) = - 2 sin (x)

f (x) = 6 + \frac{1}{5} x - \frac{1}{2} x^{2}

f (y) = x - \frac{1}{x}

What is the critical (x^4-3x^2)/((x^2-1)^2) ?
The critical (x^4-3x^2)/((x^2-1)^2) is x=0