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

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

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

critical points

f (x) = \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) = I n (x + 4)

cos (x) + 2 x

x^{4} + x^{3} + x^{2} + 1

y = x^{4} - 12 x^{3} + 48 x^{2} - 64 x

4 x^{2} - 6 x

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