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

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

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

critical points

\frac{x ^{2} - 4}{x ^{2} - 1}

x = 0

Solution steps

\frac{x ^{2} - 4}{x ^{2} - 1}

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 ^{2} - 4}{x ^{2} - 1} : x < - 1 or - 1 < x < 1 or x > 1

\frac{x ^{2} - 4}{x ^{2} - 1}

is not defined at

x = - 1, x = 1

, therefore:

x = 0

f (x) = - 4 x - 5

f (x) = - \frac{6}{( x - 1 ) ^{3}}

f (x) = - 3 x + 1

y + 6 = 2 (x - 2)

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