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

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

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

critical points

f (x) = \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

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

f (x, y) = x^{3} y + 6 x^{2} - y

f (x, y) = x^{2} - x^{3} y^{2} + y^{4}

ln (1 + x^{2})

f (x) = 3 x^{4} + 8 x^{3} - 48 x^{2}

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