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

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

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

critical points

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

x = 0

Solution steps

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

Find where

f^{'} (x)

is equal to zero or undefined

x = 0, x = - 2, x = 2

Identify critical points not in

f (x)

domain

Domain of

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

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

is not defined at

x = - 2, x = 2

, therefore:

x = 0

f (x) = 2 x^{3} + 3 x^{2} - 12 x + 7

f (x) = x^{2} - 9

f (x, y) = x - \frac{1}{2} y^{2} - \frac{1}{3} x^{3} + y + 6

f (x) = x^{2} \cdot ln (x)

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

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