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

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

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

critical points

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

x = 0

Solution steps

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

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

is not defined at

x = - 2, x = 2

, therefore:

x = 0

\frac{x ^{2} - 16}{4 x ^{2}}

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

f (x) = x^{2} + 2 x

y - 5 = - 2 (x + 5)

(\frac{2}{3})^{x} + 2