What is the critical f(x)= 6/(1-x^2) ?

The critical f(x)= 6/(1-x^2) is x=0

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critical f(x)= 6/(1-x^2)

critical points

f (x) = \frac{6}{1 - x ^{2}}

x = 0

Solution steps

\frac{6}{1 - x ^{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{6}{1 - x ^{2}} : x < - 1 or - 1 < x < 1 or x > 1

\frac{6}{1 - x ^{2}}

is not defined at

x = - 1, x = 1

, therefore:

x = 0

f (x) = (x - 3)^{2} - 9

x + 1

f (x) = 1 - 3 sin (2 x - 9)

f (x) = sin (e^{t} - 1)

f (x) = lo g_{5} (- x)