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

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

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

critical points

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

x = 0

Solution steps

\frac{1}{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{1}{1 - x ^{2}} : - 1 < x < 1

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

is not defined at

x = - 1, x = 1

, therefore:

x = 0

f (x) = x 3 - x

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

f (x) = 3 x^{4} + 4 x^{3} - 72 x^{2}

f (x) = ln (x^{2} + 4)

f (x) = - 8 - 16 x - 16 x^{2}

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