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

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

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

critical points

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

x = 0

Solution steps

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

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

is not defined at

x = - 1, x = 1

, therefore:

x = 0

f (x) = \frac{3}{x - 1}

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

f (x) = x^{2} + 6 x

\frac{x ^{2} - 24}{x - 5}

f (x) = - 6 x^{2} + 9 x - 2

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