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

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

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

critical points

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

x = 0

Solution steps

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

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

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

is not defined at

x = - 1, x = 1

, therefore:

x = 0

\frac{x ^{2}}{x ^{2} + 4}

f (x) = x^{5} - 5 x

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

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

f (x) = (cos (x) - 0.5)^{2}

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