What is the critical (2(x^2-9))/(x^2-4) ?

The critical (2(x^2-9))/(x^2-4) is x=0

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critical (2(x^2-9))/(x^2-4)

critical points

\frac{2 ( x ^{2} - 9 )}{x ^{2} - 4}

x = 0

Solution steps

\frac{2 ( x ^{2} - 9 )}{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{2 ( x ^{2} - 9 )}{x ^{2} - 4} : x < - 2 or - 2 < x < 2 or x > 2

\frac{2 ( x ^{2} - 9 )}{x ^{2} - 4}

is not defined at

x = - 2, x = 2

, therefore:

x = 0

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

f (x) = (x - 9) e^{x}

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

x^{\frac{3}{2}} (3 x + 10)

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

What is the critical (2(x^2-9))/(x^2-4) ?
The critical (2(x^2-9))/(x^2-4) is x=0