What is the critical f(x)=(6x)/(x^2-9) ?

The critical f(x)=(6x)/(x^2-9) is None

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critical f(x)=(6x)/(x^2-9)

critical points

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

None

Solution steps

\frac{6 x}{x ^{2} - 9}

Find where

f^{'} (x)

is equal to zero or undefined

x = - 3, x = 3

Identify critical points not in

f (x)

domain

Domain of

\frac{6 x}{x ^{2} - 9} : x < - 3 or - 3 < x < 3 or x > 3

\frac{6 x}{x ^{2} - 9}

is not defined at

x = - 3, x = 3

, therefore:

None

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

\frac{3 x}{x ^{2} - 4}

x + ln (x^{2} - 3)

2 x^{3} + 9 x^{2} + 12 x + 1

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