What is the critical ((x^2-9))/(x^3+3x^2) ?

The critical ((x^2-9))/(x^3+3x^2) is x=6

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critical ((x^2-9))/(x^3+3x^2)

critical points

\frac{( x ^{2} - 9 )}{x ^{3} + 3 x ^{2}}

x = 6

Solution steps

\frac{x ^{2} - 9}{x ^{3} + 3 x ^{2}}

Find where

f^{'} (x)

is equal to zero or undefined

x = 6, x = 0

Identify critical points not in

f (x)

domain

Domain of

\frac{x ^{2} - 9}{x ^{3} + 3 x ^{2}} : x < - 3 or - 3 < x < 0 or x > 0

\frac{x ^{2} - 9}{x ^{3} + 3 x ^{2}}

is not defined at

x = 0

, therefore:

x = 6

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

f (x) = 0.002 x^{3} + 5 x + 6.244

f (x) = - 5 x - 4

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

1 - x^{2}

What is the critical ((x^2-9))/(x^3+3x^2) ?
The critical ((x^2-9))/(x^3+3x^2) is x=6