What is the critical f(x)=(x^3)/(x^2-4) ?

The critical f(x)=(x^3)/(x^2-4) is x=-2sqrt(3),x=0,x=2sqrt(3)

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critical f(x)=(x^3)/(x^2-4)

critical points

f (x) = \frac{x ^{3}}{x ^{2} - 4}

x = - 23, x = 0, x = 23

Solution steps

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

Find where

f^{'} (x)

is equal to zero or undefined

x = - 23, x = 0, x = 23, x = - 2, x = 2

Identify critical points not in

f (x)

domain

Domain of

\frac{x ^{3}}{x ^{2} - 4} : x < - 2 or - 2 < x < 2 or x > 2

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

is not defined at

x = - 2, x = 2

, therefore:

x = - 23, x = 0, x = 23

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

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

f (x) = x^{2} - 4 x + 5

f (x) = (- 4.1) (- 4.5)

f (x) = lo g_{4} (x + 1)

What is the critical f(x)=(x^3)/(x^2-4) ?
The critical f(x)=(x^3)/(x^2-4) is x=-2sqrt(3),x=0,x=2sqrt(3)