What is the critical f(x)=(x^3-8)/(x^2-4) ?

The critical f(x)=(x^3-8)/(x^2-4) is x=-4,x=0

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critical f(x)=(x^3-8)/(x^2-4)

critical points

f (x) = \frac{x ^{3} - 8}{x ^{2} - 4}

x = - 4, x = 0

Solution steps

\frac{x ^{3} - 8}{x ^{2} - 4}

Find where

f^{'} (x)

is equal to zero or undefined

x = - 4, x = 0, x = - 2

Identify critical points not in

f (x)

domain

Domain of

\frac{x ^{3} - 8}{x ^{2} - 4} : x < - 2 or - 2 < x < 2 or x > 2

\frac{x ^{3} - 8}{x ^{2} - 4}

is not defined at

x = - 2

, therefore:

x = - 4, x = 0

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

f (x) = 4 x + 2

f (x) = 3 x^{3} + 16 x

(3 x + 4 x^{3}) (y^{2} + 2 y)

\frac{2}{x - 3}

What is the critical f(x)=(x^3-8)/(x^2-4) ?
The critical f(x)=(x^3-8)/(x^2-4) is x=-4,x=0