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

The critical f(x)=(x^3)/((x-4)) is x=0,x=6

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

critical points

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

x = 0, x = 6

Solution steps

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

Find where

f^{'} (x)

is equal to zero or undefined

x = 0, x = 6, x = 4

Identify critical points not in

f (x)

domain

Domain of

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

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

is not defined at

x = 4

, therefore:

x = 0, x = 6

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

f (x) = 9 x - 7

f (x) = arctan (x)

\frac{2 x ^{2} + 1}{x + 3}

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

What is the critical f(x)=(x^3)/((x-4)) ?
The critical f(x)=(x^3)/((x-4)) is x=0,x=6