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

The critical f(x)=(x^2)/(x-6) is x=0,x=12

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

critical points

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

x = 0, x = 12

Solution steps

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

Find where

f^{'} (x)

is equal to zero or undefined

x = 0, x = 12, x = 6

Identify critical points not in

f (x)

domain

Domain of

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

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

is not defined at

x = 6

, therefore:

x = 0, x = 12

f (x) = - x^{3} + 6 x + 3

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

\frac{3 x + 8}{2 x - 3}

f (x) = \frac{x + 5}{x ^{2}}

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

What is the critical f(x)=(x^2)/(x-6) ?
The critical f(x)=(x^2)/(x-6) is x=0,x=12