What is the critical f(x)=(x^2)/(x-9) ?

The critical f(x)=(x^2)/(x-9) is x=0,x=18

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critical f(x)=(x^2)/(x-9)

critical points

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

x = 0, x = 18

Solution steps

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

Find where

f^{'} (x)

is equal to zero or undefined

x = 0, x = 18, x = 9

Identify critical points not in

f (x)

domain

Domain of

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

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

is not defined at

x = 9

, therefore:

x = 0, x = 18

f (x) = \frac{36}{x}

f (x) = 5 x^{4} + 20 x^{3}

f (x) = 2 x

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

ln (\frac{x + 3}{2 - x})

What is the critical f(x)=(x^2)/(x-9) ?
The critical f(x)=(x^2)/(x-9) is x=0,x=18