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

The critical f(x)=((x^2))/(x-3) is x=0,x=6

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

critical points

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

x = 0, x = 6

Solution steps

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

Find where

f^{'} (x)

is equal to zero or undefined

x = 0, x = 6, x = 3

Identify critical points not in

f (x)

domain

Domain of

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

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

is not defined at

x = 3

, therefore:

x = 0, x = 6

f (x) = x^{2} - 10 x + 8

f (x) = 20 - x

f (x) = \frac{x}{2 x}

y = \frac{1}{2} x - 4, (9, - 6)

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

What is the critical f(x)=((x^2))/(x-3) ?
The critical f(x)=((x^2))/(x-3) is x=0,x=6