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) = 6 + 3

5 x - 5 y = - 10 (- 4.1)

f (x) = \frac{3 x - 7}{4}

f (x) = 4 x - \frac{196}{x}, 1 \leq x \leq 9

\frac{x - 1}{x ^{2} + 2 x + 6}

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