What is the critical f(x)=((x^2))/(x-1) ?

The critical f(x)=((x^2))/(x-1) is x=0,x=2

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critical f(x)=((x^2))/(x-1)

critical points

f (x) = \frac{( x ^{2} )}{x - 1}

x = 0, x = 2

Solution steps

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

Find where

f^{'} (x)

is equal to zero or undefined

x = 0, x = 2, x = 1

Identify critical points not in

f (x)

domain

Domain of

\frac{x ^{2}}{x - 1} : x < 1 or x > 1

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

is not defined at

x = 1

, therefore:

x = 0, x = 2

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

f (x) = x^{2} e^{9 x}

y = x^{\frac{4}{5}} (x - 4)^{2}

f (x) = 4 x^{3} - 36 x^{2} + 96 x - 64

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

What is the critical f(x)=((x^2))/(x-1) ?
The critical f(x)=((x^2))/(x-1) is x=0,x=2