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

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

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

critical points

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

x = 0, x = 1

Solution steps

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

Find where

f^{'} (x)

is equal to zero or undefined

x = 0, x = 1, x = \frac{1}{2}

Identify critical points not in

f (x)

domain

Domain of

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

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

is not defined at

x = \frac{1}{2}

, therefore:

x = 0, x = 1

(x^{2} - 3)^{3}

y = x^{2} - 4 x + 7

x^{2} - 6 x

f (x) = \frac{2}{3 x ( x - 1 ) ( x + 5 )}

f (x) = - 14 x^{2} - 16 x + 128

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