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

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

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

critical points

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

x = \frac{1}{2}

Solution steps

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

Find where

f^{'} (x)

is equal to zero or undefined

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

Identify critical points not in

f (x)

domain

Domain of

\frac{2}{x ^{2} - x} : x < 0 or 0 < x < 1 or x > 1

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

is not defined at

x = 0, x = 1

, therefore:

x = \frac{1}{2}

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

2 x^{3} + x y^{2} + 5 x^{2} + y^{2}

f (x, y) = \frac{1}{3} x^{3} + y^{3} + 2 x^{2} - 12 x - 3 y

f (x, y) = 4 x^{2} - 12 x + y^{2} + 2 y - 10

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

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