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

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

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

critical points

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

x = \frac{1}{2}

Solution steps

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

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

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

is not defined at

x = 0, x = 1

, therefore:

x = \frac{1}{2}

y = cos (x)

f (x) = 3 x^{4} + 20 x^{3} - 6 x^{2}

x^{3} + 3 x^{2} - 24 x + 12

f (x) = - 7 (x + 1)^{2} + 3 x + 7, (- 1, 3)

f (x) = (x^{2} + 11) (4 - x^{2})