What is the critical f(x)= 5/((1-2x)^2) ?

The critical f(x)= 5/((1-2x)^2) is None

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critical f(x)= 5/((1-2x)^2)

critical points

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

None

Solution steps

\frac{5}{( 1 - 2 x ) ^{2}}

Find where

f^{'} (x)

is equal to zero or undefined

x = \frac{1}{2}

Identify critical points not in

f (x)

domain

Domain of

\frac{5}{( 1 - 2 x ) ^{2}} : x < \frac{1}{2} or x > \frac{1}{2}

\frac{5}{( 1 - 2 x ) ^{2}}

is not defined at

x = \frac{1}{2}

, therefore:

None

f (x) = x^{3} - \frac{3}{2} x^{2} - 6 x + 1

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

x^{3} + 2 x^{2} - 15 x - 20

f (x) = 5 - ∣ x - 5 ∣

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

What is the critical f(x)= 5/((1-2x)^2) ?
The critical f(x)= 5/((1-2x)^2) is None