What is the critical f(x)= 4/((1-4x^2)^2) ?

The critical f(x)= 4/((1-4x^2)^2) is x=0

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critical f(x)= 4/((1-4x^2)^2)

critical points

f (x) = \frac{4}{( 1 - 4 x ^{2} ) ^{2}}

x = 0

Solution steps

\frac{4}{( 1 - 4 x ^{2} ) ^{2}}

Find where

f^{'} (x)

is equal to zero or undefined

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

Identify critical points not in

f (x)

domain

Domain of

\frac{4}{( 1 - 4 x ^{2} ) ^{2}} : x < - \frac{1}{2} or - \frac{1}{2} < x < \frac{1}{2} or x > \frac{1}{2}

\frac{4}{( 1 - 4 x ^{2} ) ^{2}}

is not defined at

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

, therefore:

x = 0

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

f (x, y) = 6 x y - 4 x^{3} - 3 y^{2}

\frac{x + 3}{x + 9}

f (x) = x^{\frac{3}{5}} (x - 3)

x^{2} y + y^{3} - 48 y

What is the critical f(x)= 4/((1-4x^2)^2) ?
The critical f(x)= 4/((1-4x^2)^2) is x=0