What is the critical f(x)= 1/((2x-3)^3) ?

The critical f(x)= 1/((2x-3)^3) is None

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critical f(x)= 1/((2x-3)^3)

critical points

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

None

Solution steps

\frac{1}{( 2 x - 3 ) ^{3}}

Find where

f^{'} (x)

is equal to zero or undefined

x = \frac{3}{2}

Identify critical points not in

f (x)

domain

Domain of

\frac{1}{( 2 x - 3 ) ^{3}} : x < \frac{3}{2} or x > \frac{3}{2}

\frac{1}{( 2 x - 3 ) ^{3}}

is not defined at

x = \frac{3}{2}

, therefore:

None

f (y) = x^{3} - 3 x^{2} - 45 x + 2

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

f (x) = 6 - x

\frac{4 x ^{2}}{x ^{2} + 3}

f (x) = 5 x - 19 y

What is the critical f(x)= 1/((2x-3)^3) ?
The critical f(x)= 1/((2x-3)^3) is None