What is the critical f(x)=(6-4x)/((x-1)^3) ?

The critical f(x)=(6-4x)/((x-1)^3) is x= 7/4

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critical f(x)=(6-4x)/((x-1)^3)

critical points

f (x) = \frac{6 - 4 x}{( x - 1 ) ^{3}}

x = \frac{7}{4}

Solution steps

\frac{6 - 4 x}{( x - 1 ) ^{3}}

Find where

f^{'} (x)

is equal to zero or undefined

x = \frac{7}{4}, x = 1

Identify critical points not in

f (x)

domain

Domain of

\frac{6 - 4 x}{( x - 1 ) ^{3}} : x < 1 or x > 1

\frac{6 - 4 x}{( x - 1 ) ^{3}}

is not defined at

x = 1

, therefore:

x = \frac{7}{4}

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

f (x) = \frac{1}{3} x^{3} + x^{2} - 3 x - 8

f (x) = x^{3} - 9 x^{2} + 24 x + 1

\frac{x ^{2} - 5}{x - 3}

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

What is the critical f(x)=(6-4x)/((x-1)^3) ?
The critical f(x)=(6-4x)/((x-1)^3) is x= 7/4