What is the critical f(x)=4x^3-6x^2-6x ?

The critical f(x)=4x^3-6x^2-6x is x=(1-sqrt(3))/2 ,x=(1+sqrt(3))/2

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critical f(x)=4x^3-6x^2-6x

critical points

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

x = \frac{1 - 3}{2}, x = \frac{1 + 3}{2}

Solution steps

4 x^{3} - 6 x^{2} - 6 x

Find where

f^{'} (x)

is equal to zero or undefined

x = \frac{1 - 3}{2}, x = \frac{1 + 3}{2}

Identify critical points not in

f (x)

domain

Domain of

4 x^{3} - 6 x^{2} - 6 x : - \infty < x < \infty

All critical points are in domain

x = \frac{1 - 3}{2}, x = \frac{1 + 3}{2}

f (x) = 5 (3 x - 1)^{2} + 25

f (x) = \frac{- x ^{2} + 9}{4 x - 12}

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

\frac{5}{2}

\frac{4 x}{x ^{2} + 9}

What is the critical f(x)=4x^3-6x^2-6x ?
The critical f(x)=4x^3-6x^2-6x is x=(1-sqrt(3))/2 ,x=(1+sqrt(3))/2