What is the critical f(x)=x^3-6x^2+9x-4 ?

The critical f(x)=x^3-6x^2+9x-4 is x=1,x=3

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critical f(x)=x^3-6x^2+9x-4

critical points

f (x) = x^{3} - 6 x^{2} + 9 x - 4

x = 1, x = 3

Solution steps

x^{3} - 6 x^{2} + 9 x - 4

Find where

f^{'} (x)

is equal to zero or undefined

x = 1, x = 3

Identify critical points not in

f (x)

domain

Domain of

x^{3} - 6 x^{2} + 9 x - 4 : - \infty < x < \infty

All critical points are in domain

x = 1, x = 3

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

(- 8, 1), (- 4, - 9)

f (x) = - 3 x - 3 - 2

f (x) = - 5 x - 1

f (x) = 3 (x + 1)

What is the critical f(x)=x^3-6x^2+9x-4 ?
The critical f(x)=x^3-6x^2+9x-4 is x=1,x=3