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

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

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critical f(x)=x^3-6x^2+9x+1

critical points

f (x) = x^{3} - 6 x^{2} + 9 x + 1

x = 1, x = 3

Solution steps

x^{3} - 6 x^{2} + 9 x + 1

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 + 1 : - \infty < x < \infty

All critical points are in domain

x = 1, x = 3

(\frac{1}{2}, \frac{1}{2}), (- 2, 3)

6 x^{2} + 1

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

- 2

8 x - 4 ln (x)

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