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

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

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critical f(x)=3-3x+3x^2-x^3

critical points

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

x = 1

Solution steps

3 - 3 x + 3 x^{2} - x^{3}

Find where

f^{'} (x)

is equal to zero or undefined

x = 1

Identify critical points not in

f (x)

domain

Domain of

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

All critical points are in domain

x = 1

y = - \frac{1}{4} x + 3, (3, - 3)

f (x) = - \frac{31}{( 6 + t ) ^{2}}

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

y = \frac{x + 8}{x}

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

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