Home > Calculus > Identifying Stationary Points (Critical Points) for a Function

How do you find the critical points for # f(x)=x^3-6x^2+12x-6#?

Answer 1

William Abdalla

This is a polynomial , so all derivatives exist . Next, find any x's such that #f'(x)=0#

#f'(x) = 3(x^2-4x+4)=0#

#x=2# is the only critical point .

hope that helped

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Answer 2

Ezra Adams

To find the critical points of ( f(x) = x^3 - 6x^2 + 12x - 6 ), we first need to find the derivative ( f'(x) ) and then solve for ( x ) where ( f'(x) = 0 ) or ( f'(x) ) is undefined.

First, calculate the derivative ( f'(x) ) using the power rule:

[ f'(x) = 3x^2 - 12x + 12 ]

Next, set ( f'(x) = 0 ) and solve for ( x ) to find the critical points:

[ 3x^2 - 12x + 12 = 0 ] [ x^2 - 4x + 4 = 0 ] [ (x - 2)^2 = 0 ]

The equation ( (x - 2)^2 = 0 ) has a repeated root at ( x = 2 ).

Therefore, the critical point of ( f(x) = x^3 - 6x^2 + 12x - 6 ) is ( x = 2 ).

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.