find critical numbers of:
x^3-12x^2

Question

find critical numbers of:
x^3-12x^2

anonymous · Answer

f'(x)=3x^2-24x and then what?

anonymous · Answer

set it to zero and solve for x values. 3x( x -8) so your points are x = 0 and x = 8

nincompoop · Answer

critical numbers are where the slope of the tangent is zero

anonymous · Answer

First, find the derivative of the function. If your function is F(x)=x^3-12x^2 then your first derivative would be F'(x)=3x^2-24x.  Then, set the derivative equal to zero and solve for the x values. I would factor out a 3 to make it easier to solve. 
F'(x)=3x(x-8)
Therefore, by setting each variable to zero you get x-8=0 as well as 3x=0
Once you solve, you'll get your critical numbers, x=0 and x=8.