SOLUTION: y = x^3 – 3x^2 The critical value I got is: 2 The inflection value I got is: 1 Could you tell me if I figured these answers out correctly? Thank you so much!

Algebra ->  Rational-functions -> SOLUTION: y = x^3 – 3x^2 The critical value I got is: 2 The inflection value I got is: 1 Could you tell me if I figured these answers out correctly? Thank you so much!      Log On


   

Question 126813: y = x^3 – 3x^2
The critical value I got is: 2
The inflection value I got is: 1
Could you tell me if I figured these answers out correctly?
Thank you so much!
Answer by stanbon(75887) About Me  (Show Source):
You can put this solution on YOUR website!
y = x^3 – 3x^2
-----------
Definition: x=c is a critical point of the function f(x) if
1st f(c) exists
AND
f'(c)=0 OR f'(c) does not exist
---------------------------
Your Problem:
f'(x) = 3x^2-6x
----------------
Check f'(x)=0
3x^2-6x = 0
3x(x-2)=0
x = 0 or x = 2
These are potential critical values.
Check f(0) to see if it exists; it does
Check f(2) to see if it exists; it does
---------------
Conclusion: x=0 and x=2 are critical values of f(x)
===================
Cheers,
Stan H.