calculusderivatives
posted by mary .
Hello, please help me. I needed to find the derivative of f(x)= x^4 2x^3 + x +1. I got f'(x)= 4x^3 6x^2 +1. I would like to figure out the critical numbers so I can figure out where it increases and decreases, but I have no idea how to factor it... Is it even possible?
Thank you for your time.

calculusderivatives 
Reiny
tried x = ± 1 , no good
tried ± 1/2 and ahhh! x = 1/2 is a root
so (2x1) is a factor
long division:
4x^3 6x^2 +1 = (2x1)((2x^2  2x1)
2x^2  2x1 = 0 gave me
x = (1 ± √3)/2
so there are 3 critical points to worry about.
take it from here. 
calculusderivatives 
Steve
this factors into
2(2x1)(x^2  2x  1)
x = 1/2 or (1±√3)/2
Having that, you can find where f'<0 or f'>0 
calculusderivatives 
mary
Oooooh! Thank you two so much :) I'm more confident in regards to my test now, thank you
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