anonymous
  • anonymous
relative extrema of x^3(x+1)^2
Mathematics
jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
first differentiate it to get f'. then equate f'=0 to get the critical points. differentiate again to get f". check which point that will yield f">0. then the x is the relative minima.
anonymous
  • anonymous
for relative extrema, check which x u got yield f"<0
anonymous
  • anonymous
product rule right?

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anonymous
  • anonymous
yes.
anonymous
  • anonymous
for the (x+2)^2 part, do i do (x+1)(x+1) and derive the product of that?
anonymous
  • anonymous
no. use this: if f(x)= (g(x))^n then f'(x) = n*g'(x)*(g(x))^(n-1)
anonymous
  • anonymous
3x(x+1)^2 + (2x+2)(x^3)?
anonymous
  • anonymous
Yes except you forgot the square on your first x
anonymous
  • anonymous
3x^2*(x+1)^2 + (2x+2)(x^3). u forgot the square at front.
anonymous
  • anonymous
how about 2x+3x^(2/3)?
anonymous
  • anonymous
i end up getting 2x^(-1/3) +2
anonymous
  • anonymous
thats correct
anonymous
  • anonymous
from there i get 2/cube root x +2?
anonymous
  • anonymous
yes.
anonymous
  • anonymous
how do you do absolute value problems?

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