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First Derivative Test

  1. Mar 3, 2010 #1
    1. The problem statement, all variables and given/known data
    I have to get the following:
    - relative extrema of f
    - values of f at which the relative extrema occurs
    - intervals on which f is increasing
    - intervals on which f is decreasing

    when f(x) = (1-x)2 (1+x)3


    2. Relevant equations
    Now when get to have the first derivative by multiplication rule f'(x) = g(x)*h'(x)+h(x)*g'(x):
    f'(x) = ((1-x)2)(3(1+x)2)+((1+x)3)(2(1-x))
    is it correct to say that f'(x)=0 when x=1 or x=-1?

    and if it is, by substituting 1 and -1 to f(x), i'll arrive on ordered pairs' (1,0),(-1,0) which are on a vertical line. when i checked if the interval -1 < x < 1 is increasing or decreasing, i arrived at an answer that it is increasing which is not possible considering the locations of the two critical points.

    Where did I go wrong?
     
  2. jcsd
  3. Mar 3, 2010 #2
    first, you need to differentiate correctly. the second differentiated term needs a what by composition...?
     
  4. Mar 3, 2010 #3
    you mean this differentiation: f'(x)=((1-x)2)(3(1+x)2)+((1+x)3)(2(1-x))

    I arrived at that considering f(x)=g(x)*h(x) such that g(x)=(1-x)2 and h(x)=(1+x)3

    so applying the multiplication rule, i should have that answer.
    Do I still need to simplify it further?? will the factors vary by then?
     
  5. Mar 4, 2010 #4
    no...more like f(x)=g(h(x))*m(n(x))
     
  6. Mar 4, 2010 #5

    vela

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    You need to apply the chain rule to get the second term correct.

    You are correct that the function has zeros as x=1 and x=-1, but there's another zero that you won't see until you simplify the expression.
     
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