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Math Help - First Derivative Test

  1. #1
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    Mar 2010
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    First Derivative Test

    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


    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?
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  2. #2
    Junior Member
    Joined
    Mar 2010
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    When deriving y=\; \left( 1-x \right)^{2}\left( 1+x \right)^{3} you have to keep in mind that the derivative of (1-x)^2 is -2(1-x) which is negative and instead of
    ((1-x)^2)(3(1+x)^2)+((1+x)^3)(2(1-x))
    you should get this:
    ((1-x)^2)(3(1+x)^2)-((1+x)^3)(2(1-x))

    Also, you are correct that -1 and 1 are both zeros but you can't just assume those are the only ones.

    ((1-x)^2)(3(1+x)^2)=((1+x)^3)(2(1-x)) which simplifies to:
    3 x^4-6 x^2+3 = -2 x^4-4 x^3+4 x+2

    and eventually simplifies to reveal a third zero of x = 1/5.

    You should find that the function is increasing from x = {-infinity,1/5},{1,infinity} and decreasing from x = {1/5,1} As shown in the graph of f.
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