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Very simple question

  1. May 5, 2005 #1
    i'm trying to do analysis, but the stupid case by case part is confusing me and it's keeping me away from doing calculus.

    h(x) = 3x^2 - 3
    h'(x) = 3(x+1)(x-1)

    how did they break down h'x to analyze it?

    like, for x <= -1, --> h'x >= 0 --> h is increasing
    -1<x<1 --> h'x <= 0 --> h is decreasing..

    etc etc....

    stuff like that...how did they get h'x is greater or lesser then 0 and increasing decreasing?


    |x| + |x+1|

    how did they get x < -1 implies |x| = -x and |x + 1| = -x - 1?

    i'm sure it is really simple, but i can't see it.
  2. jcsd
  3. May 5, 2005 #2
    If [itex] h(x) = 3x^2 - 3 [/tex]

    [tex] h'(x) = 6x [/tex]

    Critical point is at x = 0

    Decreasing at x < 0, Increasing at x > 0.
  4. May 5, 2005 #3
    What does [itex]|x|[/itex] mean?

    As whozum noted, your derivative is wrong for the first question.
  5. May 5, 2005 #4
    There has to be some typo, if [tex]h(x) = 3x^2-3[/tex], then [tex]h'(x)=6x[/tex]
    The [tex]3(x+1)(x-1)[/tex] is just a factored form of h(x).

    EDIT: It seems the other posters already cleared up this question. Fix your derivative.
  6. May 5, 2005 #5

    sorry, I meant h'x = 3x^2 - 3.

    and factor that to get 3(x+1)(x-1)...how did all the cases come up?
  7. May 6, 2005 #6
    well, when is x-1 positive? When is x+1 positive? From that, can you tell when their product is positive?
  8. May 6, 2005 #7


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    Or simply plot that parabola.You know where its vertex is and you can find its intercepts with the Ox axis...

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