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Using the method shown in the details, how does this method prove that

  1. Mar 16, 2013 #1
    1. The problem statement, all variables and given/known data

    Solution:

    Let f(x) = x^3 - 3x^2 + 4x + 1
    f'(x) = 3x^2 - 6x + 4
    = 3(x-1)^2 + 1 > 0

    Therefore f(x) is always monotone increasing.
    From f(0) = 1,
    x> 0 and f(x) > 1

    and therefore proves the inequality.



    2. Relevant equations



    3. The attempt at a solution

    I understand how they got the derivative and how they got it into vertex form, but I don't know how by putting it into vertex form it proves that f(x) is always monotone increasing? Can someone please help me understand this method?
     
  2. jcsd
  3. Mar 16, 2013 #2

    tiny-tim

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    hi ybhathena! :smile:

    (try using the X2 button just above the Reply box :wink:)
    the fact that it's the vertex (as you call it) is irrelevant

    all that matters is that 3(x-1)2 + 1 is a square plus 1,

    so it must always be ≥ 1, and so it's always > 0 :wink:

    (this is called "completing the square")
     
  4. Mar 16, 2013 #3
    But what do you mean by it is always positive. Isnt it, the slope of the tangent line ? So then what would it mean that the slope is always positive?
     
  5. Mar 16, 2013 #4

    Ray Vickson

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    Exactly! That is what YOU need to figure out.
     
  6. Mar 16, 2013 #5
    Oh I think I get it, if the slope is always increasing that means the function is never negative hence monotone increasing right?
     
  7. Mar 16, 2013 #6

    Ray Vickson

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    All you need is that the slope is > 0; it does not matter if the slope is increasing, decreasing, or constant. Many strictly increasing functions have slopes that are decreasing; in such cases the slopes are always > 0 but are numerically smaller as x is increasing. (That means that for larger x the function increases at a smaller rate, but it still increases.)
     
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