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Homework Help: Finding a function

  1. Dec 10, 2008 #1
    1. The problem statement, all variables and given/known data
    The function goes through the points -3,1 -1,3 and 1,4
    f'(-3)=0 and f'(1)=0
    f''(x)>0 when x<-1 and f''(x)<0 when x>-1
    What is the function?
    2. Relevant equations


    3. The attempt at a solution
    I take it this is a third degree equation?
    f(x) = ax^3 + bx^2 + cx + d

    When I plug in the values I get
    From derivative
    0 =27a– 6b+ c
    0 =3a+ 2b+ c
    From f(x)
    1 =-27a +9b− 3c+ d
    3 =-a+ b− c+ d
    4 =a+ b+ c+ d

    I substitute and get rid of c and d and then get
    a=-5/48
    and from this i get
    b=-0.3125 c= 0.9375 d ≈ 3.48

    but this is wrong, as the graph doesn't go through -3,1
    What did I do wrong? ?
    Thanks for help
     
  2. jcsd
  3. Dec 10, 2008 #2

    tiny-tim

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    Welcome to PF!

    Hi Olle.s! :smile:Welcome to PF! :smile:
    erm … you had five equations and only four variables! :redface:

    Try a fourth degree equation. :wink:
     
  4. Dec 10, 2008 #3

    Defennder

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    Hmm, why did you assume it was a 3rd degree polynomial? There doesn't appear to be any soution to this system of equations; you can check it online via a linear solver:
    http://wims.unice.fr/wims/en_home.html
     
  5. Dec 10, 2008 #4

    gabbagabbahey

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    Re: Welcome to PF!

    I don't think a fourth degree equation will work; the fact that [itex]f''(x)<0 \text{ if } x>-1[/itex] and [itex]f''(x)>0\text{ if } x<-1[/itex] says to me that the graph of f(x) has only one inflection point, and its at x=-1. A quartic won't accomplish that.
     
  6. Dec 10, 2008 #5

    tiny-tim

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    oops!

    oooh … I never noticed they were both at -1. :redface:

    hmmm … that gives a sixth equation, f''(-1) = 0 (or ∞?) …

    so maybe a fifth degree equation will do it?

    Thanks, gabbagabbahey! :smile: It's a good job other people check up on me! :wink:
     
  7. Dec 10, 2008 #6

    gabbagabbahey

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    Are you sure it doesn't pass through (-1,2) instead? :wink:
     
  8. Dec 10, 2008 #7

    gabbagabbahey

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    Re: oops!

    Actually I don't think there is a solution to the problem written as-is (if it passes through (-1,2) instead of (-1,3) then it's a different story)

    Try sketching a graph of the function using the fact that it must be concave up for x<-1 and concave down for x>-1....it looks almost cubic, but a little fishy at x=-1.

    -EDIT- I just realized that the problem doesn't claim that f is continuous....that changes things a little :smile:
     
  9. Dec 10, 2008 #8

    tiny-tim

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    hmm … never thought of sketching it …

    yes, (-1,2) does look a lot more likely! :smile:
    uhh? what's wrong with looking fishy?? o:)
     
  10. Dec 10, 2008 #9

    gabbagabbahey

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    The accompanying golden hue distracts one from the mathematical tasks at hand of course. :uhh: (-grasps at straws in search of witticisms-)
     
  11. Dec 10, 2008 #10

    tiny-tim

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    :biggrin: i'll take that as a compliment! :biggrin:
     
  12. Dec 10, 2008 #11
    Hi,
    I'm certain it's -1, 3,
    yeah I've sketched it, it's a third degree function with a negative value in front of x^3, I think.
    And yeah, I also had some problems when juggling with the variables..
    But perhaps there's a typo somewhere in the question that makes it unsolvable?
    Because I think this should be failry easy!:frown:
     
  13. Dec 10, 2008 #12

    gabbagabbahey

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    Well; it's not unsolvable. The function may or may not be continuous.

    Given that you are told f''(x)<0 for all x>-1 and f''(0) for all x<-1, does that tell you that f''(x) exists everywhere? If a function is twice differentiable everywhere, then can you say whether or not it is continuous?
    What places, if any might f(x) be discontinuous?
     
  14. Dec 11, 2008 #13
    Hm... I'm not sure I follow you.
    Why is the graph discontinuous?
     
  15. Dec 11, 2008 #14

    gabbagabbahey

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    I haven't said that it is.

    Is there any reason to assume otherwise though?
     
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