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Getting Cubic Function

  1. Jan 22, 2008 #1
    1. The problem statement, all variables and given/known data

    From James Stewart's Essential Calculus Early Trancendentals, p.21 #5.

    Find an expression for a cubic function f if f(1)=6 and f(-1)=f(0)=f(2)=0

    2. Relevant equations

    Used zeros of the function.

    3. The attempt at a solution

    I understand that the values of x (assuming it's f(x)) which make the value 0 are the zeros of the function. We want these x values to make the whole statement 0, so

    since f(-1) = 0, then (x+1) is one of the zeros. Similarly (x-0) and (x-2) are zeros as well.

    If I put them together, I get (x+1)(x)(x-2), but then I don't know what to do with the f(1) = 6.

    The answer guide gives: f(x) = -3x(x+1)(x-2)

    For some reason, the (x-0) just dropped off, and somehow f(1)=6 equates to -3x, or maybe just -3, and the x is from the (x-0).
     
  2. jcsd
  3. Jan 22, 2008 #2

    Dick

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    Putting the roots together gives you K*(x+1)(x)(x-2) where K is any constant. That still has the right roots. To determine K, put x=1 and adjust K so you get 6.
     
  4. Jan 22, 2008 #3
    Now, f(x) = x(x + 1)(x - 2) is a function that you have arrived at from the fact that x, x + 1 and x - 2 are factors for f(x). However, these three monomials are also factors of.. let's say 2x(x + 1)(x - 2) or 4x(x + 1)(x - 2). Basically, any function of the form: f(x) = ax(x + 1)(x - 2) has those three monomials as it's factors.

    so.. can you use the fact that f(1) = 6 to get the value of 'a'?? If done properly, you should get a = -3 which will match your answer...

    EDIT:

    goddammit.. dick beat me to it.. [no pun intended]
     
  5. Jan 22, 2008 #4
    Thanks to both of you. Actually, before that was assigned to us, I went through some of the problems on my own, and tried that one. I asked the teacher about it, and she didn't seem to know how to do it. Then she assigned it to us that evening. It was the only problem in this week's homework I've had trouble with.
     
  6. Jan 22, 2008 #5

    HallsofIvy

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    Well, of course. If you don't know how to solve a problem, give it to your students!
     
  7. Jan 22, 2008 #6
    i'm gonna go out and get myself a student...
     
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