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Derive the form of a function

  1. Feb 27, 2007 #1
    1. The problem statement, all variables and given/known data
    Find a,b,c, and d such that the cubic f(x)=[tex]ax^{3}+bx^{2}+cx+d [/tex]satisfies the indicated conditions.

    Relative maximum (3,3)
    Relative minimum (5,1)
    Inflection point (4,2)

    2. Relevant equations



    3. The attempt at a solution


    I am so lost as to how to do this :/.

    Its a polynomial so f ' (x) must = 0 at x=3 and x=5 (can't not exist), and I also know that the derivative of f(x) will be a function of degree 2, which can have at most two roots. Thus the function must be of the form a(x-3)(x-5)=f ' (x), right?

    I know that the second derivative is defined for all x (can't have negative exponents, they would become constants before that point). And that f '' (x)=0 at x=4...

    I just can't see how to piece it all together. Can someone help me out?
     
  2. jcsd
  3. Feb 27, 2007 #2

    cristo

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    Work out the explicit form of the first and second derivatives of f. Then plug in x=3 and x=5 into the first derivative, which you said must be zero. Plug in x=4 into the second derivative, which again, you said must be zero. This will give you 3 equations in 3 unknowns, which you can solve.
     
  4. Feb 27, 2007 #3

    dextercioby

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    [tex] f'(x)=3a(x-3)(x-5)\equiv 3ax^2 +2bx +c [/tex]

    is a place to start.
     
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