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Puzzling solution

  1. May 4, 2010 #1
    We were given a problem:
    "The slope of f(x) at point x is twice the x value. f(2) = 3. Find f(3)."

    I did this the conventional way:

    f(x) = [tex]\int[/tex] 2x dx
    = x^2 + c
    Solving with the initial condition of f(2) = 3 gives me f(x) = x^2 - 1. So f(3) = 8.

    My class did it a different way. They found f'(3) and f'(2), took the average and added that to the initial value 3, to get eight.

    So can someone prove:
    (f'(b)+f'(a))/2+f(a) = f(b) given |b-a| = 1 ?
  2. jcsd
  3. May 4, 2010 #2


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    Homework Helper

    Well, no, because it's not true in general. Try it for say, y=x3.

    It can be proven for all quadratics though. Use the fact that if |b-a|=1 then your 2 function values are of the form f(a) and b=a+1 so f(b)=f(a+1).

    So just prove for all quadratics f(x)=ax2+bx+c

  4. May 4, 2010 #3
    That's what I thought, but I wasn't sure.

    I would go the way you did, OP. I would just find the antiderivative and go from there, which is what you exactly did.
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