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Using integral calculus to find the equation of the quartic

  1. Oct 11, 2016 #1
    1. The problem statement, all variables and given/known data
    The question states Use integral calculus to find the euation of the quartic that has (1,23) and (3, 15) and a y-intercept of 24.

    2. Relevant equations
    The previous part of the question was A quartic has stationary points of inflection at x=1 and x=3. Explain why f"(x)=k(x-1)(x-3) where k doesn't =0

    3. The attempt at a solution
    Part A:

    A quartic has a non-stationary points of inflection at x=1 and x=3. Explain why f”(x)=k(x-1)(x-3) k≠0

    A quartics points of inflection can be found by equating its second derivative to zero.

    as f”(x)=k(x-1)(x-3) if f”(x)=0

    0=k(x-1)(x-3)

    if k≠0

    0=(x-1)(x-3)

    and hence x=1 and x=3


    Part B:

    Use integral calculus to find the equation of the quartic that has (1,23) and (3,15) and a y-intercept of 24.

    f”(x)=k(x-1)(x-3)
     
  2. jcsd
  3. Oct 11, 2016 #2
    Hi Jaimee, have you tried integrating the second derivative? Or are you unsure how to do the integration?
     
  4. Oct 11, 2016 #3
    Currently i have this
    f”(x)=k(x-1)(x-3)

    ∫f”(x)=∫k(x-1)(x-3)

    f’(x)=k∫(x−3)(x−1)dx

    f’(x)=((k(x−3)^2 (x))/3)+C

    but f(x)=24 x=0

    f’(0)=0

    0=((k(0−3)^2 (0))/3)+C

    0=0+C

    C=0
    therefore f'(x)=((k(x−3)^2 (x))/3)
     
  5. Oct 11, 2016 #4

    Mark44

    Staff: Mentor

    Thread moved -- if the title includes "calculus" the thread doesn't belong in the Precalculus section.
     
  6. Oct 12, 2016 #5
    I'm not sure I follow the third line: f’(x)=((k(x−3)^2 (x))/3)+C
    Maybe try expanding (x-3)(x-1) before integrating. Also you want a quartic (order 4) so does that mean your answer can be found with the first derivative? Or do you need to do an extra step?

    EDIT: I mean the 4th line not the third. In future its good practice to number your equations, it will help people help you! :).
     
  7. Oct 12, 2016 #6

    Mark44

    Staff: Mentor

    As already noted, the line above is wrong. I can't tell what you did. You should expand (x - 3)(x - 1) before atempting to integrate.
    From the given information about the y-intercept, f(0) = 24, not f(x) = 24.

    Starting with f''(x), you need to integrate twice to get f(x). At each integration there will be a constant of integration.
    When you get to a formula for f(x), there should be two constants of integration. Use the fact that f(1) = 23, f(3) = 15, and f(0) = 24 to determine those constants of integration and the constant k.

    Your function should be a fourth-degree polynomial with all coefficients determined.
     
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