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Quartic with two stationary points of inflection

  1. Oct 1, 2012 #1
    Hey everyone!
    Recently got a question in maths which asks:
    "Use integral calculus to find the equation of the quartic that has stationary points of inflection at (1, 23) and (3, 15) and a y-intercept of 24"
    This means that the second derivative has the form (as inflection points are x-intercepts in the second derivative):
    f''(x) = k(x-1)(x-3)
    I integrate this and get an answer for f'(x), all fine and dandy. But then I say, since there are two stationary points, f'(1) = 0, and f'(3) = 0, and it all breaks down!

    Is it even possible to have a quartic with TWO stationary points of inflection, or am I just screwing something up (haha)?

    Cheers.
     
  2. jcsd
  3. Oct 1, 2012 #2

    Mentallic

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    This should be in the homework section.

    And sorry, but what is a stationary point of inflection? There are inflection points (f '' = 0), there are stationary points (f ' = 0), but a stationary point of inflection?

    Anyway, I'll assume you mean just an inflection point. Of course quartics can have two stationary points, because clearly if you take the second derivative of a general quartic [itex]f(x)=ax^4+bx^3+cx^2+dx+e[/itex] you'll end up with a quadratic, which can have anywhere from 0 to 2 roots based on its discriminant.
     
  4. Oct 1, 2012 #3
    Thankyou, and sorry, I should have explained. a stationary point of inflection is a point of inflection at which the rate of change is zero. Like this: http://mathworld.wolfram.com/images/eps-gif/InflectionPoint_700.gif

    I've been sitting here trying to sketch a quartic with two of these points (of course a quartic with two points of inflection is possible :smile:) and the only thing I can come up with is one which looks like a quadratic, but really fat. Like this: http://www.math.brown.edu/UTRA/polynomials/quartic.gif
    where the central max/min forms a line with the other two, but I still cannot get the math to work.
    [itex]f'(x) = \int k(x-1)(x-3) = k(\frac{x^{3}}{3} - 2x^{3} + 3x) + c[/itex]
    Then, f'(3) = 0 which results in:
    [itex]k(9-54+9) + c = 0[/itex]
    [itex]∴ c = 54k[/itex]
    and f'(1) = 0
    results in c = (4/3)k

    How can c = both things, i dont understand! If k = 0, then I have nothing to integrate to get f(x)

    Have I done something, or they have the question wrong?
     
    Last edited: Oct 1, 2012
  5. Oct 1, 2012 #4

    pwsnafu

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    Show your working. Why would it break down?

    Edit: argh cross post.

    Why aren't you starting with a "generic" quartic as Mentallic post above, and differentiate twice, and solve for a and b?
     
    Last edited: Oct 1, 2012
  6. Oct 1, 2012 #5

    pwsnafu

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    Err, no. An inflection point means that the second derivative changes sign. It's not enough that the second derivative vanish there.
     
  7. Oct 1, 2012 #6
    Either way I do it, if I work back from the general form, I have unknowns on both sides, the k in front of one lot, and the a, b, and c in front of the second derivative of the general form.
    Please help me

    Solving for A, b, and c I get:
    k = 12a
    k = -1.5b
    k = (2/3)c
     
  8. Oct 1, 2012 #7

    pwsnafu

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    What's k?
    You really need to show your working. We aren't mind readers.
     
  9. Oct 2, 2012 #8
    [itex]f''(x) = k(x-1)(x-3) = kx^{2} - 4kx + 3k[/itex]
    and
    [itex]f''(x) = 12ax^{2} + 6bx + 2c[/itex] (differentiating from general formula)

    [itex]∴12a = k[/itex]
    [itex]∴6b = 4k[/itex]
    [itex]∴2c = 3k[/itex]
     
  10. Oct 2, 2012 #9

    pwsnafu

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    Okay so substitute back into ##f(x) = ax^4 + bx^3 + cx^2+dx+e## and go from there.
     
  11. Oct 2, 2012 #10

    Mentallic

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    Ahh so it is. It's such a warm feeling knowing that I've had that wrong all these years.
     
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