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How to prove b^2 > 24c for a cubic with 1 max/1 min

  1. Oct 9, 2015 #1
    1. The problem statement, all variables and given/known data
    The cubic curve [itex] y = 8x^3 + bx^2 + cx + d[/itex] has two distinct points P and Q, where the gradient is zero.

    Show that [itex] b^2 > 24c[/itex]
    2. Relevant equations
    None that I can think of.

    3. The attempt at a solution
    There's two distinct points where the gradient is zero, since it's third degree these must be the local maximum and minimums points. I graphed the equation using an online graphing tool and some sliders, and saw that it was in fact true that [itex] b^2 [/itex] has to be greater than [itex]24c[/itex] for there to be these points with zero gradient, but I'm completely lost on how to show this mathematically.

    What direction should I go in to start myself off? Thanks for any tips!
     
  2. jcsd
  3. Oct 9, 2015 #2

    Svein

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    How do you find the equation for a tangent to a curve?
     
  4. Oct 9, 2015 #3
    I could get the gradient of the tangent from the first derivative [itex] \frac{dy}{dx} = m = 24x^2 + 2bx + c[/itex], and I believe the equation of the line would then be [itex] y - y_1 = m(x - x_1)[/itex], right? Sorry, I'm not picking up on where to go next.
     
  5. Oct 9, 2015 #4

    SteamKing

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    If you want to find where the gradient (slope) is zero, and the slope of a tangent to the polynomial is m = 24x2 + 2bx + c, then what must you do?
     
  6. Oct 9, 2015 #5
    You must solve for [itex]0 = 24x^2 + 2bx + c[/itex], right? How do I get an inequality out of that?
     
  7. Oct 9, 2015 #6

    SteamKing

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    Re-read the question from the OP.
     
  8. Oct 9, 2015 #7

    RUber

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    In order for the quadratic to have 2 (real) solutions, something has to be true about (2b)^2 - 4(24)(c).
     
  9. Oct 9, 2015 #8

    Mark44

    Staff: Mentor

    Questions that involve derivatives do not belong in the Precalc subsection, so I have moved this thread to the Calculus subsection..
     
  10. Oct 9, 2015 #9
    Sorry I'm not replying until now, been busy.

    Thanks! Guess I just needed to clear my head, now it makes sense. I appreciate the help!

    Thanks a bunch! I don't know why I missed this the entire time, but it is indeed the solution. Of course, due to my organization on paper, I put b in instead of 2b and was faced with 10 more minutes of confusion.

    Thanks, sorry for posting in the wrong section.
     
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