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Can you draw a cubic function with one real root without using calculus?

  1. Oct 30, 2012 #1
    Such as f(x)=(x^2+1)(x+1)?
  2. jcsd
  3. Oct 30, 2012 #2
    I don't understand the question? Why would you need calculus to draw something? Can't you just plug in x's and make a table of points?

    Can you elaborate on what you mean?
  4. Oct 30, 2012 #3
    Without using a table of values.
  5. Oct 30, 2012 #4


    Staff: Mentor

    Clearly the only real root is at x = -1. For values of x close to -1, but less than -1, x + 1 < 0, and x2 + 1 ≥ 1 (since x2 ≥ 0 for any real x). This means that the function values are going to be negative for x to the left of -1. Since the only root is at x = -1, all function values are negative when x < -1.

    You can continue this kind of analysis for x > -1.
  6. Oct 30, 2012 #5
    But that isn't enough information to graph the function. My original question was referring to the change of concavity of f(x) at x = 0.
  7. Oct 30, 2012 #6


    Staff: Mentor

    It's enough information to get a rough graph of this function.
    You didn't mention anything about concavity in your first post. In fact, this is the first you've mentioned concavity in this thread.

    You asked whether you could graph a cubic function without using calculus. Answer: yes.

    If the question now is can you include information about the concavity, then no, you can't do that without some ideas from calculus.
  8. Oct 30, 2012 #7
    Concave up in an interval [a,b] means that (1-t)f(a)+tf(b) > f((1-t)a+tb) for all t between 0 and 1. If you were persistent and clever, you could in principal use that definition to figure out where f(x) is concave up without resorting to differentiation.

    Lets say you have an inkling that the change of concavity takes place at 0. We could try to verify that by checking
    (1-t)f(0)+tf(b)>f(tb) for all t between 0 and 1 and all positive b:
    After simplifying it all down, this reduces to 1>t which is true since t is between 0 and 1.
    Similarly you could verify that the concavity is negative for x<0.

    With more work, you can also identify the inflection point using the same definition (rather than simply verify the point that is given to you).
    Last edited: Oct 30, 2012
  9. Oct 31, 2012 #8
    Sorry for not being terribly clear. I am using my phone to type. I'm going to explain myself better once I get home.
  10. Nov 1, 2012 #9
    Okay, I'm on a desktop now, so hopefully, I can be clear. Cubic functions can be roughly sketched by using the real roots. Note: my definition of a rough sketch means you get the concavities right and the end behaviour of the function right.

    If you have a real root with multiplicity 1, then the function clearly passes through the x-axis. Multiplicity 2 means that it does something similar to x^2 does at x=0. Multiplicity 3 means the function does a little wiggle.

    Cubic functions either have 1 or 3 real roots (is this correct?). If they have 3 real roots, then you can roughly sketch it by using the multiplicity thing I was talking about earlier. If it has 1 real root, then you run into a small issue. You get a real root with multiplicity 1 at a certain x value, so you know the function goes right through the x-axis at that point. However, you also can get a complex root with multiplicity 2 (e.g. f(x) in the OP). Because it is complex, it still does the wiggle, but the wiggle isn't on the x-axis. I was wondering if there's a way to find where the coordinates (x, y) of the wiggle is in f(x).
  11. Nov 1, 2012 #10


    Staff: Mentor

    Yes. If there is 1 real root, there are 2 complex roots.
    For this case where there are three roots, there are a couple possibilities:
    • 3 distinct roots (e.g., y = (x - 1)(x - 2)(x - 3)
    • 2 distinct roots where one of the roots is of multiplicity 2 (e.g., y = x2(x - 1)
    No, complex roots aren't repeated. They always come in conjugate pairs. For example, if z = a + bi is a root of a cubic polynomial, the other complex root will be z = a - bi.
    If you are graphing a cubic polynomial on the real plane, all you need to be concerned with are the real roots. There is no "wiggle" due to complex roots.
  12. Nov 1, 2012 #11
    Aren't the complex roots the only thing keeping the function from being a straight line?
  13. Nov 1, 2012 #12
    Anyways, thanks for answering. I understand now.
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