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Curve Questions

  1. Mar 25, 2004 #1
    I have a few questions

    1) Can a point on a continuous function be both a point of inflection, and a peak point (a local maximum or minimum).

    2) Do all points on a continuous curve have to be differentiable as either dy/dx or dx/dy.
     
  2. jcsd
  3. Mar 25, 2004 #2

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    My intuition says "No" to both of your questions, but that's not a very good guide, is it?
     
  4. Mar 25, 2004 #3

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    As to Q2

    One such "curve" would be a straight line going from (0,0) to (1,1), then another straight line going from (1,1) to (2,0). That is surely continuous over the interval from x=0 to x=2, but I don't believe either dy/dx or dx/dy exists at (1,1).
     
  5. Mar 25, 2004 #4

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    As to Q1

    For a point on a curve to be a local maximum, the second derivative surely has to be negative. Likewise, for it to be a local minimum, the second derivative has to be positive.
     
  6. Mar 26, 2004 #5

    Zurtex

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    Any easy example of a curve not differentiable at 1 or more points would be

    [tex]y = |x^2 - 1|[/tex]

    At points x=1 and x=-1 the curve is 0 but there is numerical solution for [tex]\frac{dy}{dx}[/tex] at these points.

    If you are unfamiliar with the || signs, they mean I am taking the modulus of x squared minus one in other words I am always taking the positive value for real numbers, e.g | -1 | = 1, | -2 | = 2 and | 3 | = 3.
     
  7. Mar 26, 2004 #6

    matt grime

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    1. No, by definition.
    2. No, by the counter examples above. There are functions that are continuous but almost nowhere differentiable (almost nowhere has a rigorous meaning).
     
  8. Mar 26, 2004 #7
    Thanks guys...

    What if for #2 it wasn’t just a continue curve, but a function with no absolute values in it? Would it then be differentiable as everywhere as dy/dx or dx/dy?
     
  9. Mar 26, 2004 #8

    matt grime

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    The absolute value isn't important. It's just that you are conditioned to believe that functions are nice because that is what you are used to.

    The modulus is useful in these situations, but it's hard to decide what you mean about not using absoute values. Arguably every function can involve absolute values. Eg, let f be any function defined on R, then there is a function g(x) where g(x) = f(|x|) if x is positive and f(-|x|) if x is negative which agrees with f. f may or may not be differentiable.

    x|x| is differentiable for instance.

    Modulus is just a red herring here.
     
  10. Mar 27, 2004 #9
    Not so. f(x) = x^4 has a local (and global) minimum at x=0, but the second derivative at x=0 is zero.

    If the first derivative is zero and second derivative is negative (positive) the point the point must be a local maximum (minimum). First and second derivative equal to zero, could be min, max, inflection point, or not.
     
  11. Mar 27, 2004 #10
    you mean something like f(x) = sqrt(x^2)?

    What? There's no absolute value in that function!
     
  12. Mar 27, 2004 #11

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    Dr. Matrix,

    You are absolutely correct in saying that a point can be an extremum and still have a vanishing second derivative. I posted without thinking deeply enough about the problem. This was one of those cases where my answer ("No") was right, but for the wrong reason.

    Also, in regard to the other question, I maybe would have been better off using as an example a straight line going from (0,0) to (1,1), and then a straight line going from (1,1) to (2,8). This curve would have been a function x=x(y) as well as a function y=y(x), unlike the example I gave first, which was a function y(x) but not a function x(y) since it was double-valued when looked at in that way.
     
    Last edited: Mar 27, 2004
  13. Mar 27, 2004 #12

    ShawnD

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    woops i screwed up
     
  14. Mar 29, 2004 #13

    HallsofIvy

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    The definition of inflection point, in my calculus book, is that the graph is concave upward on one side and concave downward on the other. Of course, that means that the second derivative must change sign at the point and so the second derivative must be 0 but that is not a sufficient condition. It also follows from the definition that an inflection point cannot be an extremum.

    As an example of a function that is not differentiable at, say x= 1 but does not involve absolute value, let f(x)= x if x<= 1, f(x)= 2x-1 if x> 1.

    If you are now going to ask "are there examples not using absolute value and not defined in separate formulas", I would have to respond "exactly what is your idea of a formula?". It happens that differentiable functions (indeed, analytic functions) are very nice and so our ways of writing formulas have developed so that those are particularly easy to write. That's a happenstance of our way of writing functions, not any thing "mathematical". In fact, "almost all" continuous functions are no-where differentiable.
     
  15. Mar 29, 2004 #14

    NateTG

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    Yes, but the function can't be differentiable at the inflection point. Plenty of examples, here's one:
    [tex]f(x)=x^2[/tex] for [tex]x\leq 0[/tex] and [tex]f(x)=\sqrt{x}[/tex] for [tex]x \geq 0[/tex].

    No, consider, for example, [tex]f(0)[/tex] in the example above.
     
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