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Question about differential and discontinuity

  1. Feb 11, 2010 #1
    Hello! I got something that do not understand
    Here is a picture:

    Why this function is not differentiable in x=0 ?

    It can be seen that the slopes of both sides of the function are same? Why it is not differentiable at x=0?

    Thanks in advance.
  2. jcsd
  3. Feb 11, 2010 #2
    The slope on both sides are decidedly NOT the same.

    On the left side, the graph slopes downward. It has a negative derivative.

    On the right side, the graph slopes upward. It has a positive derivative.

    At zero, approaching from one direction gives you a different result than approaching from the other direction. Therefore, there is no limit.

    Another way to think about it is to use tangent lines.

    Pick a point on the left hand side. What's the tangent line to the point you chose? (It should be pretty obvious). You can do the same for any point on the right hand side, too. But at zero, you have an infinite number of tangent points because of the sharp corner in the graph.
  4. Feb 11, 2010 #3
    Thank you for the post and sorry I post wrong diagram. Here is the correct one:


    The blue dot is "hole" in the graphic of the function.

    The slopes are same. Why the it is not differentiable at x=0 ?
  5. Feb 11, 2010 #4
    That one is easier. Because the function doesn't include 0 in the domain :)

    The definition of the derivative is [tex]\frac{f(x + h) - f(x)}{h}[/tex] as h approaches 0. Well, if f(x) isn't defined, the derivative has an undefined term in it, and is itself, undefined.
  6. Feb 11, 2010 #5
    Thank you.

    And what if I use the mean-value theorem?

    For example.

    Let me define [tex]y=x^2[/tex] on [0,1] and there is gap at 1/2 .

    Let ignore the continuity that is necessary condition of mean-value theorem.

    Then f'(c)=1 and 2c = 1, c=1/2.

    But c is not defined.
  7. Feb 11, 2010 #6
    That is a foolish thing to do :p

    If you defined y = x^2 on [0, 1], then there is no gap at 1/2. You would instead say you define y = x^2 for on [0, 1/2) and (1/2, 1].
  8. Feb 12, 2010 #7
    And what about this:

    [tex]F(x)=\frac{|2x - 3|*x*(x - 3)}{2x - 3}[/tex]


    How come that F'(3/2)=0 and the differential exists when there is discontinuity at x=3/2 on F(x)?
  9. Feb 12, 2010 #8
    Did you write the problem down right? I'm not getting that result.

    Regardless of the result, though, the problem is the derivative is defined using f(x). If for some value x, f(x) isn't defined, then f'(x) isn't defined.

    For example, if I have f(x) = x^2 / x, my function looks exactly like y = x, but isn't defined at x = 0. If you sloppily do some calc, you might even get that f'(x) = 1. This is the derivative for every single point except x=0. At some point in your work, you are dividing by that x, which adds an implicit requirement that x can't be zero.
  10. Feb 12, 2010 #9


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    I get for the derivative of [tex] F [/tex]


    Since [tex] (2x-3)^2 = |2x-3|^2[/tex], the derivative does simplify to [tex] |2x-3| [/tex].

    However, as Tac-Tics points out, this derivative is not defined at 3/2, since the original function is not defined there. If the OP graphs this function, it will be seen to consist of two separate curves, ``torn apart'' (to use a phrase uttered by one of my students) at 3/2, so it doesn't make sense to speak of any slope at that value of x.

    (Duh: just saw this: I need my coffee) The derivative can also be found this way: if x is larger than 3/2, the function is

    \frac{|2x-3|x(x-3)}{2x-3} = \frac{(2x-3)x(x-3)}{2x-3} = x(x-3) = x^2-3x
    and the derivative is [tex] 2x-3 = |2x-3| [/tex] (since x > 3/2 here)

    If x is smaller than 3/2, similar steps show that the function is [tex] -(x^2-3x) [/tex], and
    this derivative is [tex] -(2x-3) = |2x-3| [/tex], since x < 3/2 in this work.
  11. Feb 12, 2010 #10
    I just plugged it into wolfram alpha ;) The problem wasn't his algebra, so I didn't look into it too hard.
  12. Feb 12, 2010 #11


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    Ok, since you started, my first version came from wxmaxima. so, is technology like this an aid or a hindrance? (rhetorical question)
  13. Feb 12, 2010 #12
    Thanks for the replies.

    Ok, so the actual derivative is:

    F'(x)=|2x-3|, for x not equal to 3/2, right?
  14. Feb 12, 2010 #13


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