Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Discontinuous split function problem

  1. Sep 21, 2004 #1
    im not really sure how to do this one because I have forgotten most of the things about trig functions from last year, but here is the problem:
    (im going to say Ħ=pi because I cant find the character for pi)
    f(x)= tan((Ħx)/4) when |x|<1
    x when |x| =>(greater than or equal to) 1

    Im supposed to find the x values, if any, at which f is not continuous, and find which of the discontinuities are removable.

    I havent had a lot of problems with my split function questions, I just dont know how to handle it when trig functions are involved, can anyone explain?

  2. jcsd
  3. Sep 21, 2004 #2


    User Avatar
    Science Advisor

    First of all, you can create a π by using & p i ; but without the spaces- it doesn't look like a very good "pi" to me- the other Greek letters using & ...; are better!

    Secondly, the way to handle a "piece-wise" function (what you are calling a "split" function) is to look at the individual "pieces" and then look carefully at the point where they meet.

    -1< x< 1 f(x)= tan(π x) Okay, where is tan(π x) NOT continuous. That's really the same as asking "where is tan(&pi x) not defined. All "elementary" functions are continuous wherever they are defined.

    For x< -1 or x> 1, f(x)= 1. That's easy, that's a constant and so it is always defined and always constant.

    Now, what about x= -1 or x= 1? as x-> 1, πx-> &pi. What is the value of tan(π)? As x-> 1, 0-> 0, of course, so that limit is 0. IF limit tan(πx)= tan(π)= 0 then the function is continous at x= 1.

    As x-> -1, π x-> -&pi. What is the value of tan(-π)? As x-> 0 0-> 0, of course, so that limit is 0. IF limit tan(-πx)= tan(π)= 0, then the function is continous at x= -1.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook