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

Continuity with the following function

  1. Dec 3, 2007 #1
    Define [tex]h : \mathbb{R} \rightarrow \mathbb{R}[/tex]

    [tex]h(x) = \begin{cases} 0 &\text{if\ }\ x \in \mathbb{Q}\\ x^3 + 3x^2 &\text{if\ }\ x \notin \mathbb{Q} \end{cases}[/tex].

    a.) Determine at what points [tex]h[/tex] is continuous and discontinuous. Prove results.

    b.) Determine at what points [tex]h[/tex] is differentiable and non-diff'able. Prove results.

    My work:

    [tex]h[/tex] is obviously cont. when [tex]0=x^3 + 3x^2[/tex] as when approaching x, lim is 0 by rationals and lim is [tex]x^3 + 3x^2[/tex] by irrationals. So, it's cont. at [tex]x=0,-3[/tex] and discont. everywhere else.

    Not sure how to prove this.

    And for differentiability.. not sure.. diff'able nowhere?
  2. jcsd
  3. Dec 3, 2007 #2


    User Avatar
    Science Advisor
    Homework Helper

    maybe diffble at 0?
  4. Dec 3, 2007 #3
    use sequences to prove (a), remember the theorem that f is continuous on a set E if for every sequence x_n in E that converges to x_0 in E has the property that f(x_n) converges to f(x_0).

    for (b) u can use sequences to find out where lim[(f(x)-f(y))/(x-y)] as x goes to y, where y is fixed exists.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook