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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

    mathwonk

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    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.
     
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