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Math Help - Continuity and differentiation

  1. #1
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    Question Continuity and differentiation

    Hi

    I have this problem I haven been trying to solve for a while:

    "Check if the following function is continuous and/or differentiable :"

    / (x^2-1) /2 , |x|=< 1
    f(x) = \ |x| -1 , |x| > 1

    So I managed to prove it is continuous for all x by checking the limits as x -> 1 from both directions = 0
    and the limit as x -> 0 from both directions = -1/2 (is that necessary?)
    from that point it's continuous for all x as a polynomial in either branch.

    is that correct so far?

    now the problem starts with the derivative check...

    I get that the f'(x) = x , |x| < 1
    or f'(x) = x/|x| , |x| > 1

    so does that alone means the function isn't differentiable in x = 0 ?

    Thank you for your help!
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  2. #2
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    Re: Continuity and differentiation

    Quote Originally Posted by ryu1 View Post
    Hi

    I have this problem I haven been trying to solve for a while:

    "Check if the following function is continuous and/or differentiable :"
    Code:
             / (x^2-1)  /2      , |x|=< 1
    f(x) = \ |x| -1             , |x| > 1
    So I managed to prove it is continuous for all x by checking the limits as x -> 1 from both directions = 0
    and the limit as x -> 0 from both directions = -1/2 (is that necessary?)
    from that point it's continuous for all x as a polynomial in either branch.

    is that correct so far?

    now the problem starts with the derivative check...

    I get that the f'(x) = x , |x| < 1
    or f'(x) = x/|x| , |x| > 1

    Look at the graph.

    Doesn't the derivative exist everywhere?
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  3. #3
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    Re: Continuity and differentiation

    it is but the abs in the domain(is that what it's called? where you write which x's can "go in the function") confused me, how to prove it ?
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  4. #4
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    Re: Continuity and differentiation

    Quote Originally Posted by ryu1 View Post
    it is but the abs in the domain(is that what it's called? where you write which x's can "go in the function") confused me, how to prove it ?
    Your function is
    f(x) = \left\{ {\begin{array}{rl}{x-1,}&{x > 1}\\{\frac{{{x^2} - 1}}{2},}&{ - 1 \le x \le 1}\\{ - x-1,}&{x <  - 1}\end{array}} \right.

    So
    f'(x) = \left\{ {\begin{array}{rl}{1,}&{x > 1}\\{x,}&{ - 1 \le x \le 1}\\{ - 1,}&{x <  - 1}\end{array}} \right.

    What about that can't you understand?
    Last edited by Plato; April 26th 2013 at 08:18 AM.
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  5. #5
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    Re: Continuity and differentiation

    how you branched the |x| > 1 domain and function?
    I kind of understand but it is slightly vague to me ...

    Thanks for your help.
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  6. #6
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    Re: Continuity and differentiation

    Quote Originally Posted by ryu1 View Post
    how you branched the |x| > 1 domain and function?
    I kind of understand but it is slightly vague to me ...

    It is simply a matter of definition.

    |x| is simply the distance x is from 0.

    Thus \text{If }|x|>1\text{ then }x>1\text{ or }x<-1~.
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  7. #7
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    Re: Continuity and differentiation

    Quote Originally Posted by Plato View Post
    It is simply a matter of definition.

    |x| is simply the distance x is from 0.

    Thus \text{If }|x|>1\text{ then }x>1\text{ or }x<-1~.
    I remember that, it just that the branch there gave me a small headache because it has an abs in the domain and the function itself...
    it is clearer (it is also correct?) to write it like this:

    |x| -1 , |x| >1 <=> / x-1 , x > 1
    ...........................\-x-1 , x < -1

    Thanks a lot!
    Last edited by ryu1; April 26th 2013 at 09:56 AM.
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