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Math Help - f'(x) - Exists / Continous

  1. #1
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    f'(x) - Exists / Continous

    In many textbooks I have read a statement like this -
    if f'(x) exists at x=x0 and is continuous at x=x0 then => some follow-up logic

    My question is

    Doesn't the existence of f'(x) (first derivative) at x=x0 imply it is continuous at x=x0? I say that because of the way f'(x) is defined at x=x0.

    Also if f'(x) is continuous at x=x0 then it obviously it exists at x=x0.

    Hence the two statements: 1. f'(x) exists at x=x0 2. f'(x) is continuous at x=x0 are equivalent

    Am I correct? Or I am missing something?
    Thanks
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  2. #2
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    Quote Originally Posted by aman_cc View Post
    In many textbooks I have read a statement like this -
    if f'(x) exists at x=x0 and is continuous at x=x0 then => some follow-up logic

    My question is

    Doesn't the existence of f'(x) (first derivative) at x=x0 imply it is continuous at x=x0? I say that because of the way f'(x) is defined at x=x0.

    Also if f'(x) is continuous at x=x0 then it obviously it exists at x=x0.

    Hence the two statements: 1. f'(x) exists at x=x0 2. f'(x) is continuous at x=x0 are equivalent

    Am I correct? Or I am missing something?
    Thanks
    You are PARTIALLY correct.

    If a function is differentiable at a point, then that implies it is continuous at that point.

    HOWEVER

    If a function is continuous at a point, it may OR MAY NOT be differentiable at that point.

    E.g. f(x) = |x| at the point x = 0 is continuous but NOT differentiable.
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  3. #3
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    Quote Originally Posted by aman_cc View Post
    My question is
    Doesn't the existence of f'(x) (first derivative) at x=x0 imply it is continuous at x=x0? I say that because of the way f'(x) is defined at x=x0.
    Hence the two statements: 1. f'(x) exists at x=x0 2. f'(x) is continuous at x=x0 are equivalent
    Am I correct? Or I am missing something?
    You are missing a simple fact: If f'(x_0) exists it does not mean that derivative, f', itself is continuous at x_0.
    Of course it implies that f is continuous at x_0 but not necessarily f'.
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    @Plato - I disagree
    From the definition of how f'(x) is defined
    Let g(x) = f(x) - f(x0) / x - x0
    f'(x0) exists => the limit x->x0 [g(x)] exists and f'(x0) = that limit.
    this makes f'(x) continous at x=x0.

    Is there an example where f'(x0) exists but f'(x) is not continous at x = x0.

    Please note I am not talking of continuity of f(x), existence of which has no relevance on existence of f'(x). @Prove It: You seem to have got confused here
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    Quote Originally Posted by aman_cc View Post
    @Plato - I disagree
    Is there an example where f'(x0) exists but f'(x) is not continous at x = x0.
    This is the standard example.
    f(x) = \left\{ {\begin{array}{cl}   {x^2 \sin \left( {\frac{1}{x}} \right),} & {x \ne 0}  \\   {0,} & {x = 0}  \\ \end{array} } \right.
    f has a derivative everywhere.
    But the derivative is not continuous.
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    @Plato - Really appreciate your example. Need time to understand it though. If you have any pointers which I could read-up to understand this better, I would appreciate that. Thanks
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