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Difinition of a derivative

  1. Apr 14, 2005 #1
    let f be a function such that [tex]lim_{ h->x} \frac{f(2+h)-f(2)}{h} = 5[/tex]

    Which of the following must be true?

    I. f is continuous at x=2.
    II. F is differentiable at x=2.
    III. The derivative of f is continuous at x=2.

    I know (I) is true because it can be differentiated
    I know that (II) is ture to because the derivative was found so it can be differentiable.
    I dont know if III is true because the it doesnt tell me the limit of f'(x)

    are these correct?
  2. jcsd
  3. Apr 14, 2005 #2
    That isn't the limit definition for a derivative unless x = 0, or maybe you typed it in wrong.
  4. Apr 14, 2005 #3


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    It's okay...At least,for me...I edited your typo and advise u to use the code

    \lim_{...} for the limit and \rightarrow for the "->"...

  5. Apr 14, 2005 #4

    Andrew Mason

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    I think this must be:

    [tex]\lim_{h\rightarrow{0}} \frac{f(2+h)-f(2)}{h} = 5[/tex] (ie. 0, not x)

    I agree, with one caveat: We don't know the domain of h. If h can be positive or negative, f would be differentiable at 2 and, therefore, continous.

    True, subject to the above.
    I think that if a function f(x) is differentiable at x, its derivative must be continuous at x.

  6. Apr 14, 2005 #5
    Not true. Counterexample: Let f(x) = x2sin(1/x) for x != 0, and f(x) = 0 when x = 0. Then f is differentiable at zero, but the derivative is not continuous at 0.
  7. Apr 14, 2005 #6
    wait, you just said that x!=0, how can "f(x) = 0 when x = 0" when you just said x cant equal zero?
  8. Apr 14, 2005 #7
    This is a piecewise function. f(x) has one rule when x is not zero (!=0), and another rule when x=0.
  9. Apr 15, 2005 #8


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    No, he didn't say that. He said that as long as x is NOT 0, you define
    f(x) to be x2 sin(1/x) (which obviously can't be correct for x=0) but that if x IS 0, f(0)= 0.
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