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    (Original post by Zacken)
    No, you're quite correct.



    No. You say it is non-existent.
    The graph of f(n) has a discnotinuity but for f'(n) it is just the line of 2n? How do you show 2n for n<1 at the point am I supposed to draw a hollow circle :I
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    (Original post by Mihael_Keehl)
    The graph of f(n) has a discnotinuity but for f'(n) it is just the line of 2n? How do you show 2n for n<1 at the point am I supposed to draw a hollow circle :I
    Are you sketching f(n) or f'(n)?
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    If the derivative of the function exists at 1 then \displaystyle f'(1)= \lim_{h\to 0}\frac{f(1+h)-f(1)}{h} must exist.

    Now such a limit exists if and only if both it's left derivative and right derivative exists and are equal.

    That is \displaystyle \lim_{h\to 0+}\frac{f(1+h)-f(1)}{h}=\lim_{h\to 0-}\frac{f(1+h)-f(1)}{h}.

    Now working with your function the left hand limit is
    \displaystyle \lim_{h\to 0-}\frac{f(1+h)-f(1)}{h}.

    Now \displaystyle 1+h&lt;1 for \displaystyle h&lt;0 (since we are approaching from the left). So try and evaluate the left hand limit and see what you get.

    Remember that if the function is differentiable at 1 then we must have this left hand limit actually exists.
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    (Original post by poorform)
    If the derivative of the function exists at 1 then \displaystyle f'(1)= \lim_{h\to 0}\frac{f(1+h)-f(1)}{h} must exist.

    Now such a limit exists if and only if both it's left derivative and right derivative exists and are equal.

    That is \displaystyle \lim_{h\to 0+}\frac{f(1+h)-f(1)}{h}=\lim_{h\to 0-}\frac{f(1+h)-f(1)}{h}.

    Now working with your function the left hand limit is
    \displaystyle \lim_{h\to 0-}\frac{f(1+h)-f(1)}{h}.

    Now \displaystyle 1+h&lt;1 for \displaystyle h&lt;0 (since we are approaching from the left). So try and evaluate the left hand limit and see what you get.

    Remember that if the function is differentiable at 1 then we must have this left hand limit actually exists.
    So for 1+h > 1 for h>0.

    Does the right hand limit take the same form of the left hand one?

    thnx btw

    I suppose my reasoning is weaker but there is a discontinuity in the dx part of f(n or x) so we cannot take a gradient,

    I would say that for the right hand limit it does the same thing right? The graph of f'(x) has a constant gradeint of the form 2x
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    The left side and ride side of the function derivative are not equal if you follow?
 
 
 
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