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Derivatives and Increments -help

  1. Aug 4, 2010 #1
    1. The problem statement, all variables and given/known data

    Find f'(c) and the error estimate for the limit:

    [tex] f'(c) = \lim_{x \to 0^{+}} \frac {f(c+\Delta x) - f(c)}{\Delta x} [/tex]

    I just included that to show that we are working with one (right) sided limit the actual problem is:

    [tex] f(x) = \frac {1}{x} \;\; with\;\; c = 3[/tex]

    2. Relevant equations

    The error is given by:

    [tex] E(\Delta x) = \frac {1}{2}M \Delta x [/tex]

    and

    [tex] |f''(c)| \leq M [/tex]

    3. The attempt at a solution

    So this really quite simple, if:

    [tex] f(x)= \frac {1}{x} [/tex]

    then

    [tex] f'(x) = -\frac{1}{x^{2}} [/tex]

    and

    [tex] f''(x) = \frac {2}{x^{3}}[/tex]

    so

    [tex] f'(c) = -\frac{1}{9} [/tex]

    and

    [tex] f''(c) = \frac {2}{27} \Delta x[/tex]

    but according to the book the error is suppose to be 1/27 which doesnt really make sense to me. I got f''(x) like so:

    [tex] \frac {(x^{2})(0)-(-1)(2x)}{(x^{2})^{2}} [/tex]

    [tex] \frac {2x}{x^{4}} [/tex]

    [tex] \frac {2}{x^{3}} [/tex]

    where did I go wrong?
     
  2. jcsd
  3. Aug 4, 2010 #2

    Mark44

    Staff: Mentor

    What are you using for [itex] \Delta x[/itex]? I looked pretty closely and didn't see it anywhere.
     
  4. Aug 4, 2010 #3
    Ah yeah, that is probably a little confusing, right now [tex]\; \Delta x \; [/tex] does not have a value as the first problems of the chapter only ask me to find f'(c) and the error. [tex] \Delta x \; [/tex] will only take on a value in the later problems which as for an estimated value for something like:

    [tex] \sqrt{65} [/tex]
     
  5. Aug 4, 2010 #4

    Mark44

    Staff: Mentor

    I guess you could take [itex]\Delta x[/itex] to be 1, so that x is in the interval [3, 4]. (You said in your OP that you were interested in the right-hand limit.) In this interval, the largest value of f''(x) = 2/x3 is 2/27, which occurs when x = 3. This is because f''(x) is a decreasing function on this interal.

    Then E([itex]\Delta x[/itex]) = (1/2)M [itex]\Delta x[/itex] = ?
     
  6. Aug 4, 2010 #5
    Oh ok I see now mark44, I simply forgot to times M by 1/2 so yeah:

    [tex] E(\Delta x) = \frac {1}{2} \frac{2}{27} \Delta x = \frac {1}{27} \Delta x [/tex]

    Note that [tex] \; \Delta x \; [/tex] is not 1, it does not have a value at this point in the exercises. The correct answer in the back is written as:

    [tex] f'(c) = -\frac {1}{9},\;\;\; error= \frac {1}{27} \Delta x [/tex]

    Thanks, I just needed to pay more attention.
     
  7. Aug 4, 2010 #6

    Mark44

    Staff: Mentor

    Also, you said in the first post that the error was given in the back of the book as 1/27, rather than 1/27 * delta x. That threw me off.
     
  8. Aug 4, 2010 #7
    Yeah I realized that... I apologize for any confusion do to that.
     
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