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Derivatives and Increments -help, again

  1. Aug 5, 2010 #1
    1. The problem statement, all variables and given/known data

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

    [tex] f(x)= \sqrt{x^{2}+1} [/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 the first derivative is:

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

    and for the second derivative I got:

    [tex] f''(x) = (x^{2}+1)^{-3/2} [/tex]

    So for f'(c) I got:

    [tex] f'(c) = \frac {2}{\sqrt{5}} [/tex]

    which is correct in the back of the book but for the error i got:

    [tex] f''(c) = 5^{-3/2} [/tex]

    so

    [tex] E(\Delta x) = \frac {1}{2} (5)^{-3/2} \Delta x \; \approx \; .04472 \Delta x[/tex]

    but according to the book it is:

    [tex] E(\Delta x) = \frac {1}{4\sqrt{2}} \Delta x \; \approx \; .17678 \Delta x [/tex]

    I have done the second derivative over now a couple of times and still get the same answer but here is how i got f''(x):

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

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

    which then boils down to:

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

    Next lets combine the terms on the top to get something like:

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

    so x^2 will cancel on the top fraction to get:

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

    and that can be rewritten as:

    [tex] \frac {1}{\sqrt{x^{2}+1}} * \frac {1}{x^{2}+1} [/tex]

    which can further be rewritten as:

    [tex] (x^{2}+1)^{-1/2} * (x^{2}+1)^{-1} = (x^{2}+1)^{-3/2} [/tex]

    so:

    [tex] f''(x) = (x^{2}+1)^{-3/2} [/tex]

    where did I go wrong?
     
  2. jcsd
  3. Aug 5, 2010 #2

    Mark44

    Staff: Mentor

    Where is M? For your error estimate you need a number M such that f''(x) <= M for all x in some interval.

    What is c? You show values for f'(c) and f''(c) without ever saying what c is.
    Your result for f''(x) look fine.
     
  4. Aug 5, 2010 #3
    Oh my bad, c = 2 and since:

    [tex] f''(x) \leq M [/tex]

    f''(c) = M so in this problem (as shown):

    [tex] M = (x^{2}+1)^{-3/2} \; and \; with \; c=2, \; M=(5)^{-3/2} [/tex]
     
  5. Aug 6, 2010 #4

    Mark44

    Staff: Mentor

    There are two parts to this problem:
    1. Find f'(2)
    2. Find the error estimate for f(x) = sqrt(x^2 + 1).

    The first part you have already done.
    For the second part, we can estimate f(x) by f(2) + f'(2)[itex]\Delta x[/itex]. The error in this estimate is E([itex]\Delta x[/itex]), the absolute value of which is equal to (1/2)f''([itex]\xi[/itex])[itex]\Delta x[/itex], for some number [itex]\xi[/itex] between 2 and x.

    Since we don't know [itex]\xi[/itex] (the Greek letter xi), the best we can do is find the largest value of f''([itex]\xi[/itex]), which we'll call M.

    As it turns out, the global maximum value for f''(x) = 1/(x^2 + 1)^(3/2) is 1, which is attained when x = 0. Since the answer in your book doesn't seem to be using that value, but seems to be using f''(1) instead, I suspect that the problem is assuming that [itex]\Delta x[/itex] is less than 1.

    You neglected to include the fact that c was given as 2. Did you forget to mention another piece of given information?
     
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