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Analysis of a function(relatively easy)

  1. Oct 15, 2009 #1
    1. The problem statement, all variables and given/known data

    Given the function:
    [tex]f(x)=\left\{
    \begin{array}{cl}
    x+2x^2\sin\left(\frac{1}{x}\right), & x\neq 0 \\
    0, & x=0
    \end{array}\right.[/tex]
    & [tex] f'(0)=1[/tex]

    "Show that all intervals containing x=0, will also have a point where f'(x) < 0. (This shows that even if f'(0) > 0, f is not increasing on any interval containing x=0"

    2. Relevant equations

    Well, this is part b of a problem. Part a was to find f'(0)=1, which I did by the mean-value theorem.

    3. The attempt at a solution

    If I could just get a hint on how to solve this, part b, I will attempt a solution.
     
  2. jcsd
  3. Oct 15, 2009 #2

    Dick

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    Write out f'(x) and think about what values it takes for x=1/(n*pi) where n is an integer.
     
  4. Oct 15, 2009 #3

    HallsofIvy

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    How did you "use the mean value theorem" to find f'(0)= 0?
     
  5. Oct 15, 2009 #4
    Well, now you are making me unsure. But this is what I did:
    [tex] f'(0)= \frac {f(1)-f(-1)}{1-(-1)} = \frac {(1+2 sin(1))-(-1 +2 sin(-1))}{2} = \frac {2+2 sin(1) - sin(-1)}{2} = \frac {2}{2} = 1[/tex]
     
  6. Oct 15, 2009 #5

    Dick

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    Aside from being numerically wrong, that's not the definition of the derivative f'(0). Don't you have to take a limit to find f'(0)?
     
  7. Oct 15, 2009 #6
    Well. By the definition of the derivative, we have got [tex]f'(x) = {\lim}\limits_{h \to 0} \frac {x + 2(x+h)^2 \cdot sin( \frac{1}{x+h}) -(x + 2x^2 sin (\frac {1}{x}))}{h}[/tex]

    Edit: flaw in the expression. Thanks to Dick for pointing it out.
     
    Last edited: Oct 15, 2009
  8. Oct 15, 2009 #7

    Dick

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    That getting there. But you want f' when x=0. That makes (f(x+h)-f(x))/h just (f(h)-f(0))/h. Since f(0)=0, that's just f(h)/h. BTW don't forget the x term in the definition of f(x).
     
  9. Oct 15, 2009 #8
    Did I forget something?
    Yes, I see it now. The reason I didn't do this at first, was that I filled in, then substituted x with 0, and therefore got some undefined trigonometric identities.

    Edit: Which I now see is defined by the definition of the function. My head is not top notch today.
     
    Last edited: Oct 15, 2009
  10. Oct 15, 2009 #9

    HallsofIvy

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    My point was that knowing that (f(1)- f(-1))/(1-(-1))= 1 only tells you that there is SOME x between -1 and 1 such that f'(p)= 1, not that that point is x=0!

    The reason I asked instead of simply saying that the mean value theorem wouldn't work is that if you could find a sequence of points [itex]\{x_n\}[/itex] such that [itex](f(x_n)- f(-x_n))/(x_n- (-x_n))= 1[/itex] then you could "squeez" that point to x= 0, but I couldn't see how to do that.

    In any case, as Dick told you, the basic formula [itex]f'(0)= lim (f(h)- f(0))/h[/itex] works.
     
  11. Oct 15, 2009 #10
    Yes, I get it now. The second task is still a bit diffuse to me though.
     
  12. Oct 15, 2009 #11

    Dick

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    You are probably overlooking post 2.
     
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