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Homework Help: Mean value theorem!

  1. Dec 4, 2003 #1
    oks...im having problems applying the mean value theorem...i understand the concept behind it but whenever i try 2 do a question i have no idea wat 2 do...heres 1 question i've been trying:

    Showing all your work, apply the Mean Value Theorem to show
    that the function arctan x - x is equal to zero only at x = 0 .

    where do i start??? thx.
  2. jcsd
  3. Dec 4, 2003 #2


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    The mean value theorem says that if f is differentiable on (a,b) then for some c between a and b, f'(c)= (f(b)- f(a))/(b-a).

    Certainly tan(x) is differentiable for x between -pi/2 and pi/2 so also is tan(x)- x.

    Suppose tan(b)- b= 0 for some b (between -pi/2 and pi/2). Apply the mean value theorem between 0 and b.
    There must be a point c between 0 and b so that f'(c)= (f(b)- f(a))/(b-a) = 0. But for f(x)= tan(x)- x, f'(x)= sec2(x)- 1.
    f'(x)= 0 means sec2(x)= 1 or sec(x)= +/- 1. That only happens at 0 and pi/2 NOT between 0 and 1.
  4. Dec 5, 2003 #3


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    Huh. That's not the mean value theorem that I'm familiar with. (In fact, it's not true unless the derivative is contiuous.)

    The version I'm familiar with is more like:

    If [tex]f[/tex] is continuous on [tex][a,b][/tex] then there is some point [tex]c; a \leq c \leq b[/tex] so that [tex]f(c)=\frac{f(a)+f(b)}{2}[/tex]

    Now, you know that
    is continuous.
    If you can show that it is always less than zero for [tex]x>0[/tex] and alwauys greather than zero for [tex]x<0[/tex], and that there must be some point where it is zero (mean value theorem might be usefull here), then you're done.
  5. Dec 5, 2003 #4


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    I think you're thinking of the intermediate value theorem:

    If [itex]f(x)[/itex] is continuous on the interval [itex][a, b][/itex], then for any [itex]x[/itex] between [itex]f(a)[/itex] and [itex]f(b)[/itex], there exists a [itex]c \in (a, b)[/itex] such that [itex]f(c) = x[/itex].
    Last edited: Dec 5, 2003
  6. Dec 5, 2003 #5


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    I guess that's what getting too much higher math does to my brain. ...
    Not to mention that it's easy prove that |arctan x| < |x| for (x &neq; 0) without resorting to calculus.
    Last edited: Dec 5, 2003
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