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Math Help - Mean Value Theorem

  1. #1
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    Mean Value Theorem

    Well, I honestly have no clue how to do this and have doing it based on examples of normal non-trigonometric functions...
    I've gotten up to the point of merely differentiating f(x) before not knowing where to go....

    Prove using the Mean Value Theorem:

    If x>0, then arctan(x) > 1/ (1+x^2)





    Thanks in advance
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  2. #2
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    Hmm, is this correct?

    f'(c)(b-a)= ∫[a,b] f(t) dt

    Let f(t) = 1/(1+t^2)

    b=x and a=0

    f'(c)(x-0)= ∫[0,x] 1/(1+t^2) dt = arctan(x)

    For 0<c<x,

    arctan(x) = x/(1+c^2)

    since c<x and since numbers are positive,

    x/(1+c^2) > x/(1+x^2) and thus arctan(x)>x/(1+x^2)
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  3. #3
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    Quote Originally Posted by storage1k View Post
    Well, I honestly have no clue how to do this and have doing it based on examples of normal non-trigonometric functions...
    I've gotten up to the point of merely differentiating f(x) before not knowing where to go....

    Prove using the Mean Value Theorem:

    If x>0, then arctan(x) > 1/ (1+x^2)

    Thanks in advance
    Here's an outline of how I'd do it:

    Consider f(x) = \arctan x and note that f'(x) = \frac{1}{1 + x^2}.

    From the Mean Value Theorem: f'(c) = \frac{f(b) - f(a)}{b - a}, where a < c < b.

    Apply the MVT to the interval [0, x] and let c = v:

    f'(v) = \frac{f(x) - f(0)}{x - 0}

     \Rightarrow \frac{1}{1 + v^2} = \frac{\arctan x - \arctan 0}{x - 0} = \frac{\arctan x}{x}

    \Rightarrow \frac{x}{1 + v^2} = \arctan x.

    Now note that \frac{x}{1 + v^2} > \frac{x}{1 + x^2} since 0 < v < x.

    Therefore \arctan x = \frac{x}{1 + v^2} > \frac{x}{1 + x^2}.



    You should now continue and do the following:

    1. Prove \frac{x}{1 + x^2} < \arctan x < x.

    2. Use part 1. to show that if 0 < \theta < \frac{\pi}{2} then \cos \theta < \frac{\theta}{\sin \theta} < \frac{1}{\cos \theta}.

    3. Use part 2. to prove that \lim_{\theta \rightarrow 0} \frac{\sin \theta}{\theta} = 1.
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