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Math Help - please help...

  1. #1
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    Question please help...



    thanks.
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  2. #2
    Grand Panjandrum
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    Prove that between two zeros of e^x\, \sin(x)=1 there is a zeros of e^x\,\cos(x)=-1

    Let: f(x) = e^x\, \sin(x) - 1

    First observe that:

    e^x\,\cos(x)+1=f'(x)-f(x),

    Now if x_1 and x_2 are consecutive zeros of f(x) then the signs of f'(x_1) and f'(x_2) would be different (the other possibility that one or both of these is zero does not occur as that would imply that for one ,or both, of the zeros that \cos(x)=\sin(x) \Rightarrow x=\pi/4+n\pi, but such a point cannot be a zero of f(x)).

    But the sign of f'(x_1) and f'(x_2) being different, implies that e^x\,\cos(x)+1 is positive at one root and negative at the other, so as e^x\,\cos(x)+1 is continuous, by the intermediate value theorem must be zero at some point between x_1 and x_2. which proves the result.

    RonL
    Last edited by CaptainBlack; December 17th 2006 at 06:40 AM.
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  3. #3
    Grand Panjandrum
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    Find where the function:

    f(x)=\left( x+\frac{1}{x} \right) ^x

    is an increasing and where it is a decreasing function.

    First we restrict attention to x>0, as f(x) is complex or undefined otherwise.

    Rewrite f(x) as: f(x)=e^{x\,\ln(x+1/x)}

    Now f(x) is differentiable on (0,\, \infty)and its derivative is:

    \frac{df}{dx}=\left[\frac{x^2}{x^2+1}+\ln(x+1/x) \right] e^{x\,\ln(x+1/x)}.

    But the term in the square brackets is always positive, as is the term with the exponential. Therefore f(x) is increasing on (0, \infty).

    RonL
    Last edited by CaptainBlack; December 17th 2006 at 01:17 AM.
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  4. #4
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    First you probably mean,
    \lim_{x\to 0^+}x^{x^x}
    You can write,
    x^{\exp(x\ln x)}
    Thus,
    \exp(\ln x\exp(x\ln x))
    Let us look at the term inside,
    \ln x\exp(x\ln x)
    Now,
    \lim_{x\to 0^+}\ln x=-\infty
    And,
    \lim_{x\to 0^+}\exp (x\ln x)=1
    By using the L'Hopital rule inside and then applying the limit composite rule.
    Thus,
    \lim_{x\to 0^+}\ln x\exp (x\ln x)=-\infty by the extended reals.

    Thus,*)
    \lim_{x\to -\infty}\exp x=0
    If we subtract 1 from this limit we have -1 as the answer to the limit question.


    *)I am note sure about this. Because I employed the limit composite rule for infinite limits (limits that do not exists). Thus, it might not work.
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