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Math Help - Using derivative and curve to solve an equation

  1. #1
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    Using derivative and curve to solve an equation

    I need to solve for all values of a the number of solutions to the equation

    e^{x}=ax

    So far, and I'm very, very unsure about this, I've got:

    e^x=ax \Leftrightarrow \frac {e^x}{x}=a
    We set f(x) = e^x / x and derive, ending up with \frac {xe^x-e^x}{x^2} = \frac {e^x(x-1)}{x^2}
    meaning that the curve changes at x = 1.

    We put up a value? table
    x | 0 1
    f'(x) | ~ - 0 +
    f(x) | ~ e

    We look at the limits
    \lim_{x\to 0^+}f(x)=\infty
    \lim_{x\to +\infty}f(x)=\infty

    So we have a curve with its minimum point at (1,e).
    The solutions are:
    a < e has 0 solutions
    a = e has 1 solution
    a > e has 2 solutions

    Correct?

    Edit:
    I forgot to add that we need to check for asymptotes. Since f(x) / x approaches infinity as x->infinity there aren't any asymptotes.
    Last edited by Sabo; March 27th 2011 at 04:07 PM.
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  2. #2
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    Quote Originally Posted by Sabo View Post
    I need to solve for all values of a the number of solutions to the equation

    e^{2}=ax
    is this really the equation, or am I missing something?

    y = e^2 is a constant function

    y = ax is a linear function for all a \in \mathbb{R}

    the two lines intersect only once for all values of a \ne 0

    if a = 0, the two lines are parallel ... no solution.
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  3. #3
    Forum Admin topsquark's Avatar
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    Is this supposed to be
    e^{2x}=ax

    \displaystyle e^{2x}=ax \Leftrightarrow \frac {e^{2x}}{x}=a

    If I am right, then there's a problem with your derivaitive:
    We set f(x) = e^2 / x and derive, ending up with \displaystyle \frac {x \cdot 2 e^{2x} - e^{2x}}{x^2}
    etc.

    -Dan
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  4. #4
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    Woa!

    I'm sorry. The equation is e^x = ax!
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  5. #5
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    Quote Originally Posted by Sabo View Post
    The equation is e^x = ax!
    The solutions are:
    a < e has 0 solutions
    a = e has 1 solution
    a > e has 2 solutions
    for 0 \le a < e , no solutions

    for a < 0 ... one solution

    the rest is ok
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  6. #6
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    Yeah, seems like I missed the second part of the curve. Thanks.
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