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Math Help - Equation has exactly 2 solutions

  1. #1
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    Equation has exactly 2 solutions

    m(x+1)=e^{|x|}

    Find m\in\mathbb{R} so that the equation has exactly two solutions. I really don't know where to start... but for that equation to have any solutions m(x+1) has to be positive and non-zero.

    One answer from below is corect:

    A) m\in(1,\infty)
    B) m\in(-\infty,-e^2)\cup(1,\infty)
    C) m\in(-\infty,-e^2]\cup[1,\infty)
    D) m\in(-\infty,-e^2)\cup(0,1)
    E) m\in{\O}
    F) none of the above
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  2. #2
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    Quote Originally Posted by Utherr View Post
    m(x+1)=e^{|x|}
    Well m(x+1) is linear and this looks like e^x for x\in (0,\infty) and then reflected again in the y-axis giving a fat looking parabola.


    Quote Originally Posted by Utherr View Post
    m(x+1)=e^{|x|}
    but for that equation to have any solutions m(x+1) has to be positive and non-zero.
    do you mean m has to be positive? If so think again, when m=-9 you have two solutions
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  3. #3
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    Quote Originally Posted by Utherr View Post
    m(x+1)=e^{|x|}

    Find m\in\mathbb{R} so that the equation has exactly two solutions. I really don't know where to start... but for that equation to have any solutions m(x+1) has to be positive and non-zero.

    One answer from below is corect:

    A) m\in(1,\infty)
    B) m\in(-\infty,-e^2)\cup(1,\infty)
    C) m\in(-\infty,-e^2]\cup[1,\infty)
    D) m\in(-\infty,-e^2)\cup(0,1)
    E) m\in{\O}
    F) none of the above
    Draw the graph of y = e^{|x|}. Now draw possible graphs of y = m(x + 1), taking into account each given option. Come to a conclusion.
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