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Math Help - range for e^|f(x)|

  1. #1
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    range for e^|f(x)|

    Hi all, just looking at this function

    y = -e^{|-x-1|}+2

    I have found the y-intercept to be 2-e and x- intercepts to be -1\pm\ln(2)

    These are correct.

    I have also found a cusp type maximum at (-1,1) . Therefore  y \in (-\infty,1]

    This range was found using trial and error, is there a way to find this range other way for this type of function?
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  2. #2
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    Quote Originally Posted by Bushy View Post
    Hi all, just looking at this function

    y = -e^{|-x-1|}+2

    I have found the y-intercept to be 2-e and x- intercepts to be -1\pm\ln(2)

    These are correct.

    I have also found a cusp type maximum at (-1,1) . Therefore  y \in (-\infty,1]

    This range was found using trial and error, is there a way to find this range other way for this type of function?
    First, of course, |-x-1|= |x+ 1|. Since that is never negative, e^{|x+1|} is never less than 1, -e^{x+1} is never larger than -1 and so y= 2- e^{|x+1|} is never larger than 1. Of course when x= -1, |-x-1|= 0, e^{-x-1}= 1 and y= 1. Since |x+1| can be arbitrarily large and e^x is an increasing function, y can be any negative number. That gives the range as (-\infty, 1]
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  3. #3
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    Thank you, this method seems obvious now it has been presented.
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