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Math Help - [SOLVED] Natural Log Help

  1. #1
    Junior Member Morphayne's Avatar
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    [SOLVED] Natural Log Help

    I have to find the max. and min. values of the following function:

    y=3xe^{-x}+x

    When I take the first derivative, I get:

    y'=3e^{-x}-3xe^{-x}+1

    Though I'm not supposed to, I took a peak at the graph of the first derivative on my trusty graphing calculator and found that there are no critical values. So my question is, how do I show that there are no critical values using only calculation? (We are not allowed to use graphing calculators.)

    (Sorry if I posted this in the wrong section. )
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  2. #2
    Moo
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    Hello,


    The sign of y' is not obvious from a first sight.

    So let's see what the second derivative is...

    y''=\dots=3e^{-x}(x-2)

    It is 0 if x=2. If x<2, y''<0, if x>2, y''>0

    It means that y' decreases if x<2 and increases if x>2.
    There is a minimum at x=2 for y'.

    But y'(2)=3e^{-2}-6e^{-2}+1=-\frac{3}{e^2}+1

    e^2 \approx 2.6^2 and is obviously >3.
    So \frac{3}{e^2}<1

    Hence -\frac{3}{e^2}+1=m>0
    m is a minimum for y.


    From this, you can deduce that y' is always positive, since for any x, we have y' \geq m>0.


    Therefore, y is always increasing.

    Now, study the limits of y as x goes to -infinity and +infinity, and see if you can have y=0
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