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ok so I did all the revision problems and noted the ones I couldn't do today and Im posting them together, hope thats not a problem with the power that be?

I have exhibit A:
$e^{-x} -x + 2 $
So I differentiate to find where the derivative hits $0:$
$-e^{-x} -1 = 0 $

Now HOW do I figure when this hits zero!?
$-1 = e^{-x} $
$\ln(-1) = \ln(e^{-x})$ ???

More to come ... as one day rests between me and my final exam/attempt at math!

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    ``"This method works by drawing a graph, seeing where it crosses the axis, then re-plotting the graph" ``2010-10-29

3 Answers 3

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e^{-x}>0 for all real $x$. Hence $-e^{-x}<0$ for all real $x$, whence $-e^{-x} - 1 <-1$ for all real $x$. So it nevers "hits" zero. Look at a graph of your function.

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    ugh! I did sometime similar but with a very stupid mistake! =(2010-10-30
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HINT $\rm\ e^{-x}\:$ and $\rm\: -x\: $ are both strictly descreasing on $\:\mathbb R\:$, hence so is their sum + 2.

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I don't understand why the book wants you to find the max and min. You have correctly deduced that the derivative is never zero, which says there isn't a max or min. Looking for the root of this function is not hard. If you graph it over a reasonable range, say -5 to 5, you will find it close enough to get N correctly

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    It is true that if there are maxima and minima they help you understand the graph. But you can understand the graph even if it doesn't have any. In this case, you know that the graph is decreasing to the right, starts from $+\inf $ and goes to $-\inf $. That's the general shape, but there are lots of graphs like that.2010-10-29