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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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I think there is a problem. $e^{-x}$ can never take negative values – anonymous Oct 25 '10 at 17:19
darn book! OK, its part of a larger problem actually that says $e^{-x} -x + 2=0$ has one root a. Find an integer N such that N < a < N + 1 using the sign-change rule. The book says >> begin by finding the minima and maxima of the function, then you get an idea about where it would pass the x axis, then you can use decimal search. – gideon Oct 25 '10 at 17:34
I've heard of (or to be truthful, have extensively used) "binary search" (alias "binary chop", alias "bisection"), but "decimal search" sounds like something completely different... – J. M. Oct 25 '10 at 23:56
<quote>"This method works by drawing a graph, seeing where it crosses the axis, then re-plotting the graph" </quote> – gideon Oct 29 '10 at 5:03
up vote 3 down vote accepted

$e^{-x}>0$ 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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That is $f'(x)=-e^{-x}-1<-1$ --- what does it tell you giddy? – AD. Oct 25 '10 at 17:44
hmm not sure! I think that > $ e^{-x}>0 $ – gideon Oct 29 '10 at 5:06
A similar question showed up in my exam(which overall did not go well! :'( ) Q1>> ln(x+2) = 2ln(x) and it said Solve and answer to 2 decimal spaces? I solved ^this^ question that I posted by the iterative search method (convert the function into sequence $x_{r +1} = F(x_{r})$ and then see where it converges.) BUT it did not work for Q1 in the exam.. why?? – gideon Oct 29 '10 at 5:14
@giddy: I don't know what you mean. $\log{x+2}=2\log{x}$ can be solved simply by taking exponentials, whence $x+2=x^2$. – Weltschmerz Oct 30 '10 at 1:24
That was log(x+2) above. – Weltschmerz Oct 30 '10 at 1:57

HINT $\rm\ e^{-x}\:$ and $\rm\: -x\: $ are both strictly descreasing on $\:\mathbb R\:$, hence so is their sum + 2.

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+1 I hope you be back asap. :-) – Babak S. Jul 2 '13 at 10:57

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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The book says finding a minima and maxima helps you understand the graph better. This is when they dont give you graph paper so you kinda have to draw a rough outline of the graph so the min max helps there. – gideon Oct 29 '10 at 5:04
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. – Ross Millikan Oct 29 '10 at 14:02

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