document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=953598>Question 953598</A>: <I><SPAN STYLE=\"word-break: break-all;\"> Decreasing or increasing ? an=ne^-n<BR>\n" );
document.write( "I need not only answer but explanation too. </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #582397 by </B><A HREF=/tutors/aboutme.mpl?userid=Theo><B>Theo(13342)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=Theo><IMG SRC=http://www.algebra.com/images/aboutme-small.gif BORDER=0 alt=\"About Me\" VALIGN=MIDDLE></A>&nbsp;<IMG SRC=http://www.algebra.com/cgi-bin/show-face.mpl?user=Theo><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=582397\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> an = ne^-n\r<BR>\n" );
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document.write( "I believe you might mean:\r<BR>\n" );
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document.write( "An = n * e^-n\r<BR>\n" );
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document.write( "An meaning the nth term in the progression.\r<BR>\n" );
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document.write( "The fact that the exponent is -n, I would assume that the sequence is decreasing.\r<BR>\n" );
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document.write( "You can take a few values of n and show what is happening.\r<BR>\n" );
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document.write( "since it doesn't tell you that n has to be > 0, you should assume that n can be negative or positive or 0.\r<BR>\n" );
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document.write( "so, let's take some values of n between -3 and 3 and see what happens.\r<BR>\n" );
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document.write( "the following table shows what happens:\r<BR>\n" );
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document.write( "n           n * e^(-n)         \r\n" );
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document.write( "-3          -3 * e^(-(-3)) = -3 * e^(3) = -60.25661...\r\n" );
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document.write( "-2          -2 * e^(-(-2)) = -2 * e^(2) = -14.77811...\r\n" );
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document.write( "-1          -1 * e^(-(-1)) = -1 * e^(1) = -2.71828...\r\n" );
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document.write( "0           0 * e^(0) = 0\r\n" );
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document.write( "1           1 * e^(-1) = .367879...\r\n" );
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document.write( "2           2 * e^(-2) = .270670...\r\n" );
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document.write( "3           3 * e^(-3) = .149361...\r\n" );
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document.write( "100         100 * e^(-100) = 3.720075... * 10^(-42)\r\n" );
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document.write( "it's clear that the function gets very small the further negative you go with the value of n and that the function stays positive and approaches 0 the further positive you get.\r<BR>\n" );
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document.write( "it does not appear that there is a vertical asymptote.\r<BR>\n" );
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document.write( "in order for there to be a vertical asymptote, the denominator of a rational equation needs to be equal to 0 at some point.\r<BR>\n" );
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document.write( "n * e^(-n) is the same as n / e^n\r<BR>\n" );
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document.write( "the denominator is e^n, but the denominator can never be equal to 0 because e^n can never be equal to 0.\r<BR>\n" );
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document.write( "when n = 0, e^n is equal to e^0 which is equal to 1.\r<BR>\n" );
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document.write( "so, there is no vertical asymptote, even though the function gets extremely negative as n gets more negative.\r<BR>\n" );
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document.write( "in fact, it gets so negative that you won't be able to see it on the graph after n becomes more negative than -10 or so unless you scale into the millions.\r<BR>\n" );
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document.write( "for example:\r<BR>\n" );
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document.write( "when n = -1000, n * e^(-n) becomes -1000 * e^1000 which becomes so large a negative value that the calculator can't handle it.\r<BR>\n" );
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document.write( "when n = -100, n * e^(-n) becomes -100 * e^100 which becomes -2.688117... * 10^45.\r<BR>\n" );
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document.write( "that's a very large negative number, but the function does not approach infinity as n approaches a certain number, so there is no vertical asymptote.\r<BR>\n" );
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document.write( "is there a horizontal asymptote?\r<BR>\n" );
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document.write( "that can be found by increasing n as large as it can get.\r<BR>\n" );
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document.write( "as n approaches infinity, the function approaches 0.\r<BR>\n" );
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document.write( "you get n * e^-n becomes n / e^n.\r<BR>\n" );
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document.write( "as n approached infinity, the function gets smaller and smaller.\r<BR>\n" );
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document.write( "you can see this clearer by dividing numerator and denominator by n to get:\r<BR>\n" );
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document.write( "(n/n) / ((e^n)/n) which results in 1 / ((e^n)/n)\r<BR>\n" );
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document.write( "(e^n)/n  will get larger and larger as n gets larger.\r<BR>\n" );
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document.write( "the numerator stays the same at 1.\r<BR>\n" );
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document.write( "the function gets smaller and smaller and approaches zero as n grows larger and larger and approaches infinity.\r<BR>\n" );
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document.write( "for example, when n = 1000, n/e^n is equal to 1000 / e^1000 which is equal to 0.\r<BR>\n" );
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document.write( "it's not really equal to 0, but the calculator can't handle a number that small so the calculator shows 0.\r<BR>\n" );
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document.write( "when n = 100, n/e^n is equal to 100 / e^100 which is equal to 3.72007... * 10^(-42).\r<BR>\n" );
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document.write( "that's a very small number that's pretty close to 0.  \r<BR>\n" );
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document.write( "as n gets larger, it gets close to 0 but never reaches 0.\r<BR>\n" );
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document.write( "so you have a function that approaches minus infinity as n gets more negative and you have a function that approaches 0 as n gets more positive.\r<BR>\n" );
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document.write( "a graph of the equation is shown below:\r<BR>\n" );
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document.write( "in the graph, n is replaced by x in order to satisfy the requirements of the graphing software.\r<BR>\n" );
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document.write( "otherwise the equation is the same and when i talk about n, i'm also talking about x because they represent the same thing.\r<BR>\n" );
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document.write( "there are 2 graphs to show you how the function behaves.\r<BR>\n" );
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document.write( "the first graph is a view from a distance to see what happens on the negative side of the graph as n gets more negative.\r<BR>\n" );
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document.write( "the second graph is a close up view to see what happens on the positive side of the graph as n gets more positive.\r<BR>\n" );
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document.write( "don't forget that x represent n in the graph.\r<BR>\n" );
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document.write( "so what is the answer to your question?\r<BR>\n" );
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document.write( "the question was:\r<BR>\n" );
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document.write( "Decreasing or increasing ? an=ne^-n<BR>\n" );
document.write( "I need not only answer but explanation too.\r<BR>\n" );
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document.write( "the function increases to a maximum point and then decreases thereafter as it approaches the horizontal asymptote of y = 0.\r<BR>\n" );
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document.write( "how do you explain that?\r<BR>\n" );
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document.write( "good question.\r<BR>\n" );
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document.write( "i would say you can say that n * e^-n is the same as n / e^n and it is clear clear that e^n rises dramatically faster than n as the value of n gets larger.\r<BR>\n" );
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document.write( "as n approaches infinity, the function n / e^n therefore approaches 0.   \r<BR>\n" );
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document.write( "the function is decreasing after it reaches it's maximum point at somewhere between n = 0 and n = 2.\r<BR>\n" );
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document.write( "in fact, it appears that the maximum value is as n = 1.\r<BR>\n" );
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document.write( "when n = 1, the function is equal .3678794412...\r<BR>\n" );
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document.write( "when n = .999, the function is equal to .3678792571\r<BR>\n" );
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document.write( "when n = 1.001, the function is equal to .3678792574...\r<BR>\n" );
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document.write( "I can't swear to it, but n = 1 does appear to be the maximum point of the function.\r<BR>\n" );
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document.write( "so the function reaches a maximum at n = 1 and then decreases thereafter and approaches 0 but never quite touches it.\r<BR>\n" );
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document.write( "the short answer.\r<BR>\n" );
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document.write( "the function increases for a while and then decreases.\r<BR>\n" );
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document.write( " </SPAN>\n" );
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