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markmil2002
Average Member

USA
11 Posts

 Posted - 10/05/2007 :  20:49:59 I'm trying to determine the x value(s) at which a function has horizontal tangents. The function is:(sin 4x)/(e^(4x)) at which -2

tkhunny

USA
1001 Posts

 Posted - 10/05/2007 :  22:05:19 I think you mean 4/e^(4x) = 4*e^(-4x), but that's not particularly important, since that expression also never is zero.Do you know your double-angle formulas? You could have lots of fun.Alternatively, you could find this one:cos(x) - sin(x) = (2)cos(x + (/4))That might lead you to this...cos(4x) - sin(4x) = (2)cos(4[x + (/16)])It is an important and useful transformation.

sahsjing

USA
2399 Posts

 Posted - 10/06/2007 :  15:52:00 y = (e^-(4x))(sin 4x)y' = (e^-(4x))(4cos 4x - 4sin 4x) = 0=> cos 4x = sin 4x = cos (/2 - 4x 2n) x = pi/16 n/4------Just another way... Edited by - sahsjing on 10/06/2007 15:53:19

HallsofIvy

USA
78 Posts

 Posted - 10/11/2007 :  09:43:31 I would have just noted that sin(4x)- cos(4x)= 0 is the same as sin(4x)= cos(4x). For what angles are sine cosine the same? Of course, that is the same as saying sin(4x)/cos(4x)= tan(4x)= 1. That is why it is important that tangent is positive in the first and third quadrants.
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