Author 
Topic 

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<x<2
For the derivative I got: ((4(cos(4x))(e^(4x)))((4e^(4x))(sin(4x))))/(e^((4x))
that denominator in the derivative is e to the 4x squared. I also apologize for all of the parenthesis. Assuming that the derivative is correct, I set it equal to zero and solve for x right? Well I did that and I couldn't really figure out how to get x by itself. I got the solving down to: (4e^(4x))((cos (4x))  (sin (4x)))
obviously (4e^(4x)) will never equal zero because a zero in the exponent would make it 1, bit I cannot figure out a way to get x alone with cos(4x)  sin(4x) Am I doing something wrong? 


tkhunny
Advanced Member
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 doubleangle 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
Advanced Member
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
Advanced Member
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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