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 Horizontal Tan
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Average Member

11 Posts

Posted - 10/05/2007 :  20:49:59  Show Profile
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:

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?
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Advanced Member

1001 Posts

Posted - 10/05/2007 :  22:05:19  Show Profile
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.
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Advanced Member

2399 Posts

Posted - 10/06/2007 :  15:52:00  Show Profile
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
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Advanced Member

78 Posts

Posted - 10/11/2007 :  09:43:31  Show Profile
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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