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Pre-Calculus and Calculus

Horizontal Tan

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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

<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 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
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

)
x = pi/16

/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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