determine the x values at which f(x)= sin 4x /[e^4x], -2pi< x < 2pi?

Question

has horizontal tangents.

Answer 1

f´(x) = 1/e^8x *( e^4x*4cos4x-4e^4x*sin 4x)=0
so cos4x = sin 4x and tan4x=1
4x=pi/4+kpi and x = pi/16 +kpi/4 and k any integer
-8=

Answer 2

e^(4x) (-4cos4x) - (sin4x)(e^4)
------------------------- ---------------- = f'(x)
............... e^8x

e^4 (-4cos4x -sin4x)
---------------------------- = f'(x)
..........e^8x

-4 cos4x - sin4x
---------------------- = f'(x)
........ e^4x

f'(x) = 0 at the tangents...
and the denominator will never be irrational within the given domain, so you only need to concern your self with the numerator...


4 cos4x + sin4x = 0
4 cos4x = - sin4x
The answer to this is what you are looking for... I graphed it and there are many answers between -2pi and 2pi...