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Finding Maxima and Minima: Turning Points


Date: 05/10/98 at 08:53:17
From: James
Subject: Working out the turning points of function

Hello,

I am having trouble finding the turning points of the function:

   5e^(-x/2)sin(2x)

Now I have worked out the derivative to be:

   -5e^(-x/2)*(0.5sin(2x) - 2cos(2x)) 

and I also know that setting the equation equal to zero will give me 
the turning points. Because -5e^(-x/2) cannot equal zero I set:

   .5sin(2x) - 2cos(2x) = 0 
              .5sin(2x) = 2cos(2x)
              .5tan(2x) = 2
                tan(2x) = 4
                     2x = 1.3258
                      x = 0.6629 

This is one maximum turning point. How would I work out the other 
maximum and minimum turning points?
 
Thank you for your help.


Date: 05/10/98 at 09:36:45
From: Doctor Sam
Subject: Re: Working out the turning points of function

James,

Nice work on this problem! In order to find the other turning points 
you need to do several things. The first is to note that tan(2x) is a 
periodic function with period pi/2. 

When you use a calculator to find the angle whose tangent is 4, you 
are finding only the principle value, the value of x for which 
tan x = 4 on the first branch of y = tan x. 

But a horizontal line at y = 4 will intersect y = tan(2x) infinitely 
often, at  0.6629, and then at 0.6629 + pi/2, 0.6629 + pi, 
0.6629 + 3pi/2, and so on. There are also infinitely many negative 
solutions: 0.6629 - pi/2, 0.6629 - pi, and so on. In general, there 
will be solutions at 0.6629 + n(pi/2) where n is an integer.

The second thing to note is that some of these turning points are 
maxima (you found one at 0.6629) and others are minima. A sign diagram 
for the derivative of your function from x = 0 to x = pi gives:

   f'(x) [+++++++ 0 ---------- 0 ++++++++]
      x  0      0.6629     0.6629+pi/2   pi  

You can find this algebraically from your derivative equation. If:

   -5e^(-x/2)*(0.5sin(2x) - 2cos(2x)) > 0 

then:

   -(0.5sin(2x) - 2cos(2x)) > 0 

so that 

   0.5sin(2x) - 2cos(2x) < 0 or tan(2x) < 4

So the signs are positive for values of x < 0.6629 and negative 
for values of x > 0.6629. This change of signs shows that the 
turning point you identified is a maximum. The signs at the next 
turning point, however, change from negative to positive marking 
a minimum value.  

Since all periods will have a similar structure (that is a maximum 
followed by a minimum) we can now identify them all:    

   The maxima occur at 0.6629 + N(pi/2)   where N = even integer
   The minima occur at 0.6629 + M(pi/2)   where M = odd integer.

Of course, the actual maximum and minimum values change in each period 
since the wave is being damped by the exponential decay factor.

I hope that helps.

-Doctor Sam,  The Math Forum
Check out our web site! http://mathforum.org/dr.math/   
    
Associated Topics:
High School Calculus

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