The hardest and/or most important part of answering a question is making sure the real question is understood. The July 1, 2010 question Using fsolve with a dispersion relation posted to MaplePrimes seemed to be about obtaining a numeric solution of an equation. Turns out it was more a question about the behavior of an implicit function.
The equation
was given, along with the additional information that 
, and 
. Essentially, a solution for 
is sought in the particular case 
. The challenge was in the comments "On a sketch of 
vs 
, apparently 
on the lower branch, but there should be some 
on the upper branch" and "For this particular problem, I have obtained a negative 
. Am I attempting it the right way? Is there another way of attempting it?"
Making the substitutions for 
, and 
, the equations the user sent to fsolve were then 
and 
, where
When I tried the calculation, I got
In light of the comments about 
, we have to consider the possibility that there are multiple solutions, both real and complex. The real case is dispatched with the graph in Figure 1, a plot of 
against 
.
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Figure 1 Graph of  defined implicitly by 
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All real points 
satisfying 
lie in the first quadrant. There are no real values of 
that are negative. The little hitch near 
is not an aborted asymptote because 
= 
implies 
at this endpoint. Figure 2 zooms in on this end of the graph and poses two more questions.
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Figure 2 A small portion of the graph of 
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What are the coordinates of the maximum shown in the graph, and between what two values of 
is 
not a function? The second question is equivalent to asking for the rightmost solution of 
. The solution is given by
The critical point that answers the first question is found by simultaneously solving 
and 
(computed implicitly) for 
. The result is
That there is only one real solution for 
when 
is clear from Figure 1 and the answers to these two questions. Just to be sure, a graph of 
is given in Figure 3, a figure that also shows 
is the only real solution of 
=0.
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Figure 3
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All that remains is to show there are no complex solutions of 
. We do this with Figure 4, a plot of 
with 
treated as a complex variable.
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Figure 4 Graph of  with  treated as a complex variable
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The magnitude of 
is zero at only one point, so that must be the real point 
.
Figure 2 and the concomitant calculations show that for 
, there are two values of 
that satisfy 
. For 
, there is only one solution of 
, and it is positive.
Note: The tinted cell and the four figures are tables containing hidden input. To see the input, use the Table Properties dialog to select "Show input."
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