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/
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