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find any extreme values for f(x) = e-x┬Ěsin(2x) from 0< x < p/4.

Use process of optimization

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2 Answers

Do you mean f(x) = e-x sin(2x)?

f'(x) = e-x (2cos(2x) - sin(2x))

Solve f'(x) = 0 for extreme values,

tan(2x) = 1/2

x = .2318 rad , f(.2318) = .3547 <==Answer


The way they write equations and formulas sometimes is incorrect. You can write the same function in the way as Robert J. did, or as f(x) = exp(-xsin2x). If the exp(-x) is to be multiplied by sin(2x) then it is to be included in parenthesis, such as f(x) = exp(-x) sin(2x) , which is a totally different function. If we take

    f(x) = exp( -xsin(2x))  then   f'(x) = 0 if tan(2x) =-2x. This equation can be satisfied if x = 0 which is visible

without differentiation. If f(x) = exp(-x) sin (2x) then  f'(x) = exp(-x)(2cos(2x) -sin(2x)) = 0 or

                            tan(2x) = 2      x = (1/2) atan(2)  (cos(2x) = 1 and can be a divisor)

Because tan (2x) is a periodic function with the period = pi, we can write more common solution in the following way:      x = (1/2)(atan(2x) +(or -) pin)