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

Use process of optimization

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Robert J. | Certified High School AP Calculus and Physics TeacherCertified High School AP Calculus and Ph...
4.6 4.6 (13 lesson ratings) (13)

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


Grigori S. | Certified Physics and Math Teacher G.S.Certified Physics and Math Teacher G.S.

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)