
#1
Jun2306, 11:58 PM

P: 113

The question is find the maximum value of the following function
f(x) = 3cos(4*pi*x1.3) + 5cos(2*pi*x+0.5). 



#2
Jun2406, 09:22 AM

Math
Emeritus
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Thanks
PF Gold
P: 38,879

"The question"? Where, in your homework?
Looks pretty straight forward to me: differentiate and set the derivative equal to 0. Use the sin(a+b) formula to isolate [itex]sin(3\pi x)[/itex] and [itex]cos(3\pi x)[/itex]. 



#3
Jun2506, 02:06 AM

P: 113

Hey HallsofIvy,
Your suggestion was quite correct. However I dare say differentiating and equating to zero does not always help. Here f (x) = 3cos (4*pi*x1.3) + 5cos (2*pi*x+0.5) Or, f’ (x) =  [12*pi*sin (4*pi*x1.3) + 10*pi*sin (2*pi*x+0.5)] Equating this to zero, 12*pi*sin (4*pi*x1.3) + 10*pi*sin (2*pi*x+0.5)] = 0 Now if I use the sin (a+b) formula the equation gets rather complicated. I shall be thankful to you if you could please show the full solution. However the answer is 5.7811. 



#4
Jun2506, 12:16 PM

Math
Emeritus
Sci Advisor
Thanks
PF Gold
P: 38,879

maximum of...
Yes, it gets complicated. Anything wrong with that? You'll also, by the way, need to use the double angle formulas to reduce [itex]4\pi x[/itex] to [itex]2\pi x[/itex]. After you done all that, you will have an equation of the form [itex]A sin(2\pi x)+ B cos(2\pi x)= 0[/itex] with rather complicated numbers for A and B. But they are only numbers! Write [itex]tan(2\pi x)= B/A[/itex] and solve.



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