Max Value of f(x): 3cos(4πx-1.3) + 5cos(2πx+0.5)

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The discussion centers on finding the maximum value of the function f(x) = 3cos(4πx-1.3) + 5cos(2πx+0.5). Participants suggest differentiating the function and setting the derivative to zero, but note that this approach can lead to complex equations. The derivative is given as f’(x) = -[12πsin(4πx-1.3) + 10πsin(2πx+0.5)], which complicates the solution process. To simplify, using the sine addition formula and double angle formulas is recommended, ultimately leading to an equation involving A sin(2πx) + B cos(2πx) = 0. The maximum value of the function is determined to be approximately 5.7811.
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The question is find the maximum value of the following function

f(x) = 3cos(4*pi*x-1.3) + 5cos(2*pi*x+0.5).
 
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"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 sin(3\pi x) and cos(3\pi x).
 
Calculus isn't always helpful!

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*x-1.3) + 5cos (2*pi*x+0.5)
Or, f’ (x) = - [12*pi*sin (4*pi*x-1.3) + 10*pi*sin (2*pi*x+0.5)]
Equating this to zero,
12*pi*sin (4*pi*x-1.3) + 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.
 
Yes, it gets complicated. Anything wrong with that? You'll also, by the way, need to use the double angle formulas to reduce 4\pi x to 2\pi x. After you done all that, you will have an equation of the form A sin(2\pi x)+ B cos(2\pi x)= 0 with rather complicated numbers for A and B. But they are only numbers! Write tan(2\pi x)= -B/A and solve.
 

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