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

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Discussion Overview

The discussion revolves around finding the maximum value of the function f(x) = 3cos(4πx-1.3) + 5cos(2πx+0.5). Participants explore various methods for solving this problem, including calculus and trigonometric identities.

Discussion Character

  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant suggests differentiating the function and setting the derivative to zero to find critical points.
  • Another participant expresses skepticism about the effectiveness of differentiation in this case, noting that it can lead to complicated equations.
  • A specific derivative is provided, indicating the complexity of the resulting equation when using the sine addition formula.
  • There is mention of needing to apply double angle formulas to simplify the trigonometric expressions further.
  • A proposed maximum value of 5.7811 is mentioned, although it is unclear how this value was derived.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the best method to find the maximum value, with some advocating for differentiation while others highlight its complications. The discussion remains unresolved regarding the most effective approach.

Contextual Notes

The discussion includes unresolved mathematical steps and assumptions regarding the application of trigonometric identities and calculus techniques.

AlbertEinstein
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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 [itex]sin(3\pi x)[/itex] and [itex]cos(3\pi x)[/itex].
 
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 [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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