Maxima and Minima of tan(sinx + sin3x) on (-pi,pi)

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Homework Help Overview

The discussion revolves around finding the maxima and minima of the function f(x)=tan(sinx + sin3x) on the interval (-pi, pi). The subject area includes calculus, specifically the analysis of functions and their derivatives.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the derivative of the function and its critical points. There is an exploration of the conditions under which the derivative equals zero, leading to the equation cos(x) = -3cos(3x). Some participants question the correctness of the derivative and the implications of the resulting equation.

Discussion Status

The discussion is ongoing, with participants clarifying the derivative and exploring the implications of the critical points. There is a recognition of the complexity of the equation that arises, and some uncertainty about the number of solutions.

Contextual Notes

Participants are working within the constraints of the interval (-pi, pi) and are addressing potential errors in the differentiation process. The nature of the function tan(sinx + sin3x) may also introduce considerations regarding its domain and behavior.

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Homework Statement


Find the maxima and minima of the function f(x)=tan(sinx + sin3x) on the interval (-pi,pi)


Homework Equations





The Attempt at a Solution


I found the derivative as sec^2x (sinx +sin3x) (cosx + cos3x) and when i set it equal to zero i found critical points at Sin=pi/2, and cos=pi/2
 
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Your derivative is incorrect. The derivative of sin(3x) is not cos(3x)...
 
Wait i typed that part wrong. i had derivative of sin(3x) as 3Cos(3x)
 
Good, so your derivative is

\frac{\cos(x)+3\cos(3x)}{\cos^2(\sin(x)+\sin(3x))}

This can only become zero if \cos(x)+3\cos(3x). So you'll need to solve the equation \cos(x)=-3\cos(3x). I believe there will be infinitly many solutions to this, but I'm not sure...
 

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