Identifying Extrema of a Trigonometric Function on an Interval

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Homework Help Overview

The discussion revolves around identifying the maximum and minimum values of the trigonometric function f(x) = 2sin(2x) - 2sin(x) over the interval [-π, π]. Participants are exploring the critical points derived from the first derivative and evaluating the function at these points.

Discussion Character

  • Exploratory, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • The original poster attempts to find critical points by setting the first derivative to zero and evaluating the function at these points. Some participants question the completeness of the critical points identified, suggesting the inclusion of boundary values. Others propose a transformation of the function to simplify the analysis.

Discussion Status

The discussion is active, with participants providing insights and corrections regarding critical values. There is a collaborative effort to ensure all relevant points are considered, although no consensus on the maximum value has been reached yet.

Contextual Notes

Participants note the importance of including endpoints of the interval as critical values and discuss the implications of the function's behavior at these points. There is also mention of potential oversight in identifying critical points.

Glissando
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Homework Statement


Find the maximum and minimum values of the function f(x) = 2sin2x-2sinx on the interval [-pi, pi]]

Homework Equations


first derivative

The Attempt at a Solution


f'(x) = 4sinxcosx-2cosx
0 = 2cosx (2sinx-1)
0 = 2sinx-1, sinx = 1/2, x = pi/6, 5pi/6
0 = 2cosx, cosx = 0, x = pi/2

I got that MIN = -0.5 (plugged pi/6 and 5pi/6 into the original function) but I can't seem to find where MAX @4 is.

Thank you!

Thank you!
 
Last edited:
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Hi Glissando! :smile:
Glissando said:

Homework Statement


Find the maximum and minimum values of the function f(x) = 2sin2x-2sinx on the interval [-pi, pi]]

Homework Equations


first derivative

The Attempt at a Solution


f'(x) = 4sinxcosx-2cosx
0 = 2cosx (2sinx-1)
0 = 2sinx-1, sinx = 1/2, x = pi/6, 5pi/6
0 = 2cosx, cosx = 0, x = pi/2

You forgot -pi/2 here, no?

Anyway, that would give you 6 critical values (don't forget to add the boundaries of the interval as critical values!):

-\pi,-\frac{\pi}{2},\frac{\pi}{6},\frac{\pi}{2}, \frac{5\pi}{6},\pi

Now you need to calculate f of all these values and find out which one is the greatest!
 
Put y = sin(x), to get the function 2*y*(y-1) on the interval -1 <= y <= 1. The min occurs at y = 1/2. You have an "end point" max, where the derivative (with respect to y) need not equal zero.

RGV
 
Last edited:
Thank you (: Figured it out!
 

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