Finding Relative Extrema & Points Of Inflection

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Homework Help Overview

The discussion revolves around finding the relative extrema and points of inflection for the function f(x) = 1/2 x - sin(x) within the interval (0, 2π). Participants are exploring the necessary conditions for identifying these features of the function.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • The original poster attempts to find critical points by setting the first derivative f'(x) = 1/2 - cos(x) equal to zero, but expresses uncertainty about solving for x. They also try to find inflection points by setting the second derivative f''(x) = sin(x) to zero, questioning their earlier steps.
  • Some participants clarify the process of solving for critical points and inflection points, discussing the implications of the interval's boundaries and the nature of the trigonometric functions involved.
  • Others suggest using reference angles and provide insights into the symmetry of the cosine function to find additional solutions.

Discussion Status

The discussion is active, with participants providing various insights and approaches to the problem. While some guidance has been offered regarding the critical points and inflection points, there is no explicit consensus on the final answers, and multiple interpretations are being explored.

Contextual Notes

Participants note the original poster's formatting issues and the potential impact on clarity. The interval (0, 2π) is emphasized, indicating that endpoints are not included in the solutions.

Cod
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For the function below, I have to find the exact values of x for which relative extreme exist and the exact values of x for which points of inflection exist.

[tex]f(x) = 1x/2 - sin(x)[/tex] when x is in the interval [tex](0,2pi)[/tex]

Here's what I have:

[tex]f'x = 1/2 - cos(x) = 0[/tex] (I'm not sure how to solve for x in this spot)

Also, for the finding the inflection points, I have this:

[tex]f"x = sin(x) = 0[/tex]

[tex]x = 0/sin = 0[/tex] (This right? Or did I screw up before this?)



I guess my troubles are based around the basic trig functions as well as algebra. I'm in the process of moving, so my algebra book will be hard to find; however, I'm about to start looking.

Any help to point me in the right direction would be fantastic. Thanks.
 
Last edited:
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First of all, wtf is up with your post? It there's a big black area and I can see some of the page's source code.

[tex]f'(x) = \frac{1}{2} -cos(x) = 0[/tex]

[tex]cos(x) = \frac{1}{2}[/tex]

You can either cheap out and use your calculator or you can draw those goofy triangles. http://myfiles.dyndns.org/math/value_triangle1.jpg is the one that applies in this case.
Turn 60 degrees into radians by dividing by 180 then multiplying by pi.

[tex]x = \frac{\pi}{3}[/tex]


I can't even read what you have for inflection stuff because of the black area I mentioned.

[tex]f''(x) = sin(x) = 0[/tex]

[tex]x = 0, \pi , 2\pi[/tex]

That is NOT the answer though! The interval is between 0 and 2[tex]\pi[/tex] but those are rounded brackets. Round brackets mean you do NOT include the limits, that means 0 and 2[tex]\pi[/tex] are not answers.

[tex]x = \pi[/tex]
 
Last edited by a moderator:
I have a program that finds roots. First extremum x is a little more than 1 radian and second about 5.20
 
Originally posted by outandbeyond2004
I have a program that finds roots. First extremum x is a little more than 1 radian and second about 5.20

Excellent point.

cos(x) = 1/2 is a referance angle. The first correct answer is pi/3. The second answer is mirrored around the x-axis (sine is mirrored around the y axis). The second answer should be 2pi - pi/3 which is 5pi/3 (about 5.23 as beyond had said).
 
Thanks for the help fellas. I don't know why my post looks like that. I checked it about 20 times looking for a mistake in the Latex coding, but there aren't any.
 

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