Critical Numbers, local extrema, max/min

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Homework Help Overview

The discussion revolves around classifying local extrema for the function f(x) = x/2 + cos(x) within the interval [0, 2π]. Participants are exploring critical numbers and attempting to determine both local and absolute extrema.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • The original poster attempts to find critical numbers by taking the derivative and setting it to zero. They express uncertainty about solving for x when sin(x) = 1/2. Subsequent posts identify critical numbers and discuss the behavior of the function across intervals, raising questions about finding exact values for function outputs at those critical points.

Discussion Status

Participants are actively engaging in the problem, with some providing calculations and others questioning the implications of their findings. There is a mix of interpretations regarding local maxima and minima, and while some guidance has been offered, no consensus has been reached on the classification of extrema.

Contextual Notes

Participants are working under the constraints of homework rules, focusing on exact values and critical points without providing complete solutions. There is an ongoing discussion about the behavior of the function and the implications of the derived values.

davemoosehead
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Homework Statement



Classify the local extrema of f(x) = x/2 + cos(x), 0 <= x <= 2pi. Give the exact values of the critical numbers and extrema. Find the absolute (or global) maximum and minimum values of the function.

The Attempt at a Solution



So i need to find the critical numbers first right? Take the derivative..
f'(x) = 1/2 - sin(x)
set equal to zero..
0 = 1/2 - sin(x) -> 1/2 = sin(x) how do i find x?
 
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arcsin(1/2) = x = pi/6
 
Last edited:
critical numbers: pi/6, 5pi/6
I drew out this table:
Interval | Test Value | Sign | Behavior
(0, pi/6) | pi/10 | + | Inc
(pi/6, 5pi/6)| pi/2 | - | Dec
(5pi/6, 2pi) | pi | + | Inc

f(pi/6) = pi/12 + cos pi/6 = pi/12 + 6sqrt(3)/12 = pi + 6sqrt(3) / 12
f(5pi/6)= 5pi/12 + [cos 5pi/6 = ?]

how do i find an exact value for these two?
 
The exact value is right as you write it but

cos(pi/6) = sqrt(3)/2
and
cos(5pi/6) = -sqrt(3)/2.

so

f(pi/6) = sqrt(3)/2 + pi/12
f(5pi/6) = 5pi/12 - sqrt(3)/2
 
so,

pi/12 + sqrt(3)/2 is a local max and
5pi/12 - sqrt(3)/2 is a local min?

f(0) = 1 is a global min and
f(2pi) = pi + 1 is a global max?
 
When you drew the table, what does it tell you?
 

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