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Critical Numbers, local extrema, max/min

by davemoosehead
Tags: critical, extrema, local, max or min, numbers
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davemoosehead
#1
Nov21-07, 06:59 AM
P: 26
1. The problem statement, all variables and given/known data

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.

3. 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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danni7070
#2
Nov21-07, 07:08 AM
P: 92
arcsin(1/2) = x = pi/6
davemoosehead
#3
Nov21-07, 07:46 AM
P: 26
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?

danni7070
#4
Nov21-07, 07:56 AM
P: 92
Critical Numbers, local extrema, max/min

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
davemoosehead
#5
Nov21-07, 02:56 PM
P: 26
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?
danni7070
#6
Nov22-07, 03:37 AM
P: 92
When you drew the table, what does it tell you?


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