# Critical Numbers, local extrema, max/min

Tags: critical, extrema, local, max or min, numbers
 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?
 P: 92 arcsin(1/2) = x = pi/6
 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?
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
 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?
 P: 92 When you drew the table, what does it tell you?

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