# Critical Numbers, local extrema, max/min

## 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?

arcsin(1/2) = x = pi/6

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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?