Critical Numbers, local extrema, max/minby davemoosehead Tags: critical, extrema, local, max or min, numbers 

#1
Nov2107, 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? 



#2
Nov2107, 07:08 AM

P: 92

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




#3
Nov2107, 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? 



#4
Nov2107, 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 



#5
Nov2107, 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? 



#6
Nov2207, 03:37 AM

P: 92

When you drew the table, what does it tell you?



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