Calculus of a Single Variable, Chapter 3, 3.3, Section 3.3, Problem 43

Given: f(x)=sin(x)+cos(x),(0,2pi)
Find the critical values by setting the first derivative equal to zero and solving for the x value(s).
f'(x)=cos(x)-sin(x)=0
cos(x)-sin(x)=0
cos(x)=sin(x)
1=sin(x)/cos(x)
1=tan(x)
x=pi/4,x=(5/4)pi
The critical values are at x=pi /4 and x=(5pi/4) .
If f'(x)>0 the function is increasing in an interval.
If f'(x)<0 the function is decreasing in an interval.
Choose an x value in the interval (0,pi/4).
f'(pi/6)=cos(pi/6)-sin(pi/6)=.3660
Since f'(pi/6)>0 the function is increasing on the interval (0,pi/4).
Choose an x value in the interval (pi/4,(5pi)/(4)).
f'(pi/2)=cos(pi/2)-sin(pi/2)=-1
Since f'(pi /2)<0 the function is decreasing on the interval (pi/4,(5pi)/4).
Choose an x value in the interval ((5pi)/4,2pi).
f'((3pi)/2)=cos((3pi)/2)-sin((3pi)/2)=1
Since f'((3pi)/2)>0 the function is decreasing on the interval ((5pi)/4,2pi).
Since the function changed directions from increasing to decreasing there is a relative maximum. The relative maximum is at the point (pi/4,sqrt(2)).
Since the function changed direction from decreasing to increasing there is a relative minimum. The relative minimum is at the point ((5pi)/4,-sqrt(2)).

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