Question

f(x) = sin(x) + cos(x)      Consider the function on the interval (0, 2pi).  Find the open intervals on which the function is increasing or decreasing, apply first derivative test to identify all relative extrema.

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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)).

Answer

The function f(x) = sin x + cos x.
In the open interval (0, 2*pi) , the intervals in which the function is decreasing and increasing has to be identified.
For a function f(x), if f'(x) >0, the function is increasing at x; and if f'(x)<0, it is decreasing at x. At values of x where f'(x) = 0, the function f(x) has relative extrema.
For f(x) = sin x + cos x
f'(x) = cos x - sin x
cos x - sin x = 0
=> sin x = cos x
=> tan x = 1
x = tan^-1(1)
In the interval (0,2*pi) , x can take on the values {pi/4, 5*pi/4}
Now there are 3 intervals, (0, pi/4) , (pi/4, 5*pi/4 ), (5*pi/4, 2*pi)
In the interval (0, pi/4) , the value of f'(x) is positive.
In the interval (pi/4, (5*pi)/4) , the value of f'(x) is negative
In the interval ((5*pi)/4, 2*pi) , the value of f'(x) is positive.
The function f(x) = sin x + cos x is decreasing in the interval ((5*pi)/4, 2*pi) and increasing in the intervals (0, pi/4) and ((5*pi)/4, 2*pi) 
The relative extrema are at the points where x = pi/4 and x = (5*pi)/4 .

Math

`f(x) = sin(x) + cos(x)` Consider the function on the interval (0, 2pi). Find the open intervals on which the function is increasing or decreasing, apply first derivative test to identify all relative extrema.

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