Asked by sh
                Find the interval(s) where the function is increasing where and it is decreasing.
f(x)=sin(x+(pi/2)) for 0≤x≤2pi
so my derivative is f'(x)=cos(x+(pi/2))?
and my critical number is -pi/2?
            
        f(x)=sin(x+(pi/2)) for 0≤x≤2pi
so my derivative is f'(x)=cos(x+(pi/2))?
and my critical number is -pi/2?
Answers
                    Answered by
            Damon
            
    cosine is positive for x +pi/2 = 0 to pi/2 and from 3 pi/2 to 2 pi
when x+pi/2 = 0 then x = -pi/2 or 3pi/2
when x + pi/2 = pi/2, x = 0
so one interval is x = 3pi/2 to 0
when x+pi/2 = 3 pi/2 then x = pi
when x+pi/2 = 2 pi then x = 3 pi/2
so second interval is frm x = pi to 3 pi/2
so finally the entire interval from x = pi to x = 2 pi (which is 0) has a positive derivative, function increasing
    
when x+pi/2 = 0 then x = -pi/2 or 3pi/2
when x + pi/2 = pi/2, x = 0
so one interval is x = 3pi/2 to 0
when x+pi/2 = 3 pi/2 then x = pi
when x+pi/2 = 2 pi then x = 3 pi/2
so second interval is frm x = pi to 3 pi/2
so finally the entire interval from x = pi to x = 2 pi (which is 0) has a positive derivative, function increasing
                    Answered by
            sh
            
    Wow, so confusing. Thanks!
    
                    Answered by
            sh
            
    So the whole interval is increasing? The function of cosx has a section of decreasing though.
    
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