Asked by help
                f(x)= (1-sinx) interval (0≤x≤2π) find where the function is increasing decreasing..
i know that f`(x)=1-sinx
it is equal to zero when x=pi/2
im stuck there..
            
        i know that f`(x)=1-sinx
it is equal to zero when x=pi/2
im stuck there..
Answers
                    Answered by
            MathMate
            
    if f'(x) >0 (strictly increasing) between a and b, then f(x) is increasing in the interval [a,b].
if f'(x) <0 (strictly decreasing) between b and c, then f(x) is decreasing in the interval [b,c].
You will find that the point b is common to both intervals, this is normal, because increasing or decreasing at a point is defined with respect to another point on the given interval.
    
if f'(x) <0 (strictly decreasing) between b and c, then f(x) is decreasing in the interval [b,c].
You will find that the point b is common to both intervals, this is normal, because increasing or decreasing at a point is defined with respect to another point on the given interval.
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