`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...

`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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Chapter 3, 3.3 - Problem 43 - Calculus of a Single Variable (10th Edition, Ron Larson).
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mathace | (Level 3) Assistant Educator

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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=(5`pi`/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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