`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.

Expert Answers
mathace eNotes educator| Certified Educator

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)).`

justaguide eNotes educator| Certified Educator

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` .