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...

(The entire section contains 2 answers and 401 words.)

## Unlock This Answer Now

Start your **48-hour free trial** to unlock this answer and thousands more. Enjoy eNotes ad-free and cancel anytime.

Already a member? Log in here.

Posted on

Posted on