Answer:
Therefore, the function f(x) = sin x + cos x has a relative minimum at x = pi/2, and it is increasing on the interval (0, pi/2) and decreasing on the interval (pi/2, 2pi).
Explanation:
The function f(x) = sin x + cos x is increasing on the interval (0, pi/2) and decreasing on the interval (pi/2, 2pi).
To find the intervals on which the function is increasing or decreasing, we can use the derivative of the function. The derivative of f(x) is:
f'(x) = cos x - sin x
The derivative is zero at x = pi/2.
f'(pi/2) = cos(pi/2) - sin(pi/2) = 0
The derivative is positive when x is in the interval (0, pi/2). This means that the function is increasing on this interval.
The derivative is negative when x is in the interval (pi/2, 2pi). This means that the function is decreasing on this interval.
The relative extrema of the function can be found by applying the First Derivative Test. The First Derivative Test states that if the derivative of a function is positive at a point, then the function has a relative minimum at that point. If the derivative of a function is negative at a point, then the function has a relative maximum at that point.
Applying the First Derivative Test to the function f(x) = sin x + cos x, we find that the function has a relative minimum at x = pi/2. This is because the derivative of the function is positive at x = pi/2, and the function is increasing on the interval (0, pi/2).
The function has no relative maximum. This is because the derivative of the function is negative at all points in the interval (pi/2, 2pi).
Therefore, the function f(x) = sin x + cos x has a relative minimum at x = pi/2, and it is increasing on the interval (0, pi/2) and decreasing on the interval (pi/2, 2pi).
Here is a more detailed explanation of the steps involved:
We found the derivative of the function.
We set the derivative equal to zero and solved for x.
We evaluated the derivative at the critical point to see if it was positive or negative.
We concluded that the function had a relative minimum at the critical point.