Answer:
Explanation:
To find the points of inflection and discuss the concavity of the function f(x) = sin(x) + cos(x) on the interval (0, 2π), we need to determine the second derivative of the function and analyze its behavior.
First, let's find the first derivative of f(x):
f'(x) = d/dx(sin(x) + cos(x)) = cos(x) - sin(x)
Now, let's find the second derivative of f(x):
f''(x) = d/dx(cos(x) - sin(x)) = -sin(x) - cos(x)
To locate the points of inflection, we need to find where the second derivative changes sign. In other words, we want to find x-values where f''(x) = 0 or does not exist.
Setting f''(x) = 0, we have:
-sin(x) - cos(x) = 0
Rearranging the equation, we get:
sin(x) = -cos(x)
Dividing both sides by cos(x), we have:
tan(x) = -1
On the interval (0, 2π), the solutions to this equation are x = 3π/4 and x = 7π/4.
Now, let's analyze the concavity of f(x) using the sign of f''(x).
For x in the interval (0, 3π/4):
Taking a test value, let's choose x = π/2. Substituting this value into f''(x) = -sin(x) - cos(x), we get f''(π/2) = -1 - 0 = -1. Since f''(x) is negative in this interval, the function is concave downward.
For x in the interval (3π/4, 7π/4):
Taking a test value, let's choose x = π. Substituting this value into f''(x) = -sin(x) - cos(x), we get f''(π) = 0 - (-1) = 1. Since f''(x) is positive in this interval, the function is concave upward.
For x in the interval (7π/4, 2π):
Taking a test value, let's choose x = 5π/2. Substituting this value into f''(x) = -sin(x) - cos(x), we get f''(5π/2) = 0 - 0 = 0. Since f''(x) is zero in this interval, we cannot determine the concavity conclusively.
In summary:
The points of inflection for the function f(x) = sin(x) + cos(x) on the interval (0, 2π) are x = 3π/4 and x = 7π/4.
The function is concave downward on the interval (0, 3π/4) and concave upward on the interval (3π/4, 7π/4).
The concavity cannot be determined conclusively on the interval (7π/4, 2π) as f''(x) = 0 in this interval.