find the absolute maximum and minimum for f(x) = x+ sin(x) on the closed interval [0, 2pi]
f' = 1 + cos(x)
f'=0 when cos(x) = -1, or x=pi
however,
f''(x) = -sin(x) which is also 0 at x=pi.
So, (pi,pi) is an inflection point, not a max or min.
f(2pi) = 2pi
so that is the absolute max on [0,2pi]
Note that f'(x) >= 0 for all x, so there is no max or min, which require changing direction.
To find the absolute maximum and minimum of the function f(x) = x + sin(x) on the closed interval [0, 2pi], you can follow these steps:
Step 1: Find the critical points of the function by taking its derivative and setting it equal to zero.
f'(x) = 1 + cos(x)
1 + cos(x) = 0
cos(x) = -1
x = π
Step 2: Evaluate the function at the critical points, as well as at the endpoints of the interval [0, 2pi].
f(0) = 0 + sin(0) = 0
f(π) = π + sin(π) = π
Step 3: Compare the values obtained in Step 2 to determine the absolute maximum and minimum of the function.
The function f(x) has a critical point at x = π. Evaluating the function at x = π, we get f(π) = π. This is also the function's value at x = 2π. Therefore, the absolute maximum value of the function is π.
At x = 0, the function has a value of f(0) = 0. Comparing this value with the absolute maximum value, we see that the function does not have an absolute minimum on the closed interval [0, 2π].
To find the absolute maximum and minimum of a function on a closed interval, we need to check the critical points and the endpoints of the interval.
Step 1: Find the critical points by taking the derivative of the function.
f(x) = x + sin(x)
f'(x) = 1 + cos(x)
To find the critical points, set the derivative equal to zero and solve:
1 + cos(x) = 0
cos(x) = -1
The solutions for x are x = π.
Step 2: Find the values of the function at the critical points and the endpoints of the interval.
f(0) = 0 + sin(0) = 0
f(2π) = 2π + sin(2π) = 2π + 0 = 2π
f(π) = π + sin(π) = π + 0 = π
Step 3: Compare the values obtained to find the absolute maximum and minimum.
The values of the function on the closed interval [0, 2π] are:
f(0) = 0
f(2π) = 2π
f(π) = π
The absolute maximum is 2π, and it occurs at x = 2π.
The absolute minimum is 0, and it occurs at x = 0.
Therefore, the absolute maximum and minimum for f(x) = x + sin(x) on the closed interval [0, 2π] are 2π and 0, respectively.