Find the relative extrema of the function f(x) = 1/2 x - sinx on the interval (0, 2 pie)

please show me how you got this!!

To find the relative extrema of the given function f(x) = 1/2 x - sin(x) on the interval (0, 2π), we need to find the critical points by taking the derivative of the function and then determine whether those points are relative maximum or minimum points.

1. Find the derivative of the function:
f'(x) = d/dx (1/2 x - sin(x))
= 1/2 - cos(x)

2. Set the derivative equal to zero and solve for x to find the critical points:
1/2 - cos(x) = 0

Rearrange the equation:
cos(x) = 1/2

Using the unit circle, we know that the angles whose cosines are equal to 1/2 are π/3 and 5π/3:

x = π/3 or x = 5π/3

3. Check the nature of critical points by plugging the critical points into the second derivative:

f''(x) = d^2/dx^2 (1/2 - cos(x))
= sin(x)

At x = π/3:
f''(π/3) = sin(π/3) = √3/2 > 0

At x = 5π/3:
f''(5π/3) = sin(5π/3) = -√3/2 < 0

4. Determine the nature of each critical point:
- At x = π/3, f'(π/3) = 1/2 - cos(π/3) = 1/2 - 1/2 = 0
Since f'(π/3) changes sign from negative to positive, we have a relative minimum at x = π/3.

- At x = 5π/3, f'(5π/3) = 1/2 - cos(5π/3) = 1/2 + 1/2 = 1
Since f'(5π/3) changes sign from positive to negative, we have a relative maximum at x = 5π/3.

5. Check the values of f(x) at the critical points and endpoints to identify the global extrema:
- At x = 0, f(0) = 1/2(0) - sin(0) = 0
- At x = 2π, f(2π) = 1/2(2π) - sin(2π) = π

Comparing the values, we see that the function takes its minimum value at x = 0 and its maximum value at x = 2π.

Therefore, the relative extrema on the interval (0, 2π) are:
- Relative minimum at x = π/3
- Relative maximum at x = 5π/3

To find the relative extrema of the function f(x) = 1/2x - sin(x) on the interval (0, 2π), we need to follow these steps:

1. Find the derivative of the function: f'(x).
2. Set f'(x) equal to zero and solve for x. These values will give us potential locations of relative extrema.
3. Determine the second derivative, f''(x), to classify the relative extrema as maximum or minimum.
4. Plug the critical points (obtained from step 2) into f(x) to find their corresponding y-values, f(critical-point).

Let's go through these steps one by one:

Step 1: Find the derivative f'(x):
f(x) = 1/2x - sin(x)
To find the derivative, we differentiate each term of the function. The derivative of 1/2x is 1/2, and the derivative of sin(x) is cos(x).
Therefore, f'(x) = 1/2 - cos(x).

Step 2: Set f'(x) = 0 and solve for x:
1/2 - cos(x) = 0
cos(x) = 1/2

On the interval (0, 2π), the solutions to this equation are x = π/3 and x = 5π/3.

Step 3: Determine the second derivative f''(x):
To classify the relative extrema as a maximum or minimum, we need to find the second derivative. Differentiating f'(x) = 1/2 - cos(x), we get
f''(x) = sin(x).

Step 4: Find the y-values (f(critical-point)):
Plug the critical points we found in Step 2 (x = π/3 and x = 5π/3) into the original function f(x) = 1/2x - sin(x).

For x = π/3:
f(π/3) = (1/2)(π/3) - sin(π/3)

For x = 5π/3:
f(5π/3) = (1/2)(5π/3) - sin(5π/3)

Calculating these values will give us the y-coordinates of the relative extrema.

By following these steps, you can find the relative extrema of the given function f(x) = 1/2x - sin(x) on the interval (0, 2π).

same way you do it for anything else. Look for f'(x) = 0

f = x/2 - sinx
f' = 1/2 - cosx
f'=0 when cosx = 1/2, or x=pi/3, 5pi/3