To determine the absolute maximum and minimum values of the function \( f(x) = x + \cos(x) \) on the closed interval \([0, 2\pi]\), we can apply the Extreme Value Theorem, which states that a continuous function on a closed interval will attain both a maximum and a minimum.

First, we need to find the critical points of the function by calculating its derivative. The derivative of \( f(x) \) is given by:

\[f'(x) = 1 - \sin(x)\]

Next, we set the derivative equal to zero to find the critical points:

\[1 - \sin(x) = 0 \implies \sin(x) = 1\]

Within the interval \([0, 2\pi]\), the sine function equals 1 at:

\[x = \frac{\pi}{2}\]

Now, we will evaluate the function \( f(x) \) at the critical point and the endpoints of the interval, which are \( x = 0 \) and \( x = 2\pi \).

Calculating \( f \) at these points:

1. For \( x = 0 \):\[f(0) = 0 + \cos(0) = 0 + 1 = 1\]2. For \( x = \frac{\pi}{2} \):\[f\left(\frac{\pi}{2}\right) = \frac{\pi}{2} + \cos\left(\frac{\pi}{2}\right) = \frac{\pi}{2} + 0 = \frac{\pi}{2} \approx 1.57\]3. For \( x = 2\pi \):\[f(2\pi) = 2\pi + \cos(2\pi) = 2\pi + 1 \approx 6.28 + 1 = 7.28\]

Now we compare the function values:

- \( f(0) = 1 \)- \( f\left(\frac{\pi}{2}\right) \approx 1.57 \)- \( f(2\pi) \approx 7.28 \)

From these calculations, we can identify the absolute maximum and minimum values:

- The absolute maximum value is \( f(2\pi) = 2\pi + 1 \) at \( x = 2\pi \).- The absolute minimum value is \( f(0) = 1 \) at \( x = 0 \).

In conclusion, the function \( f(x) = x + \cos(x) \) has an absolute maximum of \( 2\pi + 1 \) at \( x = 2\pi \) and an absolute minimum of \( 1 \) at \( x = 0 \).