Find the maximum and minimum values of `f(x) = (-x + 2sin x) " on " [0, 2pi]`

Answer

To find the maximum and minimum values of the function \( f(x) = -x + 2\sin x \) on the interval \([0, 2\pi]\), we will follow these steps:

### Step 1: Find the derivative of the function
We first differentiate the function to find critical points.

\[ f'(x) = \frac{d}{dx}(-x + 2\sin x) = -1 + 2\cos x \]

### Step 2: Set the derivative equal to zero
To find the critical points, we set the derivative equal to zero:

\[ -1 + 2\cos x = 0 \]

Solving for \(\cos x\):

\[ 2\cos x = 1 \implies \cos x = \frac{1}{2} \]

### Step 3: Find the values of \(x\) that satisfy \(\cos x = \frac{1}{2}\)
The values of \(x\) in the interval \([0, 2\pi]\) where \(\cos x = \frac{1}{2}\) are:

\[ x = \frac{\pi}{3}, \quad x = \frac{5\pi}{3} \]

### Step 4: Evaluate the function at critical points and endpoints
Next, we evaluate \(f(x)\) at the critical points and at the endpoints of the interval \([0, 2\pi]\).

1. **At \(x = 0\)**:
\[ f(0) = -0 + 2\sin(0) = 0 \]

2. **At \(x = \frac{\pi}{3}\)**:
\[ f\left(\frac{\pi}{3}\right) = -\frac{\pi}{3} + 2\sin\left(\frac{\pi}{3}\right) = -\frac{\pi}{3} + 2 \cdot \frac{\sqrt{3}}{2} = -\frac{\pi}{3} + \sqrt{3} \]

3. **At \(x = \frac{5\pi}{3}\)**:
\[ f\left(\frac{5\pi}{3}\right) = -\frac{5\pi}{3} + 2\sin\left(\frac{5\pi}{3}\right) = -\frac{5\pi}{3} + 2 \cdot \left(-\frac{\sqrt{3}}{2}\right) = -\frac{5\pi}{3} - \sqrt{3} \]

4. **At \(x = 2\pi\)**:
\[ f(2\pi) = -2\pi + 2\sin(2\pi) = -2\pi + 0 = -2\pi \]

### Step 5: Compare the values
Now we compare the values obtained:

- \(f(0) = 0\)
- \(f\left(\frac{\pi}{3}\right) = -\frac{\pi}{3} + \sqrt{3}\)
- \(f\left(\frac{5\pi}{3}\right) = -\frac{5\pi}{3} - \sqrt{3}\)
- \(f(2\pi) = -2\pi\)

### Step 6: Determine the maximum and minimum
To find the maximum and minimum values, we need to evaluate the numerical values of \(f\left(\frac{\pi}{3}\right)\) and \(f\left(\frac{5\pi}{3}\right)\) and compare them with the others.

- \(f(0) = 0\)
- \(f\left(\frac{\pi}{3}\right) \approx -1.047 + 1.732 \approx 0.685\)
- \(f\left(\frac{5\pi}{3}\right) \approx -5.236 - 1.732 \approx -6.968\)
- \(f(2\pi) \approx -6.283\)

### Conclusion
Thus, the maximum value of \(f(x)\) on the interval \([0, 2\pi]\) is:

\[ \text{Maximum} = f\left(\frac{\pi}{3}\right) \approx 0.685 \]

And the minimum value is:

\[ \text{Minimum} = f\left(\frac{5\pi}{3}\right) \approx -6.968 \]