Find the points of local maxima and local minima, if any, of the following functions. Find also the local maximum and local minimum values :
`f(x)=-x+2sinx,0lexle2pi`

Answer

To find the points of local maxima and minima for the function \( f(x) = -x + 2\sin x \) on the interval \( [0, 2\pi] \), we will follow these steps:

### Step 1: Find the first derivative of the function
We start by differentiating \( f(x) \):
\[ f'(x) = \frac{d}{dx}(-x + 2\sin x) = -1 + 2\cos x \]

### Step 2: Set the first derivative to zero to find critical points
To find the critical points, we set the first derivative equal to zero:
\[ -1 + 2\cos x = 0 \]
\[ 2\cos x = 1 \]
\[ \cos x = \frac{1}{2} \]

### Step 3: Solve for \( x \)
The solutions to \( \cos x = \frac{1}{2} \) in the interval \( [0, 2\pi] \) are:
\[ x = \frac{\pi}{3}, \quad x = \frac{5\pi}{3} \]

### Step 4: Find the second derivative
Next, we find the second derivative to determine the nature of the critical points:
\[ f''(x) = \frac{d}{dx}(-1 + 2\cos x) = -2\sin x \]

### Step 5: Evaluate the second derivative at the critical points
1. For \( x = \frac{\pi}{3} \):
\[ f''\left(\frac{\pi}{3}\right) = -2\sin\left(\frac{\pi}{3}\right) = -2 \cdot \frac{\sqrt{3}}{2} = -\sqrt{3} < 0 \]
This indicates a local maximum at \( x = \frac{\pi}{3} \).

2. For \( x = \frac{5\pi}{3} \):
\[ f''\left(\frac{5\pi}{3}\right) = -2\sin\left(\frac{5\pi}{3}\right) = -2 \cdot \left(-\frac{\sqrt{3}}{2}\right) = \sqrt{3} > 0 \]
This indicates a local minimum at \( x = \frac{5\pi}{3} \).

### Step 6: Calculate the function values at the critical points
1. For the local maximum at \( x = \frac{\pi}{3} \):
\[ f\left(\frac{\pi}{3}\right) = -\frac{\pi}{3} + 2\sin\left(\frac{\pi}{3}\right) = -\frac{\pi}{3} + 2 \cdot \frac{\sqrt{3}}{2} = -\frac{\pi}{3} + \sqrt{3} \]

2. For the local minimum at \( x = \frac{5\pi}{3} \):
\[ f\left(\frac{5\pi}{3}\right) = -\frac{5\pi}{3} + 2\sin\left(\frac{5\pi}{3}\right) = -\frac{5\pi}{3} + 2 \cdot \left(-\frac{\sqrt{3}}{2}\right) = -\frac{5\pi}{3} - \sqrt{3} \]

### Summary of Results
- Local maximum at \( x = \frac{\pi}{3} \) with value \( f\left(\frac{\pi}{3}\right) = -\frac{\pi}{3} + \sqrt{3} \).
- Local minimum at \( x = \frac{5\pi}{3} \) with value \( f\left(\frac{5\pi}{3}\right) = -\frac{5\pi}{3} - \sqrt{3} \).