The critical points of the function `f (x) = 2 sin x, x in [0,2pi]` are :

Options

A. `0, (pi)/(2), (3pi)/(2)`
B. `0,pi , 2pi`
C. `(pi)/(2), (3pi)/(2)`
D. `pi,2pi`

Answer

To find the critical points of the function \( f(x) = 2 \sin x \) over the interval \( [0, 2\pi] \), we will follow these steps:

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

**Hint:** Remember that the derivative of \( \sin x \) is \( \cos x \).

### Step 2: Set the derivative equal to zero
To find the critical points, we set the derivative equal to zero:
\[ f'(x) = 2 \cos x = 0 \]
This simplifies to:
\[ \cos x = 0 \]

**Hint:** Setting the derivative to zero helps us find points where the function may have local maxima, minima, or points of inflection.

### Step 3: Solve for \( x \)
Next, we need to find the values of \( x \) where \( \cos x = 0 \) within the interval \( [0, 2\pi] \). The cosine function is zero at:
\[ x = \frac{\pi}{2}, \quad x = \frac{3\pi}{2} \]

**Hint:** Recall the unit circle and the angles where the cosine value is zero.

### Step 4: List the critical points
The critical points of the function \( f(x) = 2 \sin x \) in the interval \( [0, 2\pi] \) are:
\[ x = \frac{\pi}{2}, \quad x = \frac{3\pi}{2} \]

**Hint:** Make sure to check the specified interval to ensure all critical points are included.

### Conclusion
Thus, the critical points of the function \( f(x) = 2 \sin x \) in the interval \( [0, 2\pi] \) are \( \frac{\pi}{2} \) and \( \frac{3\pi}{2} \).

**Final Answer:** The critical points are \( \frac{\pi}{2}, \frac{3\pi}{2} \).