Let `f(x) = sin^(3)x - 3 sinx + 6, AA x` The `in (0, pi)`.number of local maximum/maxima of the function f(x) is

Options

A. 0
B. 1
C. 2
D. 3

Answer

To find the number of local maxima of the function \( f(x) = \sin^3 x - 3 \sin x + 6 \) in the interval \( (0, \pi) \), we will follow these steps:

### Step 1: Find the derivative of the function
We start by differentiating \( f(x) \):
\[ f'(x) = \frac{d}{dx}(\sin^3 x) - \frac{d}{dx}(3 \sin x) + \frac{d}{dx}(6) \]
Using the chain rule and the derivative of sine:
\[ f'(x) = 3 \sin^2 x \cos x - 3 \cos x \]
This simplifies to:
\[ f'(x) = 3 \cos x (\sin^2 x - 1) \]

### Step 2: Set the derivative equal to zero
To find critical points, we set \( f'(x) = 0 \):
\[ 3 \cos x (\sin^2 x - 1) = 0 \]
This gives us two cases to consider:
1. \( 3 \cos x = 0 \)
2. \( \sin^2 x - 1 = 0 \)

### Step 3: Solve for critical points
**Case 1:** \( 3 \cos x = 0 \)
\[ \cos x = 0 \implies x = \frac{\pi}{2} \]

**Case 2:** \( \sin^2 x - 1 = 0 \)
\[ \sin^2 x = 1 \implies \sin x = \pm 1 \]
In the interval \( (0, \pi) \), \( \sin x = 1 \) gives us:
\[ x = \frac{\pi}{2} \]
The solution \( \sin x = -1 \) does not provide any valid solutions in \( (0, \pi) \).

### Step 4: Determine the nature of the critical point
We have found a critical point at \( x = \frac{\pi}{2} \). To determine whether this point is a local maximum or minimum, we can use the first derivative test. We will check the sign of \( f'(x) \) around \( x = \frac{\pi}{2} \):

- For \( x < \frac{\pi}{2} \) (e.g., \( x = \frac{\pi}{4} \)):
\[ f'(\frac{\pi}{4}) = 3 \cos(\frac{\pi}{4})(\sin^2(\frac{\pi}{4}) - 1) = 3 \cdot \frac{\sqrt{2}}{2} \cdot (1 - 1) = 0 \quad (\text{not useful}) \]
Instead, we can check \( x = \frac{\pi}{3} \):
\[ f'(\frac{\pi}{3}) = 3 \cos(\frac{\pi}{3})(\sin^2(\frac{\pi}{3}) - 1) = 3 \cdot \frac{1}{2} \cdot \left(\left(\frac{\sqrt{3}}{2}\right)^2 - 1\right) < 0 \quad (\text{negative}) \]

- For \( x > \frac{\pi}{2} \) (e.g., \( x = \frac{2\pi}{3} \)):
\[ f'(\frac{2\pi}{3}) = 3 \cos(\frac{2\pi}{3})(\sin^2(\frac{2\pi}{3}) - 1) = 3 \cdot -\frac{1}{2} \cdot \left(\left(\frac{\sqrt{3}}{2}\right)^2 - 1\right) > 0 \quad (\text{positive}) \]

### Conclusion
Since \( f'(x) \) changes from negative to positive at \( x = \frac{\pi}{2} \), this indicates that \( x = \frac{\pi}{2} \) is a local minimum, not a maximum.

Thus, the number of local maxima of the function \( f(x) \) in the interval \( (0, \pi) \) is:
\[ \boxed{0} \]