The function `f(x) =4sin^(3)x-6sin^(2)x +12 sinx + 100` is strictly

Options

A. increasing in `pi, (3pi)/2`
B. decreasing in `(pi/2,pi)`
C. decreasing in `[-pi/2, pi/2]`
D. decreasing in `[0,pi/2]`

Explanation

To determine the behavior of the function \( f(x) = 4\sin^3 x - 6\sin^2 x + 12\sin x + 100 \), we need to analyze its derivative \( f'(x) \) to find the intervals where the function is increasing or decreasing.

### Step 1: Differentiate the function
We start by differentiating \( f(x) \):
\[ f'(x) = \frac{d}{dx}(4\sin^3 x) - \frac{d}{dx}(6\sin^2 x) + \frac{d}{dx}(12\sin x) + \frac{d}{dx}(100) \]

Using the chain rule and the power rule:
\[ f'(x) = 4 \cdot 3\sin^2 x \cdot \cos x - 6 \cdot 2\sin x \cdot \cos x + 12\cos x + 0 \]
\[ = 12\sin^2 x \cos x - 12\sin x \cos x + 12\cos x \]
\[ = 12\cos x (\sin^2 x - \sin x + 1) \]

### Step 2: Analyze the quadratic expression
Next, we analyze the quadratic expression \( \sin^2 x - \sin x + 1 \):
Let \( t = \sin x \). Then we have:
\[ t^2 - t + 1 \]

To determine if this quadratic is always positive, we calculate its discriminant:
\[ D = (-1)^2 - 4 \cdot 1 \cdot 1 = 1 - 4 = -3 \]
Since the discriminant is negative, the quadratic \( t^2 - t + 1 \) has no real roots and opens upwards (the coefficient of \( t^2 \) is positive). Therefore, it is always positive for all \( t \).

### Step 3: Determine the sign of \( f'(x) \)
Now we can conclude that:
\[ f'(x) = 12\cos x (\sin^2 x - \sin x + 1) \]
Since \( \sin^2 x - \sin x + 1 > 0 \), the sign of \( f'(x) \) depends solely on \( \cos x \).

### Step 4: Analyze the intervals based on \( \cos x \)
- **Increasing Intervals**: \( f'(x) > 0 \) when \( \cos x > 0 \). This occurs in the intervals:
- \( (-\frac{\pi}{2}, \frac{\pi}{2}) \)
- \( (3\frac{\pi}{2}, 2\pi) \)

- **Decreasing Intervals**: \( f'(x) < 0 \) when \( \cos x < 0 \). This occurs in the intervals:
- \( (\frac{\pi}{2}, \frac{3\pi}{2}) \)

### Conclusion
From the analysis, we can conclude that:
- The function \( f(x) \) is **increasing** in the intervals \( (-\frac{\pi}{2}, \frac{\pi}{2}) \) and \( (3\frac{\pi}{2}, 2\pi) \).
- The function \( f(x) \) is **decreasing** in the interval \( (\frac{\pi}{2}, \frac{3\pi}{2}) \).