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

Options

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

Answer

To determine the intervals where the function \( f(x) = 4 \sin^3 x - 6 \sin^2 x + 12 \sin x + 100 \) is strictly increasing or decreasing, we will follow these steps:

### Step 1: Find the derivative of the function
To analyze the behavior of the function, we first need to find its derivative \( f'(x) \).

\[ f'(x) = \frac{d}{dx}(4 \sin^3 x - 6 \sin^2 x + 12 \sin x + 100) \]

Using the chain rule and the power rule, we differentiate each term:

1. The derivative of \( 4 \sin^3 x \) is \( 12 \sin^2 x \cos x \) (using \( \frac{d}{dx}(\sin^n x) = n \sin^{n-1} x \cos x \)).
2. The derivative of \( -6 \sin^2 x \) is \( -12 \sin x \cos x \).
3. The derivative of \( 12 \sin x \) is \( 12 \cos x \).
4. The derivative of the constant \( 100 \) is \( 0 \).

Combining these, we have:

\[ f'(x) = 12 \sin^2 x \cos x - 12 \sin x \cos x + 12 \cos x \]

Factoring out \( 12 \cos x \):

\[ f'(x) = 12 \cos x (\sin^2 x - \sin x + 1) \]

### Step 2: Analyze the derivative
Next, we need to determine where \( f'(x) > 0 \) (increasing) and \( f'(x) < 0 \) (decreasing).

1. The term \( \sin^2 x - \sin x + 1 \) is always positive because its discriminant \( (-1)^2 - 4 \cdot 1 \cdot 1 = 1 - 4 = -3 \) is negative, indicating no real roots.
2. Therefore, the sign of \( f'(x) \) depends solely on \( \cos x \).

### Step 3: Determine intervals based on \( \cos x \)
- \( f'(x) > 0 \) when \( \cos x > 0 \):
- This occurs in the intervals \( (-\frac{\pi}{2}, \frac{\pi}{2}) \) (1st quadrant) and \( (3\frac{\pi}{2}, 2\pi) \) (4th quadrant).

- \( f'(x) < 0 \) when \( \cos x < 0 \):
- This occurs in the intervals \( (\frac{\pi}{2}, \frac{3\pi}{2}) \) (2nd and 3rd quadrants).

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