The interval in which `f(x)=3cos^(4)x+10cos^(3)x+6cos^(2)x-3` increases or decreases in `(0,pi)`

Options

A. decreases on `((pi)/(2),(2pi)/(3))` and increases on `(0,(pi)/(2))cup((2pi)/(3),pi)`
B. decreases on `((pi)/(2),pi)`and increases on `(0,(pi)/(2))`
C. decreases on `(0,(pi)/2)cup((2pi)/(3),pi)` and increases on `((pi)/(2),(2pi)/(3))`
D. decreases on `(0,(pi)/(2))` and increases on `((pi)/(2),pi)`

Answer

To determine the intervals in which the function \( f(x) = 3\cos^4 x + 10\cos^3 x + 6\cos^2 x - 3 \) is increasing or decreasing on the interval \( (0, \pi) \), we will follow these steps:

### Step 1: Find the derivative of \( f(x) \)

To analyze the monotonicity of the function, we first need to compute its derivative \( f'(x) \).

\[ f'(x) = \frac{d}{dx}(3\cos^4 x + 10\cos^3 x + 6\cos^2 x - 3) \]

Using the chain rule and the power rule, we get:

\[ f'(x) = 3 \cdot 4\cos^3 x \cdot (-\sin x) + 10 \cdot 3\cos^2 x \cdot (-\sin x) + 6 \cdot 2\cos x \cdot (-\sin x) \]

This simplifies to:

\[ f'(x) = -\sin x (12\cos^3 x + 30\cos^2 x + 12\cos x) \]

### Step 2: Factor the derivative

We can factor out \( -\sin x \):

\[ f'(x) = -\sin x (12\cos^3 x + 30\cos^2 x + 12\cos x) \]

### Step 3: Analyze the sign of \( f'(x) \)

The function \( f(x) \) is increasing where \( f'(x) > 0 \) and decreasing where \( f'(x) < 0 \).

1. **For \( -\sin x > 0 \)**:
- This occurs when \( \sin x < 0 \), which is not the case in the interval \( (0, \pi) \) since \( \sin x \) is positive in this interval.

2. **For \( 12\cos^3 x + 30\cos^2 x + 12\cos x > 0 \)**:
- We can analyze the polynomial \( 12\cos^3 x + 30\cos^2 x + 12\cos x \).

### Step 4: Find the roots of the polynomial

Let \( y = \cos x \). We need to solve:

\[ 12y^3 + 30y^2 + 12y = 0 \]

Factoring out \( 6y \):

\[ 6y(2y^2 + 5y + 2) = 0 \]

This gives us one root \( y = 0 \) (i.e., \( \cos x = 0 \) at \( x = \frac{\pi}{2} \)).

Now we solve the quadratic \( 2y^2 + 5y + 2 = 0 \) using the quadratic formula:

\[ y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{-5 \pm \sqrt{25 - 16}}{4} = \frac{-5 \pm 3}{4} \]

This gives us:

\[ y = -\frac{1}{2} \quad \text{and} \quad y = -2 \quad (\text{not valid since } y = \cos x \text{ must be in } [-1, 1]) \]

### Step 5: Determine intervals of increase and decrease

Now we analyze the sign of \( f'(x) \):

- **For \( 0 < x < \frac{\pi}{2} \)**:
- \( \cos x > 0 \) and \( f'(x) < 0 \) (since \( 12\cos^3 x + 30\cos^2 x + 12\cos x > 0 \)).

- **At \( x = \frac{\pi}{2} \)**:
- \( f'(x) = 0 \).

- **For \( \frac{\pi}{2} < x < \frac{2\pi}{3} \)**:
- \( \cos x < 0 \) and \( f'(x) > 0 \).

- **For \( \frac{2\pi}{3} < x < \pi \)**:
- \( \cos x < -\frac{1}{2} \) and \( f'(x) < 0 \).

### Conclusion

Thus, the function \( f(x) \) is:
- **Decreasing** on the intervals \( (0, \frac{\pi}{2}) \) and \( (\frac{2\pi}{3}, \pi) \).
- **Increasing** on the interval \( (\frac{\pi}{2}, \frac{2\pi}{3}) \).

### Final Answer

- **Increasing** on \( \left(\frac{\pi}{2}, \frac{2\pi}{3}\right) \)
- **Decreasing** on \( \left(0, \frac{\pi}{2}\right) \) and \( \left(\frac{2\pi}{3}, \pi\right) \)