The function `f(x) = 3 cos^(4)x + 10 cos^(3) x + 6 cos^(2)x - 3, (0 le x le pi)` is -

Options

A. Increasing in `((pi)/(2), (2pi)/(3))`
B. Increasing in `(0, (pi)/(2)) uu ((2pi)/(3), pi)`
C. Decreasing in `((pi)/(2), (2pi)/(3))`
D. All of the above

Answer

To determine whether the function \( f(x) = 3 \cos^4 x + 10 \cos^3 x + 6 \cos^2 x - 3 \) is monotonically increasing or decreasing in the interval \( [0, \pi] \), we need to follow these steps:

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

To analyze the monotonicity of the function, we first need to find its derivative.

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

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

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

This simplifies to:

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

Factoring out \(-6 \sin x\):

\[ f'(x) = -6 \sin x (2 \cos^3 x + 5 \cos^2 x + 2 \cos x) \]

### Step 2: Set the derivative equal to zero

To find the critical points, we set \( f'(x) = 0 \):

\[ -6 \sin x (2 \cos^3 x + 5 \cos^2 x + 2 \cos x) = 0 \]

This gives us two cases to consider:

1. \( \sin x = 0 \)
2. \( 2 \cos^3 x + 5 \cos^2 x + 2 \cos x = 0 \)

### Step 3: Solve \( \sin x = 0 \)

In the interval \( [0, \pi] \):

\[ \sin x = 0 \implies x = 0, \pi \]

### Step 4: Solve \( 2 \cos^3 x + 5 \cos^2 x + 2 \cos x = 0 \)

Factoring out \( \cos x \):

\[ \cos x (2 \cos^2 x + 5 \cos x + 2) = 0 \]

This gives us:

1. \( \cos x = 0 \implies x = \frac{\pi}{2} \)
2. For \( 2 \cos^2 x + 5 \cos x + 2 = 0 \), we can use the quadratic formula:

Let \( u = \cos x \):

\[ 2u^2 + 5u + 2 = 0 \]

Using the quadratic formula \( u = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):

\[ u = \frac{-5 \pm \sqrt{25 - 16}}{4} = \frac{-5 \pm 3}{4} \]

This gives us:

\[ u = -\frac{1}{2} \quad \text{or} \quad u = -2 \quad (\text{not possible since } -2 < -1) \]

Thus, we have \( \cos x = -\frac{1}{2} \):

\[ x = \frac{2\pi}{3} \]

### Step 5: Analyze the intervals

The critical points are \( x = 0, \frac{\pi}{2}, \frac{2\pi}{3}, \pi \). We will test the sign of \( f'(x) \) in the intervals:

1. \( (0, \frac{\pi}{2}) \)
2. \( (\frac{\pi}{2}, \frac{2\pi}{3}) \)
3. \( (\frac{2\pi}{3}, \pi) \)

**Interval Testing:**

- For \( x \in (0, \frac{\pi}{2}) \): Choose \( x = \frac{\pi}{4} \)
- \( f'(\frac{\pi}{4}) > 0 \) (function is increasing)

- For \( x \in (\frac{\pi}{2}, \frac{2\pi}{3}) \): Choose \( x = \frac{3\pi}{8} \)
- \( f'(\frac{3\pi}{8}) < 0 \) (function is decreasing)

- For \( x \in (\frac{2\pi}{3}, \pi) \): Choose \( x = \frac{5\pi}{6} \)
- \( f'(\frac{5\pi}{6}) < 0 \) (function is decreasing)

### Conclusion

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