Find the domain of each of the following functions: `f(x)=(sqrt(cosx-1/2))/(sqrt(6+35x-6x^(2))`

Explanation

To find the domain of the function \( f(x) = \frac{\sqrt{\cos x - \frac{1}{2}}}{\sqrt{6 + 35x - 6x^2}} \), we need to ensure that both the numerator and the denominator are defined and non-negative.

### Step 1: Analyze the numerator
The numerator is \( \sqrt{\cos x - \frac{1}{2}} \). For this expression to be defined and non-negative, we need:
\[ \cos x - \frac{1}{2} \geq 0 \]
This simplifies to:
\[ \cos x \geq \frac{1}{2} \]

### Step 2: Determine the values of \( x \)
The cosine function is greater than or equal to \( \frac{1}{2} \) in the intervals:
\[ x \in [-\frac{\pi}{3} + 2n\pi, \frac{\pi}{3} + 2n\pi] \quad \text{for } n \in \mathbb{Z} \]
This means the values of \( x \) that satisfy this condition are periodic with a period of \( 2\pi \).

### Step 3: Analyze the denominator
The denominator is \( \sqrt{6 + 35x - 6x^2} \). For this expression to be defined and positive, we need:
\[ 6 + 35x - 6x^2 > 0 \]
This can be rearranged to:
\[ -6x^2 + 35x + 6 > 0 \]
Multiplying through by -1 (and reversing the inequality):
\[ 6x^2 - 35x - 6 < 0 \]

### Step 4: Factor the quadratic
To find the roots of the quadratic \( 6x^2 - 35x - 6 = 0 \), we can use the quadratic formula:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
where \( a = 6, b = -35, c = -6 \):
\[ x = \frac{35 \pm \sqrt{(-35)^2 - 4 \cdot 6 \cdot (-6)}}{2 \cdot 6} \]
Calculating the discriminant:
\[ \sqrt{1225 + 144} = \sqrt{1369} = 37 \]
Thus, the roots are:
\[ x = \frac{35 \pm 37}{12} \]
Calculating the two roots:
1. \( x_1 = \frac{72}{12} = 6 \)
2. \( x_2 = \frac{-2}{12} = -\frac{1}{6} \)

### Step 5: Determine the intervals
Now we need to find the intervals where \( 6x^2 - 35x - 6 < 0 \). The roots divide the number line into intervals:
1. \( (-\infty, -\frac{1}{6}) \)
2. \( (-\frac{1}{6}, 6) \)
3. \( (6, \infty) \)

Testing a point in each interval:
- For \( x = -1 \): \( 6(-1)^2 - 35(-1) - 6 = 6 + 35 - 6 = 35 > 0 \)
- For \( x = 0 \): \( 6(0)^2 - 35(0) - 6 = -6 < 0 \)
- For \( x = 7 \): \( 6(7)^2 - 35(7) - 6 = 294 - 245 - 6 = 43 > 0 \)

Thus, the quadratic is negative in the interval:
\[ (-\frac{1}{6}, 6) \]

### Step 6: Find the intersection of intervals
Now we need the intersection of the intervals from the numerator and denominator:
1. From the numerator: \( x \in [-\frac{\pi}{3} + 2n\pi, \frac{\pi}{3} + 2n\pi] \)
2. From the denominator: \( x \in (-\frac{1}{6}, 6) \)

### Final Domain
The final domain of \( f(x) \) is:
\[ x \in \left[-\frac{\pi}{3}, \frac{\pi}{3}\right] \cap (-\frac{1}{6}, 6) \]
This gives us:
\[ \text{Domain: } \left[-\frac{\pi}{3}, \frac{\pi}{3}\right] \]