If the domain of the function `f(x) = sqrt(3 cos^(-1) (4x) - pi)` is `[a, b]`, then the value of `(4a + 64b)` is ___

Explanation

To find the domain of the function \( f(x) = \sqrt{3 \cos^{-1}(4x) - \pi} \) and subsequently calculate \( 4a + 64b \) where the domain is \([a, b]\), we will follow these steps:

### Step 1: Determine the condition for the square root to be defined
The expression inside the square root must be non-negative:
\[ 3 \cos^{-1}(4x) - \pi \geq 0 \]
This simplifies to:
\[ 3 \cos^{-1}(4x) \geq \pi \]
Dividing both sides by 3, we get:
\[ \cos^{-1}(4x) \geq \frac{\pi}{3} \]

### Step 2: Solve the inequality involving the inverse cosine
The inverse cosine function gives values in the range \([0, \pi]\). The inequality \(\cos^{-1}(4x) \geq \frac{\pi}{3}\) implies:
\[ 4x \leq \cos\left(\frac{\pi}{3}\right) \]
Since \(\cos\left(\frac{\pi}{3}\right) = \frac{1}{2}\), we have:
\[ 4x \leq \frac{1}{2} \]
Dividing by 4:
\[ x \leq \frac{1}{8} \]

### Step 3: Determine the range of \(4x\) for the inverse cosine function
The argument of the \(\cos^{-1}\) function, \(4x\), must lie within the domain of \(\cos^{-1}\), which is \([-1, 1]\):
\[ -1 \leq 4x \leq 1 \]
Dividing the entire inequality by 4 gives:
\[ -\frac{1}{4} \leq x \leq \frac{1}{4} \]

### Step 4: Combine the inequalities to find the domain
From Step 2, we have \(x \leq \frac{1}{8}\) and from Step 3, we have \(-\frac{1}{4} \leq x \leq \frac{1}{4}\). The domain of \(f(x)\) is the intersection of these two conditions:
\[ -\frac{1}{4} \leq x \leq \frac{1}{8} \]
Thus, the domain is \([a, b] = \left[-\frac{1}{4}, \frac{1}{8}\right]\).

### Step 5: Calculate \(4a + 64b\)
Here, \(a = -\frac{1}{4}\) and \(b = \frac{1}{8}\):
\[ 4a = 4 \left(-\frac{1}{4}\right) = -1 \]
\[ 64b = 64 \left(\frac{1}{8}\right) = 8 \]
Now, we can calculate:
\[ 4a + 64b = -1 + 8 = 7 \]

### Final Answer
Thus, the value of \(4a + 64b\) is \( \boxed{7} \).