The domain of the function `f(x)=log_2[log_3(log_4(x^2-3x+6)}]i s` .

Options

A. `(1, 2)`
B. `[1, 2 ] `
C. `(-oo, 1] uu (2, oo)`
D. `(-oo, 1]uu[2, oo)`

Explanation

To find the domain of the function \( f(x) = \log_2[\log_3(\log_4(x^2 - 3x + 6))] \), we need to ensure that all logarithmic expressions are defined and positive. We will proceed step by step.

### Step 1: Ensure the innermost logarithm is defined
The innermost function is \( \log_4(x^2 - 3x + 6) \). For this logarithm to be defined, the argument must be greater than 0:
\[ x^2 - 3x + 6 > 0 \]
To analyze this quadratic expression, we can find its discriminant:
\[ D = b^2 - 4ac = (-3)^2 - 4 \cdot 1 \cdot 6 = 9 - 24 = -15 \]
Since the discriminant is negative, the quadratic does not have real roots and is always positive (as the coefficient of \( x^2 \) is positive). Thus, \( x^2 - 3x + 6 > 0 \) for all \( x \).

### Step 2: Ensure the second logarithm is defined
Next, we need to ensure that \( \log_3(\log_4(x^2 - 3x + 6)) \) is defined and positive:
\[ \log_4(x^2 - 3x + 6) > 0 \]
This means:
\[ x^2 - 3x + 6 > 4 \]
Now, we solve the inequality:
\[ x^2 - 3x + 6 - 4 > 0 \implies x^2 - 3x + 2 > 0 \]
Factoring gives:
\[ (x - 1)(x - 2) > 0 \]
The roots of this quadratic are \( x = 1 \) and \( x = 2 \). To determine the intervals where the product is positive, we test the intervals:
- For \( x < 1 \): both factors are negative, so the product is positive.
- For \( 1 < x < 2 \): one factor is negative and the other is positive, so the product is negative.
- For \( x > 2 \): both factors are positive, so the product is positive.

Thus, the solution to \( (x - 1)(x - 2) > 0 \) is:
\[ x < 1 \quad \text{or} \quad x > 2 \]

### Step 3: Ensure the outermost logarithm is defined
Finally, we need to ensure that \( \log_2[\log_3(\log_4(x^2 - 3x + 6))] \) is defined. Since we have already established that \( \log_3(\log_4(x^2 - 3x + 6)) > 0 \) when \( x < 1 \) or \( x > 2 \), we do not need any additional conditions here.

### Conclusion
The domain of the function \( f(x) \) is:
\[ (-\infty, 1) \cup (2, \infty) \]

### Final Answer
The domain of the function \( f(x) \) is \( (-\infty, 1) \cup (2, \infty) \).