If `f(x) = sqrt((1)/(tan^(-1)(x^(2)-4x + 3)))`, then f(x) is continuous for

Options

A. (1, 3)
B. `(-oo, 0)`
C. `(-oo, 1) uu (3, oo)`
D. None of these

Answer

To determine the continuity of the function \( f(x) = \sqrt{\frac{1}{\tan^{-1}(x^2 - 4x + 3)}} \), we need to analyze the conditions under which this function is defined and continuous.

### Step 1: Identify the denominator
The function \( f(x) \) has a denominator given by \( \tan^{-1}(x^2 - 4x + 3) \). For \( f(x) \) to be continuous, this denominator must not be equal to zero.

### Step 2: Set the denominator to zero
We set the denominator equal to zero to find the points where \( f(x) \) is undefined:
\[ \tan^{-1}(x^2 - 4x + 3) = 0 \]

### Step 3: Solve for \( x \)
Taking the tangent of both sides, we have:
\[ x^2 - 4x + 3 = \tan(0) = 0 \]
This simplifies to:
\[ x^2 - 4x + 3 = 0 \]
Factoring the quadratic, we get:
\[ (x - 1)(x - 3) = 0 \]
Thus, the solutions are:
\[ x = 1 \quad \text{and} \quad x = 3 \]

### Step 4: Determine the intervals of continuity
The function \( f(x) \) will be continuous for all \( x \) except where the denominator is zero. Therefore, we exclude \( x = 1 \) and \( x = 3 \) from the domain.

### Step 5: Check for positivity under the square root
Next, we need to ensure that the expression under the square root is positive. For \( f(x) \) to be defined, we require:
\[ \tan^{-1}(x^2 - 4x + 3) > 0 \]
This implies:
\[ x^2 - 4x + 3 > 0 \]

### Step 6: Solve the inequality
We already factored this expression:
\[ (x - 1)(x - 3) > 0 \]
To find the intervals where this product is positive, we test the intervals determined by the roots \( x = 1 \) and \( x = 3 \):
1. For \( x < 1 \): Choose \( x = 0 \) → \( (0 - 1)(0 - 3) = 3 > 0 \)
2. For \( 1 < x < 3 \): Choose \( x = 2 \) → \( (2 - 1)(2 - 3) = -1 < 0 \)
3. For \( x > 3 \): Choose \( x = 4 \) → \( (4 - 1)(4 - 3) = 3 > 0 \)

Thus, the solution to the inequality is:
\[ x \in (-\infty, 1) \cup (3, \infty) \]

### Conclusion
The function \( f(x) \) is continuous for:
\[ x \in (-\infty, 1) \cup (3, \infty) \]