The domain of definition `f(x)=sqrt(log_(0.4) ((x-1)/(x+5)))xx1/(x^2-36)` is

Options

A. `(-oo,0)-{-6}`
B. `(0,oo)-{1,6}`
C. `(1,oo)-{6}`
D. `[1,oo)-{6}`

Answer

To find the domain of the function \( f(x) = \sqrt{\log_{0.4} \left( \frac{x-1}{x+5} \right)} \cdot \frac{1}{x^2 - 36} \), we need to ensure that all parts of the function are defined and valid.

### Step 1: Conditions for the logarithm
The logarithm \( \log_{0.4} \left( \frac{x-1}{x+5} \right) \) must be defined and non-negative. This means:
1. \( \frac{x-1}{x+5} > 0 \)
2. \( \log_{0.4} \left( \frac{x-1}{x+5} \right) \geq 0 \)

Since the base \( 0.4 \) is less than 1, the inequality for the logarithm reverses:
\[ \frac{x-1}{x+5} \leq 1 \]

### Step 2: Solve the first inequality \( \frac{x-1}{x+5} > 0 \)
This inequality holds when both the numerator and denominator are either both positive or both negative.

1. **Numerator \( x - 1 > 0 \)**:
- \( x > 1 \)

2. **Denominator \( x + 5 > 0 \)**:
- \( x > -5 \)

Thus, the first inequality gives us:
- \( x > 1 \) (since \( x > -5 \) is satisfied for \( x > 1 \))

### Step 3: Solve the second inequality \( \frac{x-1}{x+5} \leq 1 \)
Rearranging gives:
\[ \frac{x-1}{x+5} - 1 \leq 0 \]
\[ \frac{x-1 - (x+5)}{x+5} \leq 0 \]
\[ \frac{-6}{x+5} \leq 0 \]

This inequality holds when \( x + 5 > 0 \):
- \( x > -5 \)

### Step 4: Combine the results
From the first inequality, we have \( x > 1 \). From the second inequality, we have \( x > -5 \). The more restrictive condition is \( x > 1 \).

### Step 5: Conditions for the rational function
The term \( \frac{1}{x^2 - 36} \) must also be defined, which means:
\[ x^2 - 36 \neq 0 \]
This gives us:
\[ x^2 \neq 36 \implies x \neq 6 \text{ and } x \neq -6 \]

### Step 6: Final domain
Since \( x > 1 \), we only need to exclude \( x = 6 \) from our domain. Therefore, the domain of \( f(x) \) is:
\[ \text{Domain: } (1, 6) \cup (6, \infty) \]

### Summary of the domain
The domain of the function \( f(x) \) is \( (1, 6) \cup (6, \infty) \).

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