To find the domain of the function \( f(x) = \sqrt{\frac{1 - 5^x}{7^{-x} - 7}} \), we need to ensure that the expression inside the square root is non-negative and that the denominator is not zero.
### Step 1: Set up the inequality
The expression inside the square root must be greater than or equal to zero:
\[
\frac{1 - 5^x}{7^{-x} - 7} \geq 0
\]
### Step 2: Identify when the numerator is zero
The numerator \( 1 - 5^x = 0 \) gives us:
\[
5^x = 1 \implies x = 0
\]
So, \( x = 0 \) is a critical point.
### Step 3: Identify when the denominator is zero
The denominator \( 7^{-x} - 7 = 0 \) gives us:
\[
7^{-x} = 7 \implies -x = 1 \implies x = -1
\]
So, \( x = -1 \) is another critical point.
### Step 4: Analyze the intervals
The critical points divide the number line into intervals:
1. \( (-\infty, -1) \)
2. \( (-1, 0) \)
3. \( (0, \infty) \)
We will test a point from each interval to determine the sign of the expression \( \frac{1 - 5^x}{7^{-x} - 7} \).
- **Interval 1: \( x < -1 \) (e.g., \( x = -2 \))**
\[
1 - 5^{-2} = 1 - \frac{1}{25} = \frac{24}{25} > 0
\]
\[
7^{-(-2)} - 7 = 7^2 - 7 = 49 - 7 = 42 > 0
\]
Thus, \( \frac{1 - 5^{-2}}{7^{2} - 7} > 0 \).
- **Interval 2: \( -1 < x < 0 \) (e.g., \( x = -0.5 \))**
\[
1 - 5^{-0.5} = 1 - \sqrt{5} < 0
\]
\[
7^{-(-0.5)} - 7 = 7^{0.5} - 7 < 0
\]
Thus, \( \frac{1 - 5^{-0.5}}{7^{0.5} - 7} > 0 \).
- **Interval 3: \( x > 0 \) (e.g., \( x = 1 \))**
\[
1 - 5^{1} = 1 - 5 < 0
\]
\[
7^{-1} - 7 = \frac{1}{7} - 7 < 0
\]
Thus, \( \frac{1 - 5^{1}}{7^{-1} - 7} > 0 \).
### Step 5: Combine the results
From the analysis:
- The expression is non-negative in the intervals \( (-\infty, -1) \) and \( (0, \infty) \).
- At \( x = 0 \), the expression equals zero.
- At \( x = -1 \), the expression is undefined.
### Final Domain
Thus, the domain of the function is:
\[
\text{Domain} = (-\infty, -1) \cup [0, \infty)
\]
