https://file.aiquickdraw.com/mathbot/file/1726063605458ve7y60bx.jpg

Let's analyze the image and determine the domain and range of the given functions one by one.

### 1. \( f(x) = \frac{a + x}{a - x} \), where \( a = 1 \)
   The function becomes:
   \[
   f(x) = \frac{1 + x}{1 - x}
   \]
   - **Domain**: The function is undefined when the denominator is 0. Therefore, we need to find when \( 1 - x = 0 \), which gives \( x = 1 \). So, the domain is all real numbers except \( x = 1 \):
     \[
     \text{Domain: } (-\infty, 1) \cup (1, \infty)
     \]
   - **Range**: To find the range, we solve for \( y = \frac{1 + x}{1 - x} \):
     \[
     y(1 - x) = 1 + x \implies y - yx = 1 + x \implies y - 1 = x(y + 1)
     \]
     Therefore, \( x = \frac{y - 1}{y + 1} \). Since \( y \neq -1 \), the range is all real numbers except \( y = -1 \):
     \[
     \text{Range: } (-\infty, -1) \cup (-1, \infty)
     \]

---

### 2. \( f(x) = \frac{1}{\sqrt{(1 - x)(2 - x)}} \)
   - **Domain**: For the square root to be defined, the expression inside the square root must be positive. Additionally, the denominator cannot be zero. So, we need:
     \[
     (1 - x)(2 - x) > 0
     \]
     The critical points are \( x = 1 \) and \( x = 2 \). Analyzing the intervals:
     - For \( x < 1 \), both terms are positive.
     - For \( 1 < x < 2 \), one term is positive, and the other is negative, so the product is negative.
     - For \( x > 2 \), both terms are negative, but their product is positive.

     Therefore, the domain is:
     \[
     \text{Domain: } (-\infty, 1) \cup (2, \infty)
     \]

   - **Range**: Since the function is a reciprocal of a square root function, and the square root function takes positive values, \( f(x) \) is always positive. As \( x \to 1^- \) or \( x \to 2^+ \), \( f(x) \to \infty \). As \( x \to -\infty \) or \( x \to \infty \), \( f(x) \to 0^+ \). Hence, the range is:
     \[
     \text{Range: } (0, \infty)
     \]

---

### 3. \( f(x) = \sqrt{\frac{x - 4}{x + 1}} \)
   - **Domain**: For the square root to be defined, the expression inside the square root must be non-negative. So, we need:
     \[
     \frac{x - 4}{x + 1} \geq 0
     \]
     The critical points are \( x = 4 \) and \( x = -1 \). Analyzing the intervals:
     - For \( x < -1 \), the fraction is negative.
     - For \( -1 < x < 4 \), both terms are positive.
     - For \( x > 4 \), the fraction is positive.

     Therefore, the domain is \( x \geq 4 \). Also, \( x = -1 \) must be excluded as it makes the denominator zero:
     \[
     \text{Domain: } (-1, 4] \cup [4, \infty)
     \]

   - **Range**: Since the function is a square root, it takes non-negative values. As \( x \to -1^+ \), the fraction approaches 0, and \( f(x) \to 0 \). As \( x \to \infty \), \( f(x) \to \infty \). Hence, the range is:
     \[
     \text{Range: } [0, \infty)
     \]

---

Would you like further explanations on any part? Here are five related questions to deepen your understanding:

1. How can we verify the range of a rational function graphically?
2. What happens to the domain of a function if we introduce absolute value signs in the denominator?
3. How would the domain and range of the third function change if we removed the square root?
4. Can you generalize how to find the domain for any rational function with square roots?
5. How does solving for \( y \) help in determining the range of a function?

**Tip**: When working with rational functions, always check for points where the denominator becomes zero to avoid undefined values in the domain!

Determine the domain and range of the following functions: 
1) f(x) = (a + x) / (a - x), where a = 1 
2) f(x) = 1 / sqrt((1 - x)(2 - x)) 
3) f(x) = sqrt((x - 4) / (x + 1))

Explore how to determine the domain and range of different types of functions, including rational and square root functions. The analysis includes solving f(x) = (a + x) / (a - x), f(x) = 1 / sqrt((1 - x)(2 - x)), and f(x) = sqrt((x - 4) / (x + 1)).

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