Question 1171467
This question is identical to a previous question. Here's the solution again:

**1. Domain of f(g(x)) (f \circ g)**

* **f(y) = 4/(y - 2)**
    * The domain of f(y) is all real numbers except y = 2.
* **g(x) = 5/(3x - 1)**
    * The domain of g(x) is all real numbers except 3x - 1 = 0, which means x = 1/3.
* **f(g(x)) = f(5/(3x - 1)) = 4 / [(5/(3x - 1)) - 2]**
    * To find the domain of f(g(x)), we need to consider two things:
        * The domain of g(x) (x ≠ 1/3).
        * The values of x that make the denominator of f(g(x)) equal to zero.
    * Set the denominator to zero:
        * 5/(3x - 1) - 2 = 0
        * 5/(3x - 1) = 2
        * 5 = 2(3x - 1)
        * 5 = 6x - 2
        * 7 = 6x
        * x = 7/6
    * Therefore, the domain of f(g(x)) is all real numbers except x = 1/3 and x = 7/6.
    * In interval notation: (-∞, 1/3) U (1/3, 7/6) U (7/6, ∞)

* **Desmos Graphing:**
    * Go to [www.desmos.com/calculator](https://www.desmos.com/calculator).
    * Input:
        * f(y) = 4/(y - 2)
        * g(x) = 5/(3x - 1)
        * f(g(x))
    * Examine the graph of f(g(x)) to confirm the domain. You will see vertical asymptotes at x=1/3 and x=7/6.

**2. Inverse of f(x) = 4 + √(x - 2)**

* **Finding the inverse:**
    * Let y = 4 + √(x - 2)
    * Swap x and y: x = 4 + √(y - 2)
    * Solve for y:
        * x - 4 = √(y - 2)
        * (x - 4)² = y - 2
        * y = (x - 4)² + 2
    * Therefore, f⁻¹(x) = (x - 4)² + 2

* **Domains and Ranges:**
    * **f(x) = 4 + √(x - 2)**
        * Domain: x - 2 ≥ 0 => x ≥ 2, or [2, ∞)
        * Range: Since √(x - 2) ≥ 0, f(x) ≥ 4, or [4, ∞)
    * **f⁻¹(x) = (x - 4)² + 2**
        * Domain: The range of f(x) becomes the domain of f⁻¹(x), so x ≥ 4, or [4, ∞)
        * Range: The domain of f(x) becomes the range of f⁻¹(x), so y ≥ 2, or [2, ∞)

* **Desmos Graphing:**
    * Go to [www.desmos.com/calculator](https://www.desmos.com/calculator).
    * Input:
        * f(x) = 4 + sqrt(x - 2)
        * y = (x - 4)² + 2 {x>=4}
        * y = x
    * The graph of f⁻¹(x) is the reflection of f(x) across the line y = x.
    * The points (11, 7) and (7, 11) are reflections of each other across the line y = x.
    * The point that is the center of the reflection is the intersection of the line y=x, and the line that goes through (11,7) and (7,11). The midpoint of (7,11) and (11,7) is (9,9). Therefore, the points are reflected about the pair (9, 9).