Question 1171540: What can be said about the domain of the function f . g where f(y)= {4}/{y-2} and g(x)= {5}/{3x-1} ? Express it in terms of a union of intervals of real numbers. Go to www.desmos.com/calculator and obtain the graph of f , g , and f. g .
Find the inverse of the function f(x)=4+ \sqrt{x-2} .
State the domains and ranges of both the function and the inverse function in terms of intervals of real numbers.
Go to www.desmos.com/calculator and obtain the graph of f , its inverse, and g(x)=x in the same system of axes. About what pair (a, a) are (11, 7) and (7, 11) reflected about?
Answer by CPhill(1959)   (Show Source):
You can put this solution on YOUR website! Let's break this down into the requested parts.
**1. Domain of f(g(x)) (f . 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).
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