The completed table is as follows:
Function Alternative notation Domain Range
f(x) = (5x-20)/(6x-9) f(x) = 5x/6 - 20/9 x ∈ (-∞, 3/2) ∪ (3/2, ∞) y ∈ (-∞, ∞)
f(x) = √(6x-9) f(x) = √6√x - 3 x ∈ [3/2, ∞) y ∈ [0, ∞)
f(x) = 5x² - 9x + 6 f(x) = 5(x - 3/2)² x ∈ (-∞, ∞) y ∈ (-∞, ∞)
f(x) = ³√(5x-20) f(x) = (5x-20)¹/³ x ∈ (-∞, ∞) y ∈ (-∞, ∞)
f(x) = √(6-9x) f(x) = √6√1 - x x ∈ (-∞, 2/3] y ∈ [0, √6]
The domain of each of the functions in the image is all real numbers.
a) f(x) = (5x-20)/(6x-9)
The function is undefined when the denominator is 0, so the domain is all real numbers except for x = 3/2. In interval notation, this is x ∈ (-∞, 3/2) ∪ (3/2, ∞).
b) f(x) = √(6x-9)
The square root function is undefined when the radicand (the expression under the radical) is less than zero, so the domain is all real numbers such that 6x-9 ≥ 0. Solving for x, we get x ≥ 3/2. In interval notation, this is x ∈ [3/2, ∞).
c) f(x) = 5x² - 9x + 6
Quadratic functions have no holes or gaps in their graphs, so their domain is all real numbers. In interval notation, this is x ∈ (-∞, ∞).
d) f(x) = ³√(5x-20)
The cube root function is defined for all real numbers, so the domain of this function is all real numbers. In interval notation, this is x ∈ (-∞, ∞).
e) f(x) = √(6-9x)
As in part (b), the domain of this function is all real numbers such that 6-9x ≥ 0. Solving for x, we get x ≤ 2/3. In interval notation, this is x ∈ (-∞, 2/3].