Final answer:
The question about the bijective nature of a function with the given range seems to be incorrect. The quadratic function can have negative values, which do not fit the specified range. Additionally, similar concepts have been discussed within the snippets provided.
Step-by-step explanation:
The question seems to be asking about the bijectiveness of a function f(x)=9x²−6x−5 over different domains, such as real numbers, negative to positive infinity, and non-negative real numbers. To determine if a function is bijective, it should be both injective (one-to-one) and surjective (onto).
Injective means that no two different elements in the domain map to the same element in the codomain, and surjective means that every element in the codomain is an image of at least one element in the domain. However, the original function representation seems to have an error, as the output of the quadratic function can yield negative values, which would not be within the range (2, ∞) as stated. Without correcting this, we cannot determine bijectiveness over the provided domains.
Regarding the other snippets provided:
- A continuous function f(x)=20 for 0 ≤ x ≤ 20 would graph as a horizontal line between these two x-values.
- The quadratic equation x²+1.2x10-2x−6.0×10−3=0 can indeed be solved using the quadratic formula.
- For a continuous probability function f(x) equal to 12 over the range 0 ≤ x ≤ 12, P(0 < x < 12) would effectively be 1, since the entire range of x is covered.