Answers:
- Domain = {c, d}
- f/g = { (c, 1/2), (d, 1/2) }
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Step-by-step explanation:
The domain of f: {a,b,c,d} which is drawn from set A.
The domain of g: {a,c,d,e} which is drawn from set B
So far, we can see the possible inputs are: a,b,c,d,e
Let's go through each lowercase letter of the given sets
- If 'a' is plugged into f/g, then we'll have 0 in the denominator due to the ordered pair (a,0) in function g. This causes a division by zero error so the input 'a' is not allowed in the domain of f/g.
- If we tried b, then it would work for function f, but it doesn't work for function g. If we tried to compute g(b), then the result is undefined. This is because the lowercase b input is not found in set uppercase B which is where function g selects its inputs from. So b is eliminated.
- The input c does work however. f(c) = 2 because of the ordered pair (c,2) in the function f. Similarly, g(c) = 4. This means f/g = 2/4 = 1/2 when the input is c. This shows that c is a valid input in the domain of f/g.
- The input d is pretty much the same story as input c. Moreover, the input d leads to the same output as c does. This time we have f/g = 4/8 = 1/2 when the input is d. Therefore, d is another valid input in the domain of f/g.
- The input e is not valid because (e,0) will lead to a division by zero error, as it's part of the g(x) function. So we must eliminate 'e' as a valid option in the domain. Even if we didn't have a division by zero error, we still have the fact that 'e' is not in the domain of the first function f. This is another reason why f/g is not defined when the input is 'e'.
To summarize, only inputs c and d are valid in the domain.
The domain is the set {c, d}
Both inputs in the domain lead to the output 1/2. This means the range of f/g is {1/2}. Despite having just one element inside, we must use curly braces to indicate we have a set.
The function f/g is { (c, 1/2), (d, 1/2) }
We simply list all of the (x,y) ordered pairs to describe the function. In this case, there are only two such pairs.