Given and

Let us first show that f is a function.

When 0 ≤ x ≤ 3, f(x) = x².

The function x² associates all the numbers 0 ≤ x ≤ 3 to unique numbers in R.

Hence, the images of {x ∈ Z: 0 ≤ x ≤ 3} exist and are unique.

When 3 ≤ x ≤ 10, f(x) = 3x.

The function x² associates all the numbers 3 ≤ x ≤ 10 to unique numbers in R.

Hence, the images of {x ∈ Z: 3 ≤ x ≤ 10} exist and are unique.

When x = 3, using the first definition, we have

f(3) = 3² = 9

When x = 3, using the second definition, we have

f(3) = 3(3) = 9

Hence, the image of x = 3 is also distinct.

Thus, as every element of the domain has an image and no element has more than one image, f is a function.

Now, let us show that g is not a function.

When 0 ≤ x ≤ 2, g(x) = x².

The function x² associates all the numbers 0 ≤ x ≤ 2 to unique numbers in R.

Hence, the images of {x ∈ Z: 0 ≤ x ≤ 2} exist and are unique.

When 2 ≤ x ≤ 10, g(x) = 3x.

The function x² associates all the numbers 2 ≤ x ≤ 10 to unique numbers in R.

Hence, the images of {x ∈ Z: 2 ≤ x ≤ 10} exist and are unique.

When x = 2, using the first definition, we have

g(2) = 2² = 4

When x = 2, using the second definition, we have

g(2) = 3(2) = 6

Here, the element 2 of the domain is associated with two elements distinct elements 4 and 6.

Thus, g is not a function.