Define a function satisfying the given condition by defining its value on each element of the domain. Note that there are many correct answers for these questions. 1. f: {5, 15, 17, 18} → {1, 6, 7, 12, 13, 17} which is injective. 2. f: {5, 15, 17, 18} → {1, 6, 7, 12, 13, 17} which is not injective. 3. f: {5, 15, 17, 18} → {12, 13, 17} which is surjective. 4. f: {5, 15, 17, 18} - → {12, 13, 17} which is not surjective. f(5) f(15) = f(17) = f(18) = = f(5) = f(15) = f(17) = f(18) = = f(5) = f(15) = f(17) = f(18) = f(5) = f(15) = f(17) = f(18) : =

9. Let f and f-1 be functions. If the point (-4, 6) is on the graph of f-1, what point is on the graph of f? Answer=

1) Use algebraic means to show that x2 + y2 = 8 is not a function. Explain your process. 2)Is there any value(s) of the domain of x2 + y2 = 8 that passes the vertical line test? If so,name the value(s) and state whether or not the existence of this value makes this relation afunction.

Refer to the graph of the function and complete the statements. 41) As x → 0,f(x) → As x → n,f(x)→ 40 15 1-10 -15 -10 -10 -15