Algebra 3 - Test 3 - Section 2 Let X = {0, 2, 4, 6} and let Y = {0, 4, 16, 36}
Let f : X → Y
f : {(0, 0),(2, 0),(2, 4),(4, 4)(6, 16)}
1. Is f a function? If yes, state the domain and range of f. If no, explain why f is not a
function.

Let X = {5, 10, 15, 20} and Y = {−1, 0, 1}
Let g : X → Y
g : {(5, 1),(10, 0),(15, 0),(20, 1)}
2. Is g a function? If yes, state the domain and range of g. If no, explain why g is not a
function.

Let X = {−1, 0, 1, 2} and Y = {3, 4, 5, 6}
Let h : X → Y
h : {(−1, 4),(0, 5),(1, 3),(2, 3)}
3. What is the range of the function h?

395.84

step-by-step explanation:

2πrh+2πr²

2 (pie)(radius)(height)+2(pie)(radius)^2

answer

the correct answer is

explanation

the given function is

the excluded values are obtained by solving:

we factor to obtain:

the excluded values are

guess at question:

a vertical line intersects the graph of g(x,y) = 0 at two points. what conclusion follows about the function f(x) implicitly defined by g(x,f(x)) = 0?

the domain of f(x) cannot include the x coordinate of any such vertical line.

informally, f(x) is not a function, and

g(x,y)=0 cannot be solved for y uniquely.

the relation r = {(x,y) such that g(x,y) = 0} is a function if

for all (a,b) ∊ r and (c,d) ∊ r, a=c implies b=d.

step-by-step explanation: