5. Which of the following are functions?
The last two problems, i.e., b & c, are multi part relations consider all parts when determining whether or not these relations are functions. Explain your reasoning for a, b, and c.
a. f(x) = x + 3
b. f(x) = 73 if x>2 otherwise f(x) = -1
c. f(x) = 79if x>0 or f(x) = -9 if x<0 or f(x) = 9 or -9 if x = 0
To determine if a relation is a function, ask one very important question:
does the relation satisfy the one-to-one condition? That is to say does the relation maps to two values in the codomain (range) from one value in the domain?
Examples:
f(x)=sqrt(x) is not a function because sqrt(x) results in two values.
f(x)=any polynomial with real (i.e. not complex) coefficients is a function, because polynomials satisfy the ono-to-one requirement.
Now can you pick out the functions from the above list, and state why the other(s) is not?
To determine whether a relation is a function or not, we need to consider whether each input (x-value) corresponds to exactly one output (y-value).
a. f(x) = x + 3:
In this function, for every input x, there is a unique output (y = x + 3). For example, if we input x = 2, the output will be y = 2 + 3 = 5. Similarly, for any other value of x, there will be a unique output. Therefore, function a is indeed a function.
b. f(x) = 73 if x > 2, otherwise f(x) = -1:
In this function, if x is greater than 2, the output is 73. However, if x is not greater than 2, there is no specific output mentioned, it is just stated as f(x) = -1. Since there is more than one output for inputs less than or equal to 2, this relation is not a function.
c. f(x) = 79 if x > 0, f(x) = -9 if x < 0, f(x) = 9 or -9 if x = 0:
In this function, for x-values greater than 0, the output is 79, and for x-values less than 0, the output is -9. However, for x = 0, there are two possible outputs mentioned (9 or -9). Thus, for this value, there are multiple outputs. As a result, this relation is not a function.
In summary, only function a (f(x) = x + 3) is a valid function among a, b, and c because it has a unique output for every input.