How do you decide whether the relation x^2 + y = 81 defines a function?

1 Answer
May 29, 2018

y = f(x) = -x^2+81 is a function

x = f(y) = \pm\sqrt(-y+81) is not.

Explanation:

A function is a law of association between elements in two different sets.

So, to define a function, you must have a "starting" set A (the domain), a "landing" set B (the codomain), and a rule that associates to every element of A one and only one element of B.

In most cases, students deal with numerical functions, i.e. functions in which both domain and codomain are the set of real numbers \mathbb{R}.

So, does your equation describes a function? It depends on the variable we choose and independent:

Case 1: x is independent

In this case, we're looking for a rule that assigns a value to y, given a value for x as input. Subtracting x^2 from both sides of the equation, we get

y = -x^2+81

which is indeed a function: for every value x we may choose, the correponding y will be -x^2+81

Case 1: y is independent

This case is the other way around: we're looking for a rule that assigns a value to x, given a value for y as input. Subtracting y from both sides of the equation, we get

x^2 = -y+81

which is not a function: for every value y we may choose, assuming -y+81 is positive, we have ambiguity on the correponding x.

Assume, for example, y=0. The equation would be x^2 = 81, which yields x=\pm9.

So, in this case, we're not able to associate one and only one output value to our input value.