# Why is x=y^3 a function and x=y^2 not a function?

Asked by bryanking (21 ) January 3rd, 2011

confused with functions

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Neither of these is a function; they are equalities or equations. An equation can be part of a function definition, but not the whole thing.

A function is a shorthand for a set of operations to which values can be passed. For example, you could define a function f(x,y) (called “f of x and y”) such that f(x,y)=(x=y^3), or a function g(x,y) where g(x,y)=(x=y^2). Function names are arbitrary.

Leaving aside the technical details of @koanhead‘s answer, it’s not a function because there are two values of “y” that satisfy the equation.

bob_ (18235 )

@bob I’m not sure I understand what you mean. Both x=y^3 and x=y^2 have a value of x for all possible values of y. If mean the statement x=y^3 && x=y^2 (that is, “x is equal to y cubed AND x is equal to y squared” then yes, it’s true for y=1 and y=0, but I think that statement is technically not an equation since it has “AND” in the middle instead of “is equal to”.
Anyway, I’m confused. Can you clarify what you mean?

@koanhead If y=f(x), then in the first case

x = y^3
[x]^(⅓) = [y^3]^(⅓)
x^(⅓) = y
f(x) = x^(⅓)

Which is a valid function because there is only one value of “y” for each value of “x”.

In the second case,

x = y^2
x^(½) = -y OR y

And that’s where the wheels come off the wagon.

bob_ (18235 )

The (confusing) convention is that when giving definitions of functions of real numbers, one writes “y = ...x…”, where…x… is some expression involving x. Unless that is what’s happening here, then both expressions determine functions. Otherwise, the first expression is another way of writing y = x^(⅓) and the second another way of writing y = x^(½). In the first case, each value of x corresponds to exactly one value of y, while in the second each value of x corresponds to two values of y, as @bob_ indicated.

ratboy (15107 )

it goes this way in the problem which makes me confused:

Determine if the set is a function. State the domain if it is a function.
1.) {(x,y) | x= y^2}
2.) {(x,y) | x= y^3}

#1 is not a function which is found in the answer at the back of the book while #2 is a function with domain (-infinity,+infinity)

im not asking help from a homework, im doing a bit of self-study and i found this part confusing. thank you, everyone, for your input :)

bryanking (21 )

Graph them. x=y² is a parabola that opens to the right; you have two values of y for each x. For example, (-1)² and 1² both satisfy the equation. However, x=y³ is a cubic, with only one value of y for each x.

@bryanking: This is confusing. The answer given in the text assumes the domain is the real numbers, so why ask? {(x,y)|x=y^2} is a function on the domain [0,+infinity).

ratboy (15107 )

@ratboy im asking because i’m not as smart as you

bryanking (21 )

or