I have several related questions, so I'm going to label them to make sure I understand what questions that answers are referring to. I understand that a function is an expression that produces one output for each set of inputs, but also,

a) could it be said that all functions are equations, but not all equations are functions?

In the fourth post on this page (http://www.physicsforums.com/showthread.php?t=449496) the writer takes the equation x^2 + y^2 + z^2 = 1 and rewrites it as z = f(x, y) = sqrt(x^2 + y^2) and then says “You can draw a line anywhere, so long as it's parallel to the z-axis and it will intersect the surface exactly once in the case of z=f(x,y). No so if f(x,y,z)=const.”

b) Is the writer saying that when f(x,y,z)=constant is written as z=f(x,y) it becomes a function?

If so, does this mean(?) that

c) z = f(x,y) = polynomial is always a function?

d) z = f(x,y) = polynomial = constant is always a non-function/equation?

e) Then if z = f(x,y) = polynomial = constant is always a non-function/equation, does this mean it is incorrect to say that the notation f(x,y) means “function of x and y”?

f) Equations have solutions/roots and functions have outputs?

Thank you!

  • $\begingroup$ An equation is something that has an equals sign in it. A function is something like a monkey. $\endgroup$ – user856 Jun 26 '12 at 2:56
  • $\begingroup$ You might be interested in this:What is the difference between equation and formula? $\endgroup$ – draks ... Jun 26 '12 at 18:33
  • $\begingroup$ Functions and equations are entirely different sorts of things. Functions take inputs and produce outputs. Equations do neither of those things. Equations can be true or false, and they will be true or false once all the variables in them are assigned specific values. Functions cannot be true or false. $\endgroup$ – Andreas Blass May 16 '15 at 1:48

A function associates every ( except where it is discontinuous) points in co-ordinate space with a certain unique value. It can associate either a vector or a scalar. When we write $f(x,y,z)$=constant we mean the points satisfying this relation form a surface in space. There is no sense of association of points with values. But if you could modify same relation into $z=f(x,y)$ for every $(x,y)$ you could associate a value to $z$. Thus you could say $z$ is the function of $x$ & $y$. Yes $z=f(x,y)$ is a function NO, $z=f(x,y)$=constant is also a function called constant function

what makes you think that it is incorrect to say $f(x)$ is a function of $x$? Ya you can say equations have roots (points satisfying them) and functions have outputs

  • 2
    $\begingroup$ A function associates every ( except where it is discontinuous) points in co-ordinate space . . . You should omit "except where it is discontinuous". $\endgroup$ – Dave L. Renfro Feb 11 '13 at 19:35
  • $\begingroup$ I strongly agree with Dave L.Renfro. Mixing up the concept of "discontinuous" and "undefined" function is a common error. For example, the function $f\colon x\in \mathbb{R}\setminus{0} \mapsto 1/x$ is not discontinuous at 0, because it is not even defined there. $\endgroup$ – Giuseppe Negro Apr 4 '14 at 11:37

If a mapping assigns a unique element in the range for each element in the domain, then this mapping is a function.

a. In 2-dimensions if the graph passes the vertical line test, then for each $x$ in the domain there is only one element $y$ in the range (since the vertical line only crossed the graph at one point) and hence this is a function.

In 3 dimensions, the domain can be considered to be the set of all pairs of $x$ and $y$.

b. Once you write it as $z = f(x,y)$, then for each $x$ and $y$ you have a unique $z$ and hence is a function. Note that there is $z$ term on the right hand side.

c. $x = f(x,y) = $ polynomial is a function (see b).

d. $z = $ constant is a function (see b)

e. Yes, $z = f(x)$ is read as $z$ is a function of $x$.

f. A common way of thinking of a function $f(x)$ is that it takes an input $x$ and maps it to $f(x)$, the output. Solutions to equations are those values that make the equation true. So for $x + 3 = 0$, the solution is $x=-3$. It does not make sense to talk about a solution to a function, $f(x)$, but one can talk about, for instance, a solution to $f(x)=0$, which are the zeros of the function.


  • The equation $x^2 + y^2 + z^2 =1$ is not the same as $z = \sqrt{1-x^2-y^2}$, as the latter is only the top portion of the sphere (since the use of the square root implies the principal square root only, ie, $z \ge 0$).
  • $\begingroup$ I don't see how your answer for d is explained by your answer for b. Wouldn't having a polynomial equal a constant be an equation, which would mean that only some values would satisfy/solve the equation? I asked the question the way I did to try to get clarification on the wording of the text I quoted from the linked post. Specifically where he says that a line parallel with the z axis would not intersect the surface exactly once for f(x,y,z)=const., but it would for z=f(x,y). $\endgroup$ – ChrisC70 Jun 26 '12 at 3:19
  • $\begingroup$ @ChrisC70: For d, we have $z=f(x,y)=constant$, which is the same as saying $z=constant$. This is a plane parallel to the $x-y$ plane and hence a function. I think perhaps you are confusing this with the case of 2-dimensions, such as $f(x,y)=constant$, which may not be a function (ex. $f(x,y) = x^2 + y^2 = 1$). In 3-dimensions, $z = f(x,y)$ is a function. $\endgroup$ – Peter Grill Jun 26 '12 at 4:02

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