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I'm a little confused by the rule: If you draw a vertical line that intersects the graph at more than 1 point then it is not a function.

Because then a circle like $y^2 + x^2 = 1$ is not a function?

And indeed if I rewrite it as $f(x) = \sqrt(1 - x^2)$ then wolfram alpha doesn't draw a circle. I guess I'm missing the intuition as to why this is though?

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Now you got several similar answers - I hope it helps! –  AD. Oct 27 '11 at 7:32
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7 Answers

up vote 15 down vote accepted

The definition of a function is so important. In addition to the above, the picture below (taken from: What is a function) may help.

(the left hand side is your X and the right hand side is the value Y)

What is a function

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A function is a rule that assigns uniquely to a member of domain set, a member of the image set. The key word is "uniquely". So if you assign say 2 as well as -2 to number 1, then you have a rule, but not a function. That is the logic behind the vertical line test. If you draw a vertical line and it intersects the graph of the function in two distinct points, then you can see that it means I have assigned both of these points to the point where my vertical line crosses the x-axis. An example of this is the circle.

However a semi-circle is a legit function-the upper half is the positive square root (y=+$\sqrt{1-x^2}$) and the bottom half is the negative square root (y=-$\sqrt{1-x^2}$).

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Functions need to be well-defined as part of their definition, so for a given input there can only be one output.

$f(x,y)=x^2+y^2-1$ is a function of two variables, and the set of points for which this function gets $0$ is the unit circle.

However writing $y^2+x^2=1$ as a function of $x$ alone cannot be done, as $x=\dfrac12$ has two solutions ($y=\pm\sqrt{\dfrac34}$).

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If you want have a function that "draws" a circle with radius $r$ and center $P = (x_0, y_0)$ on the cartesian plane, you can use the function $f : [0, 2\pi] \rightarrow \mathbb{R} \times \mathbb{R}$ defined by $$f(\varphi) = (x_0 + r \cos \varphi, y_0 + r \sin \varphi)$$ But, of course, this is not a function from $\mathbb{R}$ to $\mathbb{R}$.

Also, you can define a curve in the plane by means of an equation of two variables $x$ and $y$. If you have a (continuous) function $f : A\subseteq \mathbb{R}\rightarrow \mathbb{R}$, you can get an equation $y = f(x)$ from it, which defines a curve. But you cannot always transform an equation containing two variables to an equivalent equation $y = f(x)$. The equation $x^2 + y^2 = r^2, r\in\mathbb{R}$ is an example of this fact.

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+1 for “…from $\mathbb{R}$ to $\mathbb{R}$”; the OP’s equation is indeed a function, just one from reals to pairs of reals, like your parametric version. –  Jon Purdy Oct 27 '11 at 14:13
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I would not say that an equation is function: an equation can define a function. Both an equation and a function can define a curve on the plane, but some equations do not have an associated function. –  Giorgio Oct 27 '11 at 15:00
    
Fair enough. It seems to me that the OP is saying $\pm y$ intuitively ought to represent the pair $(+y, -y)$. –  Jon Purdy Oct 27 '11 at 16:14
    
Ah, ok, you mean $\pm y = (+y, -y)$, so you would have a function $f:\mathbb{R}\rightarrow \mathbb{R}\times\mathbb{R}$. –  Giorgio Oct 27 '11 at 16:16
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A function $f(x_1, \ldots, x_n)$ has the property, that for one set of values $(v_1, \ldots, v_n)$ there is at most one result. If you compare, your $f(0) = 1$, but there are 2 values for $y$ s.t. $y^2 + x^2 = 1 \mid x = 0$, namely $\{ 1, -1 \}$.

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The standard definition of a function $f$ is that it takes one value $f(x)$ for each $x$ (where it is defined).

In particular, the square root is a single valued function - for a real number $x$, the square root is given by $\sqrt{x^2} =|x|$.

In your example, when solving for $y$ in the circle equation $y^2+x^2=1$ there are two possibilities $$y=\sqrt{1-x^2}\qquad \text{or}\qquad y=-\sqrt{1-x^2}$$ which are two different functions and the union of their graphs is the circle.

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$y^2+x^2=1$ is implicit definition of $y$

An equivalent explicit definition of $y$ is:

$y=\pm \sqrt{1-x^2}$ , with condition $x\in [-1,1] $

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Yes, but $y = \pm \sqrt{1 - x^2}$ is not a function. –  Giorgio Oct 27 '11 at 15:03
    
@Giorgio,$\pm\sqrt{1-x^2}$ means $y=+\sqrt{1-x^2} \lor y=-\sqrt{1-x^2}$ –  pedja Oct 27 '11 at 19:27
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