How do you decide whether the relation $x^{2} + y^{2} = 1$ $x^2 + y^2 = 1$ defines a function?

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How do you decide whether the relation $x^{2} + y^{2} = 1$ $x^2 + y^2 = 1$ defines a function?

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Answer 1 · 2018-04-23T20:01:51

$x^{2} + y^{2} = 1$ $x^2+y^2=1$ does not describe a function because there exist valid values of $x$ $x$ for which more than one value of $y$ $y$ make the equation true.

Explanation:

Let's write this equation in a different form.

$y^{2} - (1 - x^{2}) = 0$ $y^2-(1-x^2)=0$ .

Now think of it like the difference of two squares and write this as the product of two binomials.

$(y - \sqrt{1 - x^{2}}) (y + \sqrt{1 - x^{2}}) = 0$ $(y-sqrt(1-x^2))(y+sqrt(1-x^2))=0$

Note that there are TWO solutions for $y$ $y$ here, namely

$y = \sqrt{1 - x^{2}}$ $y=sqrt(1-x^2)$ , and $y = - \sqrt{1 - x^{2}}$ $y=-sqrt(1-x^2)$ .

This relation is NOT a function. In order for an equation to represent a function, every $x$ $x$ in the range of the function must only have one $y$ $y$ -value.

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How do you decide whether the relation $x^{2} + y^{2} = 1$ $x^2 + y^2 = 1$ defines a function?

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