How do you decide whether the relation $x + y^{3} = 64$ $x + y^3 = 64$ defines a function?

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How do you decide whether the relation $x + y^{3} = 64$ $x + y^3 = 64$ defines a function?

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Answer 1 · 2016-01-04T15:31:31

It can be rewritten as $y = \sqrt[3]{64 - x}$ $y = root(3)(64-x)$ , which uniquely determines $y$ $y$ for any Real value of $x$ $x$ , so yes it is a function.

Explanation:

Given $x + y^{3} = 64$ $x+y^3=64$ , subtract $x$ $x$ from both sides to get:

$y^{3} = 64 - x$ $y^3=64-x$

Take cube roots of both sides:

$y = \sqrt[3]{64 - x}$ $y = root(3)(64-x)$

This uniquely determines $y$ $y$ for any Real value of $x$ $x$ , so the relation does describe a function.

Also note that $x = 64 - y^{3}$ $x = 64 - y^3$ so it has a well defined inverse function too.

Footnote
$x + y^{3} = 64$ $x+y^3=64$ does not define a function if $x$ $x$ and $y$ $y$ are allowed to take Complex values, since every number has $3$ $3$ cube roots in the Complex plane, hence there would be three possible values of $y$ $y$ for each value of $x$ $x$ .

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How do you decide whether the relation $x + y^{3} = 64$ $x + y^3 = 64$ defines a function?

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How do you decide whether the relation $x + y^{3} = 64$ $x + y^3 = 64$ defines a function?