## Intermediate Algebra (12th Edition)

$y \text{ is NOT a function of }x \\\text{Domain: } [0,\infty)$
$\bf{\text{Solution Outline:}}$ To determine if the given equation, $x=y^4 ,$ is a function, solve first for $y$ in terms of $x.$ Then check if $x$ is unique for every value of $y.$ To find the domain, find the set of all possible values of $x.$ $\bf{\text{Solution Details:}}$ Using the properties of equality, the given equation is equivalent to \begin{array}{l}\require{cancel} x=y^4 \\\\ y^4=x \\\\ y=\pm\sqrt{x} .\end{array} If $x=16,$ then $y=2$ or $y=-2.$ That is, the ordered pairs, $\{ (16,2),(16,-2) \}$ satisfy the given equation. This means that the value of $x$ is not unique. Hence, $y$ is NOT a function of $x.$ Since $x$ appears in the radicand of a radical with an even index, then \begin{array}{l}\require{cancel} x\ge0 .\end{array} Hence, the domain is the set of all nonnegative numbers. Hence, the given equation has the following characteristics: \begin{array}{l}\require{cancel} y \text{ is NOT a function of }x \\\text{Domain: } [0,\infty) .\end{array}