If f is a relation from the set of positive r | vedclass

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(D) Given the relation $f: R_{+} \rightarrow R_{+}$ defined by $f(x) = 3x^2 - 2$. For $f$ to be a function,every element $x$ in the domain $R_{+}$ mus

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not a function

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(D) Given the relation $f: R_{+} \rightarrow R_{+}$ defined by $f(x) = 3x^2 - 2$. For $f$ to be a function,every element $x$ in the domain $R_{+}$ must have a unique image $f(x)$ in the codomain $R_{+}$. Given $x \in (0, \infty)$,we have $x^2 > 0$,so $3x^2 > 0$. Thus,$f(x) = 3x^2 - 2 > -2$. This means the range of $f$ is $(-2, \infty)$. Since the codomain is given as $R_{+} = (0, \infty)$,and the range $(-2, \infty)$ is not a subset of the codomain $(0, \infty)$ (for example,if $x=0.5$,$f(0.5) = 3(0.25) - 2 = 0.75 - 2 = -1.25$,which is not in $R_{+}$),the relation $f$ does not map every element of the domain to an element in the codomain. Therefore,$f$ is not a function.

If $f$ is a relation from the set of positive real numbers to the set of positive real numbers defined by $f(x) = 3x^2 - 2$,then $f$ is

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