If \(f:\mathbb{R}\to\mathbb{R}\) is defined by \(f(x)=\begin{cases}x+2,&x<0\\x^2,&x>0\end{cases}\), why is it not a function from \(\mathbb{R}\) to \(\mathbb{R}\)?

Correct answer: Because no value is assigned at \(x=0\)

A function from \(\mathbb{R}\) to \(\mathbb{R}\) must assign exactly one real output to every real input. The given rule covers only \(x<0\) and \(x>0\), so no value is assigned when \(x=0\). Hence its domain is not all of \(\mathbb{R}\). At \(x=1\), only \(1^2=1\) is applicable, and at \(x=-1\), the value is \((-1)+2=1\); therefore options B and C are incorrect. Exam tip: To test whether a relation is a function, check that every input in the stated domain has exactly one output.