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If the function \(f:\{1,2,3,4,5,6,7,8,9\}\to\{0,1\}\) is defined by \(f(n)=1\) when \(n\) is divisible by \(3\), and \(f(n)=0\) otherwise, what is \(f^{-1}(\{1\})\)?

Answer and explanation

Correct answer: \(\{3,6,9\}\)

The preimage of a set consists of all elements of the domain whose images belong to that set. Here, \(f(n)=1\) exactly when \(n\) is divisible by \(3\). Among the given domain elements, these are \(3,6,9\), so \(f^{-1}(\{1\})=\{3,6,9\}\). Option A contains numbers not divisible by \(3\), while option D incorrectly includes \(0\), which is not in the domain. Exam tip: To find a preimage, check which domain elements map to the specified target value or set.

Related tags

PreimageFunctionsRelationsDivisibilityDomainReal Valued FunctionsTheir Domain And RangeReal Valued Functions Their Domain And RangeRelations And Functions

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