2. Let \( f(x)=y=3 x+2 \) and \( f: A \rightarrow R \). If \( f(A)=\{-4,5,11\} \), what is \( A \) ? A) \( \{1,0,2\} \) B) \( \{-1,0,2\} \) C) \( \{1,3,6\} \) D) \( \{1,2,3\} \) E) \( \{-2,1,3\} \)

To find the set \( A \) for the function \( f(x) = 3x + 2 \) such that \( f(A) = \{-4, 5, 11\} \), we can solve for \( x \) for each value in \( f(A) \). 1. Setting \( 3x + 2 = -4 \): \[ 3x = -4 - 2 = -6 \implies x = -2 \] 2. Setting \( 3x + 2 = 5 \): \[ 3x = 5 - 2 = 3 \implies x = 1 \] 3. Setting \( 3x + 2 = 11 \): \[ 3x = 11 - 2 = 9 \implies x = 3 \] Thus, the values of \( x \) that give us \( f(x) \) as \{-4, 5, 11\} are \( -2, 1, \) and \( 3 \). Now, looking at the provided options: A) \( \{1,0,2\} \) B) \( \{-1,0,2\} \) C) \( \{1,3,6\} \) D) \( \{1,2,3\} \) E) \( \{-2,1,3\} \) The set \( A \) is indeed \( \{-2, 1, 3\} \) which corresponds to option E. So, the answer is: E) \( \{-2,1,3\} \)