∀x ∈ R define the function f (x) = x3 − x2 − 3x + 2. Find f −1({−1})

Answer 1

By definition f^{-1}({b})={x in R : f(x)=b}. In your case, b=-1, so this becomes f^{-1}({-1})={x in R : f(x)=-1}. Now, we are given f(x), so this set becomes f^{-1}({-1})={x in R : x^3-x^2-3x+2=-1}. In other words, this set is all the x values for which x^3-x^2-3x+2=-1 is true, or rather x^3-x^2-3x+3=0. Now this has 3 solutions, they are -sqrt(3), sqrt(3), and 1. Therefore, the set is ^{-1}({-1})={-sqrt(3),sqrt(3),1}.