Let f(x) be defined by
f(x) = x - 3 if x <= 3
f(x) = x^2 - 6x + 9 if x > 3
Calculate f^{-1}(0) + f^{-1}(6).
+9519 Note that the function is increasing everywhere (so when we find a solution to f(x) = k, that must be the unique solution).
When f(x) = 0, x = 3 since f(3) = 3 - 3 = 0 (solution is unique because f is increasing).
That means \(f^{-1}(0) = 3\).
For any x <= 3, f(x) = x - 3 <= 0.
Therefore, when f(x) = 6, x > 3.
Can you try to find the value of x from here? That value would be \(f^{-1}(6)\), and now you just add the results together.
