Let \(f\) be defined by\(f(x) = \left\{ \begin{array}{cl} 2-x & \text{ if } x \leq 1, \\ 2x-x^2 & \text{ if } x>1. \end{array} \right.\) Calculate \(f^{-1}(-3)+f^{-1}(0)+f^{-1}(3)\).
+128475 f-1(-3) + f-1(0) + f-1(3)
For the first term, either
2 - x = -3 or
2x - x^2 = -3
If 2 -x =-3 then x must be 5 but this is outside x ≤ 1
So....it must be that
2x -x^2 = -3
x^2 -2x - 3 =0
(x -3) ( x + 1) = 0
Set each factor to 0 and solve for x and we get that x = -1 [ reject since this value of x < 1] or
x = 3 [ accept since this value of x is > 1 ]
So f-1(-3) = 3
For the second term, either
2 - x = 0 or
2x - x^2 = 0
If 2 - x = 0 the x =2 and this is value of x is not ≤ 1
If 2x - x^2 = 0 then
x ( 2 - x) = 0
Set each factor to 0 and we have that either x = 0 [reject since this is not > 1] or
x = 2 which is > 1
So
f-1(0) = 2
For the last term either
2 - x = 3 or
2x - x^2 = 3
If 2 - x = 3 then x = -1 and this value of x is ≤ 1
So
f-1(3) = -1
So
f-1(-3) + f-1(0) + f-1(3) =
3 + 2 - 1 =
4
+6248 \(\text{it helps to start by dividing the range into portions and mapping that back to the domain}\\ \text{the range of $f(x)$ is the entire real line which can be expressed as $\\ (-\infty,\infty) = (-\infty, 1)\cup [1,\infty)$}\\~\\ \text{on $(-\infty, 1),~f^{-1}(x) = \sqrt{1-x}+1$}\\ \text{on $[1,\infty),~f^{-1}(x) = 2-x$}\\~\\ f^{-1}(-3) = 3\\ f^{-1}(0) = 2\\ f^{-1}(3) = -1\\ \text{These sum to 4}\)
.
