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 \(f^{-1}(-3)+f^{-1}(0)+f^{-1}(3)\).

Guest Aug 23, 2019

#1**+2 **

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

CPhill Aug 23, 2019

#2**+2 **

\(\text{it helps to start by dividing the range into portions and mapping that back to the domain}\\ \text{the range of $f(x)$ is the entire real line which can be expressed as $\\ (-\infty,\infty) = (-\infty, 1)\cup [1,\infty)$}\\~\\ \text{on $(-\infty, 1),~f^{-1}(x) = \sqrt{1-x}+1$}\\ \text{on $[1,\infty),~f^{-1}(x) = 2-x$}\\~\\ f^{-1}(-3) = 3\\ f^{-1}(0) = 2\\ f^{-1}(3) = -1\\ \text{These sum to 4}\)

.Rom Aug 23, 2019