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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 \(f^{-1}(-3)+f^{-1}(0)+f^{-1}(3)\).

 Aug 23, 2019
 #1
avatar+111396 
+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

 

 

cool cool cool

 Aug 23, 2019
 #2
avatar+6192 
+2

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

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 Aug 23, 2019

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