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# hey I could use some help

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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
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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   Aug 23, 2019
#2
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$$\text{it helps to start by dividing the range into portions and mapping that back to the domain}\\ \text{the range of f(x) is the entire real line which can be expressed as \\ (-\infty,\infty) = (-\infty, 1)\cup [1,\infty)}\\~\\ \text{on (-\infty, 1),~f^{-1}(x) = \sqrt{1-x}+1}\\ \text{on [1,\infty),~f^{-1}(x) = 2-x}\\~\\ f^{-1}(-3) = 3\\ f^{-1}(0) = 2\\ f^{-1}(3) = -1\\ \text{These sum to 4}$$

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Aug 23, 2019