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Let \(f\) be defined by

 \(f(x) = \left\{ \begin{array}{cl} 3-x & \text{ if } x \leq 3, \\ -x^3+2x^2+3x & \text{ if } x>3. \end{array} \right.\)

 

Calculate \(f^{-1}(0)+f^{-1}(6)\).

 

Hello, I have started on this problem, and I have found that \(f^{-1}(0)=3.\) I found the inverse of \(3-x,\) which was \(3-x.\) But I don't know how exactly to find the inverse of the second one, it looks really tricky. I tried an inverse function calculator and got a bunch of fractions, radicals, and complex numbers - yuck. 

 

Could somebody please help? 

 Jun 5, 2022
 #1
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\(f^{-1}(0) + f^{-1}(6) = 4\)

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 Jun 5, 2022
 #2
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Umm... guest, could you please explain or show your work? Thanks.

Guest Jun 5, 2022
 #3
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+1

 

 

\(f(x) = \left\{ \begin{array}{cl} 3-x & \text{ if } x \leq 3, \\ -x^3+2x^2+3x & \text{ if } x>3. \end{array} \right.\)

 

consider 

y=3-x     

when x=3 y= 0

when x<=3    y>=3

since y=3-x

x=3-y

and the inverse becomes    y=3-x   where  x>=0

 

so

\(f^{-1}(x)=3-x\qquad if \qquad x\ge0\\ f^{-1}(0)=3\\ f^{-1}(6)=3-6=-3\\ \\~\\f^{-1}(0)+f^{-1}(6)=3-3=0 \)

 Jun 5, 2022

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