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# Piecewise & Inverse Functions Problem

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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.

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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Guest Jun 5, 2022
#3
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$$f(x) = \left\{ \begin{array}{cl} 3-x & \text{ if } x \leq 3, \\ -x^3+2x^2+3x & \text{ if } x>3. \end{array} \right.$$

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