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?
+118608
\(f(x) = \left\{ \begin{array}{cl} 3-x & \text{ if } x \leq 3, \\ -x^3+2x^2+3x & \text{ if } x>3. \end{array} \right.\)
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
\(f^{-1}(x)=3-x\qquad if \qquad x\ge0\\ f^{-1}(0)=3\\ f^{-1}(6)=3-6=-3\\ \\~\\f^{-1}(0)+f^{-1}(6)=3-3=0 \)
