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 $f^{-1}(0)+f^{-1}(6)$.
+130085 \( \[f(x) = \left\{ \begin{array}{cl} 3-x & \text{ if } x \leq 3, \\ -x^3+2x^2+3x & \text{ if } x>3. \end{array} \right.\]Calculate $f^{-1}(0)+f^{-1}(6)$.\)
f-1 (0) must mean that either 3 - x = 0 or -x^3 + 2x^2 + 3x = 0
However......in the second function x = 0 is the only thing that makes this true, but that function isn't defined for x = 0
So.....it must mean that 3 - x = 0....and x = 3......so f-1 (0) = 3
And
f-1 (6) must mean that either 3 - x = 6 or that -x^3 + 2x^2 + 3x = 6
Looking at the second function we have that
-x^3 +2x^2 + 3x = 6 → x^3 -2x^2 - 3x + 6 = 0 → x^2( x - 2) - 3(x - 2) = 0 →
(x^2 - 3) (x -2) = 0 → x = ±√3 or x = 2 are the x values that make this true....but....these values are all < 3 and this function isn't defined for these values
But 3 - x = 6 when x = -3 which is valid
So
f-1 (0) = 3 and f-1 (6) = -3
So .... their sum = 0
