+367 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 \(f^{-1}(-3)+f^{-1}(0)+f^{-1}(3)\).
Thank you
+2407 Hello imdumb! (I'm sure you're actually a humble genius)
Inverses are f(a) = b, f^(-1)(b) = a. I'm sorry, I can't figure out how to get the -1 thingy on latex.
Equation 1:
Let's start with f^(-1)(-3) = x.
So f(x) = -3.
We'll go through both cases.
Case 1:
If f(x) = 2 - x when x =< 1
Then x must be 5 (2 - 5 = -3), however this isn't possible because x needs to less than or equal to 1.
Case 2:
If f(x) = 2x - x^2 when x > 1
Then x = -1 or 3 (solving the quadratic 2x - x^2 = -3), however, x > 1 so x = 3.
Answer:
Thus, f^(-1)(-3) = 3.
Equation 2:
Moving on to the next one, f^(-1)(0) = x.
So f(x) = 0.
Case 1:
If f(x) = 2 - x when x =< 1
Then x must be 2, but not possible since 2 is not less than or equal to 1.
Case 2:
If f(x) = 2x - x^2 when x > 1
Then x = 2 or x = -2, but x > 1 so x = 2.
Answer:
Thus, f^(-1)(0) = 2.
Equation 3:
Finally we have, f^(-1)(3) = x
So f(x) = 3.
Case 1:
If f(x) = 2 - x when x =< 1
Then x must be -1, which works perfectly since 1 =< 1.
If case 1 is correct, we don't have to test case 2.
Answer:
Thus, f^(-1)(3) = -1
Final Answer:
Now we just need to add everything.
3 + 2 - 1 = 4
Hopefully 4 is the correct answer.
I hope this helped. :)))
=^._.^=
