+17 For some constants a and b let \(f(x) = \left\{ \begin{array}{cl} ax + b & \text{if } x < 2, \\ 8 - 3x & \text{if } x \ge 2. \end{array} \right.\) The function f has the property that f(f(x))=x for all x. What is a+b?
+1347 We know that f(f(x)) = x for all x. If x < 2, then f(x) = ax + b. Substituting into the first equation, we get
f(ax + b) = x
Since f(ax + b) = 8 - 3(ax + b) if ax + b >= 2, then we must have ax + b >= 2. This means that x >= -2/a.
If -2/a <= x < 2, then f(x) = ax + b. Substituting into the first equation, we get
(ax + b) + b = x
This simplifies to ax + 2b = x. Since ax + b >= 2, then x must equal 2. However, we know that f(2) = 8 - 3(2) = 2, so this case is not possible.
Therefore, the only possible value of x is x = -2/a. In this case, f(x) = f(-2/a) = 8 - 3(-2/a) = 10/a. Substituting into the first equation, we get
10/a + b = -2/a
This simplifies to a + b = -10. Therefore, a + b = −10
