If
\(f(x) = \begin{cases} x^2-4 &\quad \text{if } x \ge -4, \\ x + 3 &\quad \text{otherwise}, \end{cases} \)
Then for how many values of x is f(f(x)) = 5?
+33659 First think of the inner f(x) as, say, y.
f(y) = 5
so either y2 - 4 = 5 or y + 3 = 5
If y2 - 4 = 5, then y = 3 is a possibility as 3 > -4, also y = -3 is a possibility as -3 > -4
If y + 3 = 5 then y = 2, which is not possible as 2 is not less than -4
So there are two possibilities for y, which means you have to consider two possibilities for f(x)
(1) f(x) = 3
3 = x2 - 4 means x2 = 7 or x = ±√7 both of which are > -4
3 = x + 3 means x = 0 which is not possible because 0 is not less than -4.
(2) f(x) = -3
Can you do this one?
