+865 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 $f(f(x)) = 5$?
+36916 x = 3 , -3 ( you would THINK x = 2 would work, but 2 >= -4 so f(2) = 22-4=0 )
so f(x) has to equal 3 or -3
x^2-4 = 3 then x = +-sqrt7
x^2-4 = -3 then x = +- 1 4 answers I believe........ edited
