+96 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?\(\)
+129899 f(f(x) either means that
(x^2 - 4)^2 - 4 = 5 or (x + 3) + 3 = 5
x^4 - 8x^2 + 16 - 4 = 5 x + 6 = 5
x^4 - 8x^2 + 7 = 0 x = -1
(x^2 - 7) ( x^2 - 1) = 0 This value of x is not in the domain of the function x + 3 (x must be < -4)
x = ± √7 or x = ± 1
All four values of x
are ≥ -4
So.....four values
