Restrictions on a Function 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 $f(f(x)) = 5$?
I'm not sure how to really start from this question other than bashing values. I need some help on a start, thanks!
 A: First find the values such that $f(x) = 5$:
$x^2-4=5\iff x^2=9 \iff x=\pm 3$ (both are greater than or equal to $-4$)
$x+3=5\iff x = 2$ (but this value is not less than $-4$).
Hence we must have $f(x) = \pm 3$
So now we find the values such that $f(x) = 3$
$x^2-4=3\iff x^2=7 \iff x=\pm \sqrt{7}$ (both are greater than or equal to $-4$)
$x+3=3\iff x = 0$ (but this value is not less than $-4$).
We now find the values such that $f(x) =-3$
$x^2-4=-3\iff x^2=1 \iff x=\pm 1$ (both are greater than or equal to $-4$)
$x+3=-3\iff x = -6$ (this value is less than $-4$).

Hence the values are $\pm\sqrt{7},\pm1$ and $-6$
A: Just do it in cases:
Case 1:  $f(x) < -4$ then $f(f(x)) = 5$ means
$f(x) + 3 = 5$ so $f(x) =2$ and that's a contradiction.
So it must be that:
Case 2: $f(x) \ge -4$ so $f(f(x)) = 5$ means
$f(x)^2 -4 = 5$
$f(x)^2 = 9$
$f(x) = \pm 3$.
Case 3:  $x < -4$ so $f(x) = \pm 3$ means $x+3\pm 3$ meas $x=-6, 0$ but only $-6< -4$ so $x = -6$.
Case 4: $x \ge -4$ so $f(x) = \pm 3$ means $x^2 -4 =\pm 3$ so $x^2 = 1, 7$.
So $x = \pm 1, \pm \sqrt 7$.  All of those are acceptable as the are all $< -4$
So there are five solutions.  
a) $x=-6$ and $f(f(-6)) = f(-6+3) = f(-3)= (-3)^2-4 = 9-4=5$.
b)c) $x =\pm 1$ and $f(f(\pm 1))= f((\pm 1)^2-4)=f(-3)= 5$.
d)e) $x =\pm \sqrt7$ and $f(f(\pm \sqrt 7)) = f((\pm \sqrt 7)^2-4)=f(7-4)=f(3)=3^2-4 = 9-4 = 5$.
