To determine the number of values of \( x \) for which \( f(f(x)) = 5 \), we need to analyze the given piecewise function:
\[
f(x) = \begin{cases}
x^2 - 4 & \text{if } x \geq -4 \\
x + 3 & \text{otherwise}
\end{cases}
\]
We'll consider each case separately.
### Case 1: \( x \geq -4 \)
For \( x \geq -4 \), \( f(x) = x^2 - 4 \). We need to find \( y \) such that:
\[
f(y) = 5
\]
So,
\[
y^2 - 4 = 5 \implies y^2 = 9 \implies y = \pm 3
\]
However, since \( x \geq -4 \), both solutions \( y = 3 \) and \( y = -3 \) are valid because they are within the domain \( x \geq -4 \). Hence, \( y = 3 \) and \( y = -3 \).
Next, we need \( f(x) \) such that:
\[
f(x) = 3 \quad \text{or} \quad f(x) = -3
\]
#### Subcase 1.1: \( f(x) = 3 \)
\[
x^2 - 4 = 3 \implies x^2 = 7 \implies x = \pm \sqrt{7}
\]
Since \( x \geq -4 \), both solutions \( x = \sqrt{7} \) and \( x = -\sqrt{7} \) are valid.
#### Subcase 1.2: \( f(x) = -3 \)
\[
x^2 - 4 = -3 \implies x^2 = 1 \implies x = \pm 1
\]
Both solutions \( x = 1 \) and \( x = -1 \) are valid since \( x \geq -4 \).
### Case 2: \( x < -4 \)
For \( x < -4 \), \( f(x) = x + 3 \). We need to find \( y \) such that:
\[
f(y) = 5
\]
So,
\[
y + 3 = 5 \implies y = 2
\]
However, since \( x < -4 \), the solution \( y = 2 \) does not fall within this domain. Therefore, there are no solutions from this case.
### Conclusion
Summarizing the valid solutions from both subcases under \( x \geq -4 \), we have:
- \( f(x) = 3 \): \( x = \sqrt{7}, -\sqrt{7} \)
- \( f(x) = -3 \): \( x = 1, -1 \)
Thus, we find a total of \( 4 \) values of \( x \):
\[
x = \sqrt{7}, -\sqrt{7}, 1, -1
\]
Therefore, the number of values of \( x \) for which \( f(f(x)) = 5 \) is \( \boxed{4} \).
