PieceWise Functions

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 $\boxed{4}$.