\(If, f(x) = \begin{cases} x^2-4 &\quad \text{if } x \ge -4, \\ x + 3 &\quad \text{otherwise}, \end{cases} then for how many values of x is f(f(x)) = 5?\)

wassyo Jun 8, 2024

#1**0 **

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 \( \boxed{4} \).

eramsby1O1O Jun 9, 2024