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# PieceWise Functions

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

Jun 8, 2024

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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 $$\boxed{4}$$.

Jun 9, 2024