The domain of the function $$ f(x)=\left\{\begin{array}{l}x+2, \quad x \geq 1 \ \frac{1}{x^{2}-4}, \quad x

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The domain of the function $$ f(x)=\left\{\begin{array}{l}x+2, \quad x \geq 1 \ \frac{1}{x^{2}-4}, \quad x<1\end{array}ight. $$

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To determine the domain of the function

$$
f(x) = \begin{cases} 
x + 2, & x \geq 1 \ 
\frac{1}{x^2 - 4}, & x < 1 
\end{cases}
$$

we need to analyze each piece of the piecewise function separately.

1. **For the first piece** $$ x + 2 $$ where $$ x \geq 1 $$:
   - This expression is defined for all $$ x $$ that satisfy $$ x \geq 1 $$. Therefore, the domain for this part is $$ [1, \infty) $$.

2. **For the second piece** $$ \frac{1}{x^2 - 4} $$ where $$ x < 1 $$:
   - This expression is defined as long as the denominator is not zero. We need to find when $$ x^2 - 4 = 0 $$:
     $$
     x^2 - 4 = 0 \implies x^2 = 4 \implies x = 2 \text{ or } x = -2.
     $$
   - However, since this piece is only valid for $$ x < 1 $$, we only consider $$ x = -2 $$ as a potential issue. Thus, we need to exclude $$ x = -2 $$ from the domain of this piece.
   - Therefore, the domain for this part is $$ (-\infty, -2) \cup (-2, 1) $$.

Now, we combine the domains from both pieces:

- From the first piece, we have $$ [1, \infty) $$.
- From the second piece, we have $$ (-\infty, -2) \cup (-2, 1) $$.

Thus, the overall domain of the function $$ f(x) $$ is:

$$
(-\infty, -2) \cup (-2, 1) \cup [1, \infty).
$$
The domain of the function is $$ (-\infty, -2) \cup (-2, 1) \cup [1, \infty) $$.

The domain of the function

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