27) The domain of the function \( f: f(x)=\left\{\begin{array}{lll}x^{2}+2 & ,-2 \leq x<0 \\ x-1, & x>0\end{array}\right. \) \( \begin{array}{llll}\text { (a) } \mathbb{R} & \text { (b) }[-2,0[ & \text { (c) }[-2, \infty[-\{0\} & \text { (d) }[-2, \infty[\end{array} \)

To determine the domain of the function \( f(x) \), we need to analyze the piecewise definition provided: \[ f(x) = \begin{cases} x^2 + 2 & \text{if } -2 \leq x < 0 \\ x - 1 & \text{if } x > 0 \end{cases} \] 1. **First piece**: \( x^2 + 2 \) is defined for \( -2 \leq x < 0 \). This means that for this part of the function, \( x \) can take any value from \(-2\) up to (but not including) \(0\). 2. **Second piece**: \( x - 1 \) is defined for \( x > 0 \). This means that for this part of the function, \( x \) can take any value greater than \(0\). Now, we combine the intervals from both pieces: - From the first piece, we have the interval \([-2, 0[\). - From the second piece, we have the interval \(]0, \infty[\). Putting these together, the overall domain of the function \( f(x) \) is: \[ [-2, 0[ \cup ]0, \infty[ \] This means that the function is defined for all real numbers except for \(0\). Now, let's analyze the options given: - (a) \( \mathbb{R} \) - This is incorrect because \( f(x) \) is not defined at \( x = 0 \). - (b) \( [-2, 0[ \) - This is only part of the domain. - (c) \( [-2, \infty[ \setminus \{0\} \) - This includes all values from \(-2\) to infinity, excluding \(0\), which is correct. - (d) \( [-2, \infty[ \) - This is incorrect because it includes \(0\). Thus, the correct answer is: \[ \text{(c) } [-2, \infty[ \setminus \{0\} \]