Determine whether the function defined by; \[ f(x)=\left\{\begin{array}{ll}x+k & \text { if } x<0 \\ 4-x^{2} & , \text { if } x \geq 0\end{array} \text { is continuous at } \mathrm{x}=1\right. \]

To determine whether the function \( f(x) \) is continuous at \( x = 1 \), we need to check if the following three conditions are met: 1. \( f(1) \) is defined. 2. The limit of \( f(x) \) as \( x \) approaches 1 exists. 3. The limit of \( f(x) \) as \( x \) approaches 1 equals \( f(1) \). First, we find \( f(1) \): Since \( 1 \geq 0 \), we use the second piece of the function: \[ f(1) = 4 - 1^2 = 4 - 1 = 3. \] Next, we find the limit of \( f(x) \) as \( x \) approaches 1: \[ \lim_{x \to 1} f(x) = f(1) = 3. \] Since 1 is also in the interval where \( x \geq 0 \), we continue using the second piece: \[ \lim_{x \to 1} f(x) = 4 - 1^2 = 3. \] Now we compare the limit with \( f(1) \): Since \( \lim_{x \to 1} f(x) = 3 \) and \( f(1) = 3 \), both are equal. Thus, since all three conditions (definedness, existence of limit, and equality of limit and function value) are met, the function \( f(x) \) is continuous at \( x = 1 \).