### Understanding the Problem We are given a piecewise function \( f(x) \) defined as: \[ f(x) = \begin{cases} 3 - kx, & 1 \leq x < 2 \\ \frac{x^2}{4} - 3, & x \geq 2 \end{cases} \] The questions ask: 1. For which value(s) of \( k \) is \( f \) continuous at \( x = 2 \)? 2. At which other points in the interval \([1, \infty)\) is \( f \) continuous, and why? ### Part 1: Continuity at \( x = 2 \) Definition of Continuity at a Point: A function \( f \) is continuous at \( x = a \) if: 1. \( f(a) \) is defined. 2. \( \lim_{x \to a} f(x) \) exists. 3. \( \lim_{x \to a} f(x) = f(a) \). Applying to \( x = 2 \): 1. \( f(2) \) is defined: - Since \( 2 \geq 2 \), we use the second piece: \( f(2) = \frac{2^2}{4} - 3 = \frac{4}{4} - 3 = 1 - 3 = -2 \). 2. Limit as \( x \to 2 \): - We need to check the left-hand limit (\( x \to 2^- \)) and the right-hand limit (\( x \to 2^+ \)). - Left-hand limit (\( x \to 2^- \)): - For \( x \) approaching 2 from the left (\( 1 \leq x < 2 \)), \( f(x) = 3 - kx \). - \( \lim_{x \to 2^-} f(x) = 3 - k \cdot