\( \begin{array}{l}\text { Discuss the continuity of } \\ \text { the function at } x=1\end{array} f(x)=\left\{\left.\begin{array}{ll}3 x-1 & \text { if } x<1 \\ 4 & \text { if } x=1 \\ 2 x & \text { if } x>1\end{array} \right\rvert\,\right. \)

To discuss the continuity of the function \( f(x) \) at \( x = 1 \), we need to check three conditions: 1. **The function \( f(1) \) is defined.** 2. **The limit of \( f(x) \) as \( x \) approaches 1 exists.** 3. **The limit of \( f(x) \) as \( x \) approaches 1 is equal to \( f(1) \).** Let's analyze each of these conditions step by step. ### Step 1: Check if \( f(1) \) is defined From the piecewise definition of the function: \[ f(1) = 4 \] So, \( f(1) \) is defined. ### Step 2: Calculate the limit of \( f(x) \) as \( x \) approaches 1 We need to find the left-hand limit and the right-hand limit as \( x \) approaches 1. - **Left-hand limit** (\( x \to 1^- \)): For \( x < 1 \), the function is defined as \( f(x) = 3x - 1 \). \[ \lim_{x \to 1^-} f(x) = \lim_{x \to 1^-} (3x - 1) = 3(1) - 1 = 2 \] - **Right-hand limit** (\( x \to 1^+ \)): For \( x > 1 \), the function is defined as \( f(x) = 2x \). \[ \lim_{x \to 1^+} f(x) = \lim_{x \to 1^+} (2x) = 2(1) = 2 \] ### Step 3: Check if the limits are equal and equal to \( f(1) \) Now we compare the left-hand limit and the right-hand limit: \[ \lim_{x \to 1^-} f(x) = 2 \quad \text{and} \quad \lim_{x \to 1^+} f(x) = 2 \] Since both limits are equal, we can conclude: \[ \lim_{x \to 1} f(x) = 2 \] Now we check if this limit is equal to \( f(1) \): \[ f(1) = 4 \] Since \( \lim_{x \to 1} f(x) = 2 \) and \( f(1) = 4 \), we find that: \[ \lim_{x \to 1} f(x) \neq f(1) \] ### Conclusion Since the limit of \( f(x) \) as \( x \) approaches 1 does not equal \( f(1) \), the function \( f(x) \) is not continuous at \( x = 1 \).