Let's break down the solution into **two steps** for each part (a) and (b), providing explanations at each step and then summarizing the final answer at the end. --- ## (a) **Step 1: Interpret and apply each condition** Let's analyze each condition: 1. **f(4) is not defined:** - At \(x = 4\), there should be a *hole* (no value). 2. **f(1) is defined, but \(\lim_{x \to 1} f(x)\) does not exist:** - At \(x = 1\), the function has a value (a point), but the left and right limits do **not** meet. This could be a jump. 3. **f(-2) is defined and \(\lim_{x \to -2} f(x)\) exists, but \(\lim_{x \to -2} f(x) \ne f(-2)\):** - At \(x = -2\), there's a defined value, but it's *not* equal to the limit (removable discontinuity). 4. **f is continuous for all other x-values:** - Everywhere else, the function is smooth with no breaks. **Explanation:** At this stage, we set up our function to have: - a hole at \(x = 4\), - a jump at \(x = 1\), - a removable discontinuity at \(x = -2\), - and smooth/continuous elsewhere. --- ## (a) **Step 2: Sketch the graph** - Draw a smooth curve everywhere. - At \(x = 4\): leave an open circle (hole), don't define \(f(4)\). - At \(x = 1\): make a jump (e.g., left side approaches 2, right side approaches -1), but put a filled point for \(f(1)\) at any value (e.g., 5). - At \(x = -2\): draw the curve so that the limit as \(x \to -2\) is, say, 1, but \(f(-2)\) is a different value (e.g., 3), so you put a filled dot at (−2, 3) and an open circle at (−2, 1). **Explanation:** This sketch will satisfy all the required conditions: the jump, hole, removable discontinuity, and continuity elsewhere. --- **Final Answer (a):** > - The graph is continuous everywhere except: > - Hole at \(x = 4\), > - Jump at \(x = 1\) (with \(f(1)\) defined, but different left/right limits), > - Removable discontinuity at \(x = -2\) (limit exists, but not equal to value). --- --- ## (b) **Step 1: Interpret and apply each condition** Let's analyze each condition: 1. **\(\lim_{x \to 2^-} f(x) = 1\), \(\lim_{x \to 2^+} f(x) = -1\):** - At \(x = 2\), left and right limits are different (jump discontinuity). 2. **f(2) is undefined:** - No value at \(x = 2\), so an open circle. 3. **\(\lim_{x \to -3^-} f(x) = 2\), \(\lim_{x \to -3^+} f(x) = 2\), \(f(-3) = 0\):** - Limit exists and equals 2, but function value is 0 (removable discontinuity). 4. **f is continuous for all other x-values:** - Smooth everywhere else. **Explanation:** At this stage, we plan for: - a jump at \(x = 2\) with no defined value (open circle), - a removable discontinuity at \(x = -3\) (limit ≠ value), - and continuous elsewhere. --- ## (b) **Step 2: Sketch the graph** - Draw a smooth curve everywhere. - At \(x = 2\): left side approaches 1, right side approaches -1, no point at \(x = 2\). - At \(x = -3\): left and right both approach 2 (so open circle at (−3, 2)), but define \(f(-3) = 0\) (filled dot at (−3, 0)). - Rest: smooth curve. **Explanation:** This graph will show the jump discontinuity with undefined value at 2, and removable discontinuity at −3, continuous elsewhere. --- **Final Answer (b):** > - The graph is continuous everywhere except: > - Jump at \(x = 2\) (left limit 1, right limit -1, no value at 2), > - Removable discontinuity at \(x = -3\) (limit is 2, but value is 0). --- **If you want to see how to draw these, let me know!**