Function Continuity

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Answer Key & Explanations

Function Continuity · Grade 12 · Worksheet 1

1. Charlotte, a civil engineer, is examining the graph of a function f(x) that models the stress distribution (in megapascals) along a steel girder, where x is the distance in meters from the left support. The graph shows that as x approaches 9 from the left, f(x) approaches 16, and as x approaches 9 from the right, f(x) approaches 16. However, at x = 9, the graph has an open circle at y = 16 and a separate closed dot at (9, 22). Based on this graphical information, is the function continuous at x = 9? If not, identify the type of discontinuity and explain your reasoning using the conditions for continuity. Answer: The function is not continuous at x = 9; it has a removable discontinuity (or point discontinuity) because the limit exists (16) but does not equal the function value (22). Solution: Recall the three conditions for continuity at a point x = c: 1. f(c) must be defined. 2.
   Full step-by-step solution

   Step 1: Recall the three conditions for continuity at a point x = c: 1. f(c) must be defined. 2. The limit of f(x) as x approaches c must exist. 3. The limit of f(x) as x approaches c must equal f(c). Step 2: Analyze the information at x = 9: The function is defined at x = 9 because there is a closed dot at (9, 22). So f(9) = 22. As x approaches 9 from the left, f(x) approaches 16. As x approaches 9 from the right, f(x) approaches 16. Since the left-hand limit and right-hand limit are both 16, the limit exists and is 16. Step 3: Check the third condition: Does the limit equal the function value? The limit is 16, but f(9) = 22. Since 16 ≠ 22, the third condition fails. Step 4: Therefore, the function is not continuous at x = 9. This is a removable discontinuity (also called a point discontinuity) because the limit exists, but the function value at that point is different from the limit. The discontinuity could be "removed" by redefining f(9) to be 16. The answer is: The function is not continuous at x = 9; it has a removable discontinuity (or point discontinuity) because the limit exists (16) but does not equal the function value (22).

2. Given the graph of Matiu's function f(x) with points at x=4, x=8, and x=12, determine if f(x) is continuous at each point and classify any discontinuities: f(4)=12, lim_(x→4⁻) f(x)=12, lim_(x→4⁺) f(x)=12; f(8)=20, lim_(x→8⁻) f(x)=16, lim_(x→8⁺) f(x)=16; f(12)=24, lim_(x→12⁻) f(x)=24, lim_(x→12⁺) f(x)=28. Answer: Continuous at x=4, jump discontinuity at x=8, jump discontinuity at x=12 Solution: Analyze continuity at x=4. f(4)=12, lim_(x→4⁻) f(x)=12, lim_(x→4⁺) f(x)=12. Since f(4)=lim_(x→4⁻) f(x)=lim_(x→4⁺) f(x)=12, the function is continuous at x=4.
   Full step-by-step solution

   Step 1: Analyze continuity at x=4. f(4)=12, lim_(x→4⁻) f(x)=12, lim_(x→4⁺) f(x)=12. Since f(4)=lim_(x→4⁻) f(x)=lim_(x→4⁺) f(x)=12, the function is continuous at x=4. Step 2: Analyze continuity at x=8. f(8)=20, lim_(x→8⁻) f(x)=16, lim_(x→8⁺) f(x)=16. The left and right limits are equal (16), but f(8)=20 is different. This is a jump discontinuity at x=8. Step 3: Analyze continuity at x=12. f(12)=24, lim_(x→12⁻) f(x)=24, lim_(x→12⁺) f(x)=28. The left-hand limit (24) does not equal the right-hand limit (28), so the limit does not exist at x=12. This is a jump discontinuity. Final answer: Continuous at x=4, jump discontinuity at x=8, jump discontinuity at x=12.

3. Given the graph of Emma's function f(x) with points at x=1, x=3, x=5, x=7, and x=9, determine if f(x) is continuous at each point and classify any discontinuities: f(1)=3, lim_(x→1⁻) f(x)=3, lim_(x→1⁺) f(x)=3; f(3)=7, lim_(x→3⁻) f(x)=5, lim_(x→3⁺) f(x)=5; f(5)=9, lim_(x→5⁻) f(x)=9, lim_(x→5⁺) f(x)=11; f(7)=13, lim_(x→7⁻) f(x)=13, lim_(x→7⁺) f(x)=13; f(9)=15, lim_(x→9⁻) f(x)=15, lim_(x→9⁺) f(x)=17. Answer: Continuous at x=1, x=7; jump discontinuity at x=3, x=5, x=9 Solution: Analyze continuity at x=1. f(1)=3, lim_(x→1⁻) f(x)=3, lim_(x→1⁺) f(x)=3. Since f(1)=lim_(x→1⁻) f(x)=lim_(x→1⁺) f(x)=3, the function is continuous at x=1.
   Full step-by-step solution

   Step 1: Analyze continuity at x=1. f(1)=3, lim_(x→1⁻) f(x)=3, lim_(x→1⁺) f(x)=3. Since f(1)=lim_(x→1⁻) f(x)=lim_(x→1⁺) f(x)=3, the function is continuous at x=1. Step 2: Analyze continuity at x=3. f(3)=7, lim_(x→3⁻) f(x)=5, lim_(x→3⁺) f(x)=5. The left and right limits are equal (5), but f(3)=7 is different. This is a jump discontinuity at x=3. Step 3: Analyze continuity at x=5. f(5)=9, lim_(x→5⁻) f(x)=9, lim_(x→5⁺) f(x)=11. The left and right limits are not equal (9 ≠ 11), so the limit does not exist. This is a jump discontinuity at x=5. Step 4: Analyze continuity at x=7. f(7)=13, lim_(x→7⁻) f(x)=13, lim_(x→7⁺) f(x)=13. Since f(7)=lim_(x→7⁻) f(x)=lim_(x→7⁺) f(x)=13, the function is continuous at x=7. Step 5: Analyze continuity at x=9. f(9)=15, lim_(x→9⁻) f(x)=15, lim_(x→9⁺) f(x)=17. The left and right limits are not equal (15 ≠ 17), so the limit does not exist. This is a jump discontinuity at x=9. Final answer: Continuous at x=1, x=7; jump discontinuity at x=3, x=5, x=9.

4. Given the graph of Aroha's function f(x) with points at x=3, x=7, and x=11, determine if f(x) is continuous at each point and classify any discontinuities: f(3)=9, lim_(x→3⁻) f(x)=9, lim_(x→3⁺) f(x)=9; f(7)=15, lim_(x→7⁻) f(x)=13, lim_(x→7⁺) f(x)=13; f(11)=21, lim_(x→11⁻) f(x)=21, lim_(x→11⁺) f(x)=23. Answer: Continuous at x=3, jump discontinuity at x=7, jump discontinuity at x=11 Solution: Analyze continuity at x=3. f(3)=9, lim_(x→3⁻) f(x)=9, lim_(x→3⁺) f(x)=9. Since f(3)=lim_(x→3⁻) f(x)=lim_(x→3⁺) f(x)=9, the function is continuous at x=3.
   Full step-by-step solution

   Step 1: Analyze continuity at x=3. f(3)=9, lim_(x→3⁻) f(x)=9, lim_(x→3⁺) f(x)=9. Since f(3)=lim_(x→3⁻) f(x)=lim_(x→3⁺) f(x)=9, the function is continuous at x=3. Step 2: Analyze continuity at x=7. f(7)=15, lim_(x→7⁻) f(x)=13, lim_(x→7⁺) f(x)=13. The left and right limits are equal (13), but f(7)=15 is different. This is a jump discontinuity at x=7. Step 3: Analyze continuity at x=11. f(11)=21, lim_(x→11⁻) f(x)=21, lim_(x→11⁺) f(x)=23. The left and right limits are not equal (21 ≠ 23), so this is a jump discontinuity at x=11. Final answer: Continuous at x=3, jump discontinuity at x=7, jump discontinuity at x=11.

5. A continuous function f(x) is graphed on the coordinate plane. The graph passes through points (-2, 4), (0, 1), and (3, 5). There is a removable discontinuity at x = 1 where the limit exists but f(1) is undefined, and a jump discontinuity at x = 2 where the left-hand limit is 3 and the right-hand limit is 6. At what x-values does the function fail to be continuous? Answer: x = 1 and x = 2 Solution: A function is continuous at a point \( x = c \) if: 1. \( f(c) \) is defined. 2.
   Full step-by-step solution

   Let's go step by step. --- **Step 1: Understanding continuity** A function is continuous at a point \( x = c \) if: 1. \( f(c) \) is defined. 2. \( \lim_{x \to c} f(x) \) exists. 3. \( \lim_{x \to c} f(x) = f(c) \). If any of these fail, the function is discontinuous at \( x = c \). --- **Step 2: Given points and discontinuities** The problem says: - Points on the graph: (-2, 4), (0, 1), (3, 5). These are just points the continuous parts pass through, but discontinuities are at other x-values. - **Removable discontinuity at x = 1**: - Limit exists at x = 1. - But \( f(1) \) is undefined. - Since \( f(1) \) is undefined, condition 1 fails → not continuous at x = 1. - **Jump discontinuity at x = 2**: - Left-hand limit = 3, right-hand limit = 6. - Since left-hand limit ≠ right-hand limit, the limit as x → 2 does not exist. - Condition 2 fails → not continuous at x = 2. --- **Step 3: Check other points** The function is continuous everywhere else except at these explicitly stated discontinuities. The given points (-2, 4), (0, 1), (3, 5) are on the graph, meaning at those x-values the function is defined and matches the limit (since the graph is continuous there except at x = 1 and x = 2). --- **Step 4: Conclusion** The function fails to be continuous at: - x = 1 (removable discontinuity, f(1) undefined) - x = 2 (jump discontinuity, limit does not exist) --- **Final answer:** x = 1 and x = 2

6. Given the graph of Kaia's function f(x) with points at x=1, x=3, x=5, x=7, and x=9, determine if f(x) is continuous at each point and classify any discontinuities: f(1)=7, lim_(x→1⁻) f(x)=7, lim_(x→1⁺) f(x)=7; f(3)=11, lim_(x→3⁻) f(x)=9, lim_(x→3⁺) f(x)=9; f(5)=13, lim_(x→5⁻) f(x)=13, lim_(x→5⁺) f(x)=15; f(7)=17, lim_(x→7⁻) f(x)=17, lim_(x→7⁺) f(x)=17; f(9)=19, lim_(x→9⁻) f(x)=19, lim_(x→9⁺) f(x)=21. Answer: Continuous at x=1, jump discontinuity at x=3, jump discontinuity at x=5, continuous at x=7, jump discontinuity at x=9 Solution: Analyze continuity at x=1. f(1)=7, lim_(x→1⁻) f(x)=7, lim_(x→1⁺) f(x)=7. Since f(1)=lim_(x→1⁻) f(x)=lim_(x→1⁺) f(x)=7, the function is continuous at x=1.
   Full step-by-step solution

   Step 1: Analyze continuity at x=1. f(1)=7, lim_(x→1⁻) f(x)=7, lim_(x→1⁺) f(x)=7. Since f(1)=lim_(x→1⁻) f(x)=lim_(x→1⁺) f(x)=7, the function is continuous at x=1. Step 2: Analyze continuity at x=3. f(3)=11, lim_(x→3⁻) f(x)=9, lim_(x→3⁺) f(x)=9. Since f(3)=11 but the left and right limits are both 9, the function is discontinuous at x=3. Because the left and right limits are equal (9) but not equal to f(3), this is a removable discontinuity. Step 3: Analyze continuity at x=5. f(5)=13, lim_(x→5⁻) f(x)=13, lim_(x→5⁺) f(x)=15. Since the left-hand limit (13) and right-hand limit (15) are not equal, the function is discontinuous at x=5. This is a jump discontinuity. Step 4: Analyze continuity at x=7. f(7)=17, lim_(x→7⁻) f(x)=17, lim_(x→7⁺) f(x)=17. Since f(7)=lim_(x→7⁻) f(x)=lim_(x→7⁺) f(x)=17, the function is continuous at x=7. Step 5: Analyze continuity at x=9. f(9)=19, lim_(x→9⁻) f(x)=19, lim_(x→9⁺) f(x)=21. Since the left-hand limit (19) and right-hand limit (21) are not equal, the function is discontinuous at x=9. This is a jump discontinuity. Final answer: Continuous at x=1, removable discontinuity at x=3, jump discontinuity at x=5, continuous at x=7, jump discontinuity at x=9.