Function Continuity

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Answer Key & Explanations

Function Continuity · Grade 12 · Worksheet 2

1. Given the graph of f(x) with points at (9,4), (10,4), (11,6), and a jump discontinuity at x=10 where f(10) = 4 but lim(x→10⁻) f(x) = 4 and lim(x→10⁺) f(x) = 6. Is f(x) continuous at x=10? Answer: A. no Solution: Check if the limit from the left exists: lim(x→10⁻) f(x) = 4 Check if the limit from the right exists: lim(x→10⁺) f(x) = 6 Since the left-hand limit (4) does not equal the right-hand limit (6), the overall limit does not exist at x=10 Even though f(10) = 4 exists, the function cannot be…
   Full step-by-step solution

   Step 1: Check if the limit from the left exists: lim(x→10⁻) f(x) = 4 Step 2: Check if the limit from the right exists: lim(x→10⁺) f(x) = 6 Step 3: Since the left-hand limit (4) does not equal the right-hand limit (6), the overall limit does not exist at x=10 Step 4: Even though f(10) = 4 exists, the function cannot be continuous because the limit does not exist Step 5: This is a jump discontinuity Step 6: Therefore, f(x) is not continuous at x=10

2. Given the graph of function f(x) with a jump discontinuity at x = 3, a removable discontinuity at x = 5, and an infinite discontinuity at x = 7. Which x-values represent points of discontinuity and what type is each? Answer: x=3 jump, x=5 removable, x=7 infinite Solution: At x = 3, there is a jump discontinuity where the function has different values from the left and right sides.
   Full step-by-step solution

   Step 1: At x = 3, there is a jump discontinuity where the function has different values from the left and right sides. Step 2: At x = 5, there is a removable discontinuity where there is a hole in the graph that could be filled to make the function continuous. Step 3: At x = 7, there is an infinite discontinuity where the function approaches positive or negative infinity. Step 4: The discontinuities are at x = 3 (jump), x = 5 (removable), and x = 7 (infinite).

3. A continuous function f(x) is graphed on the coordinate plane. The graph passes through points (-2, 4), (0, 1), and (3, -2). Between x = -2 and x = 3, the function is a smooth curve with no breaks, jumps, or holes. According to the Intermediate Value Theorem, what is the minimum number of times the function must cross the x-axis between these endpoints? Answer: 1 Solution: The IVT says that if a function is continuous on a closed interval [a, b], and k is any number between f(a) and f(b), then there is at least one c in (a, b) such that f(c) = k.
   Full step-by-step solution

   Step 1: Understand the Intermediate Value Theorem (IVT). The IVT says that if a function is continuous on a closed interval [a, b], and k is any number between f(a) and f(b), then there is at least one c in (a, b) such that f(c) = k. Step 2: Identify the given points and function values. The function passes through: (-2, 4) → f(-2) = 4 (0, 1) → f(0) = 1 (3, -2) → f(3) = -2 Step 3: Check the sign changes between consecutive points. From x = -2 to x = 0: f(-2) = 4 (positive) f(0) = 1 (positive) No sign change here, so IVT does not guarantee a zero crossing in (-2, 0). From x = 0 to x = 3: f(0) = 1 (positive) f(3) = -2 (negative) Since the function is continuous and goes from positive to negative, by IVT there must be at least one x in (0, 3) such that f(x) = 0. Step 4: Determine the minimum number of x-axis crossings. We found exactly one guaranteed sign change from positive to negative between x = 0 and x = 3. No other interval from the given points shows a sign change. Thus, the minimum number of times the function must cross the x-axis between x = -2 and x = 3 is 1. Final answer: 1

4. 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.

5. Is the function continuous at x=2? (check graph for break) Answer: A. no Solution: Look at the graph at x=2 Observe that there is a jump discontinuity at x=2 where the function value suddenly changes The left-hand limit and right-hand limit at x=2 are not equal Since the limits don't match, the function is not continuous at x=2 The answer is No.
   Full step-by-step solution

   Step 1: Look at the graph at x=2 Step 2: Observe that there is a jump discontinuity at x=2 where the function value suddenly changes Step 3: The left-hand limit and right-hand limit at x=2 are not equal Step 4: Since the limits don't match, the function is not continuous at x=2 The answer is No.

6. Aroha analyzes a function graph with points at x=9, x=10, and x=11. The graph shows: at x=9, f(9)=12 and limit from both sides is 12; at x=10, f(10)=15 but limit from left is 13 and from right is 13; at x=11, f(11)=18 and limit from left is 18, but limit from right does not exist. Identify which points are continuous and classify any discontinuities. Answer: x=9 continuous; x=10 removable discontinuity; x=11 jump discontinuity Solution: Check continuity at x=9 f(9) = 12 lim(x→9) f(x) = 12 (from both sides) Since f(9) = lim(x→9) f(x), the function is continuous at x=9 Check continuity at x=10 f(10) = 15 lim(x→10) f(x) = 13 (from both sides) Since f(10) ≠ lim(x→10) f(x), the function is discontinuous at x=10 This is a removable…
   Full step-by-step solution

   Step 1: Check continuity at x=9 f(9) = 12 lim(x→9) f(x) = 12 (from both sides) Since f(9) = lim(x→9) f(x), the function is continuous at x=9 Step 2: Check continuity at x=10 f(10) = 15 lim(x→10) f(x) = 13 (from both sides) Since f(10) ≠ lim(x→10) f(x), the function is discontinuous at x=10 This is a removable discontinuity because the limit exists but doesn't equal the function value Step 3: Check continuity at x=11 f(11) = 18 lim(x→11⁻) f(x) = 18 lim(x→11⁺) f(x) does not exist Since the right-hand limit doesn't exist, the overall limit doesn't exist, so the function is discontinuous at x=11 This is a jump discontinuity because the left-hand limit exists but the right-hand limit doesn't Final answer: x=9 continuous; x=10 removable discontinuity; x=11 jump discontinuity

7. 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