Function Continuity

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Answer Key & Explanations

Function Continuity · Grade 12 · Worksheet 3

1. A function f(x) is graphed on the coordinate plane. The graph consists of a smooth curve from (-3, 2) to (1, 4), but has a hole at x = 1 where the curve would pass through (1, 4) if extended. From x = 1 to x = 4, the graph is a different smooth curve starting at (1, 2) and ending at (4, 6). At x = 2, there is a vertical asymptote where the function approaches positive infinity from both sides. Identify all x-values where the function is discontinuous and classify each discontinuity. Answer: x = 1 (removable), x = 2 (infinite) Solution: Analyze the graph at x = 1. The curve from (-3, 2) to (1, 4) has a hole at (1, 4), meaning f(1) is not defined as 4. The curve from (1, 2) to (4, 6) starts at (1, 2), so there's a jump in the function values.
   Full step-by-step solution

   Step 1: Analyze the graph at x = 1. The curve from (-3, 2) to (1, 4) has a hole at (1, 4), meaning f(1) is not defined as 4. The curve from (1, 2) to (4, 6) starts at (1, 2), so there's a jump in the function values. The left-hand limit as x approaches 1 is 4, the right-hand limit is 2, and f(1) is undefined. This is a removable discontinuity. Step 2: Analyze the graph at x = 2. There is a vertical asymptote where the function approaches positive infinity from both sides. This means the limit as x approaches 2 does not exist (it goes to infinity), so this is an infinite discontinuity. Step 3: The function is continuous everywhere else since the curves are smooth with no other breaks, holes, or asymptotes. Step 4: Therefore, the function is discontinuous at x = 1 (removable) and x = 2 (infinite).

2. Sophia is analyzing the continuity of f(x) from its graph. The graph shows: f(1) = 6, lim(x→1⁻) f(x) = 6, lim(x→1⁺) f(x) = 6. Is f(x) continuous at x = 1? Answer: B. yes Solution: Check if f(1) is defined. f(1) = 6, so the function is defined at x = 1. Check if the limit exists as x approaches 1.
   Full step-by-step solution

   Step 1: Check if f(1) is defined. f(1) = 6, so the function is defined at x = 1. Step 2: Check if the limit exists as x approaches 1. lim(x→1⁻) f(x) = 6 and lim(x→1⁺) f(x) = 6, so the limit exists and equals 6. Step 3: Check if f(1) equals the limit. f(1) = 6 and lim(x→1) f(x) = 6, so they are equal. Step 4: Since all three conditions are satisfied, f(x) is continuous at x = 1. The answer is yes.

3. Given the graph of Sophia's function f(x) with points at (-1,1), (1,2), (3,1), (5,0), (6,1), identify all x-values where the function is discontinuous and classify each discontinuity type. Answer: x=1 (jump), x=3 (jump), x=5 (removable) Solution: Continuity means a function has no breaks, jumps, or holes. For a function to be continuous at a point, the limit must exist and equal the function value.
   Full step-by-step solution

   Continuity means a function has no breaks, jumps, or holes. For a function to be continuous at a point, the limit must exist and equal the function value. Discontinuities can be removable (hole), jump (sudden change), or infinite (vertical asymptote).

4. Given the graph of f(x) with points at (-2, 3), (-1, 1), (0, 0), (1, 1), (2, 3), and an open circle at (0, 2), is f(x) continuous at x = 0? Answer: A. no Solution: Examine the graph at x = 0. There is a point at (0, 0) and an open circle at (0, 2). The function value f(0) is 0 (from the closed point).
   Full step-by-step solution

   Step 1: Examine the graph at x = 0. There is a point at (0, 0) and an open circle at (0, 2). Step 2: The function value f(0) is 0 (from the closed point). Step 3: Check the limit as x approaches 0 from the left. From points (-1, 1) and (-2, 3), the function appears to approach 0 as x → 0⁻. Step 4: Check the limit as x approaches 0 from the right. From points (1, 1) and (2, 3), the function appears to approach 0 as x → 0⁺. Step 5: The limit exists and is 0, but f(0) = 0, which equals the limit. Step 6: However, the open circle at (0, 2) indicates a removable discontinuity at x = 0, meaning the function is not continuous there. The answer is no.

5. Given the graph of Noah's function f(x) with points at x=7, x=9, and x=11, determine if f(x) is continuous at each point and classify any discontinuities: f(7)=14, lim_(x→7⁻) f(x)=14, lim_(x→7⁺) f(x)=14; f(9)=18, lim_(x→9⁻) f(x)=16, lim_(x→9⁺) f(x)=16; f(11)=22, lim_(x→11⁻) f(x)=22, lim_(x→11⁺) f(x)=24 Answer: Continuous at x=7, jump discontinuity at x=9, jump discontinuity at x=11 Solution: Analyze continuity at x=7. f(7)=14, lim_(x→7⁻) f(x)=14, lim_(x→7⁺) f(x)=14. Since f(7)=lim_(x→7⁻) f(x)=lim_(x→7⁺) f(x)=14, the function is continuous at x=7.
   Full step-by-step solution

   Step 1: Analyze continuity at x=7. f(7)=14, lim_(x→7⁻) f(x)=14, lim_(x→7⁺) f(x)=14. Since f(7)=lim_(x→7⁻) f(x)=lim_(x→7⁺) f(x)=14, the function is continuous at x=7. Step 2: Analyze continuity at x=9. f(9)=18, lim_(x→9⁻) f(x)=16, lim_(x→9⁺) f(x)=16. The left and right limits are equal (16), but f(9)=18 is different. This is a jump discontinuity at x=9. Step 3: Analyze continuity at x=11. f(11)=22, lim_(x→11⁻) f(x)=22, lim_(x→11⁺) f(x)=24. The left and right limits are not equal (22 ≠ 24), so this is a jump discontinuity at x=11. Final answer: Continuous at x=7, jump discontinuity at x=9, jump discontinuity at x=11.

6. Given the piecewise function: f(x) = { x^2 - 9 for x < 3; 2x + 3 for x ≥ 3 }. Is f(x) continuous at x = 3? Answer: B. no Solution: Check if f(3) exists. f(3) = 2(3) + 3 = 6 + 3 = 9. Find the left-hand limit as x approaches 3.
   Full step-by-step solution

   Step 1: Check if f(3) exists. f(3) = 2(3) + 3 = 6 + 3 = 9. Step 2: Find the left-hand limit as x approaches 3. lim (x→3⁻) f(x) = lim (x→3⁻) (x^2 - 9) = (3)^2 - 9 = 9 - 9 = 0. Step 3: Find the right-hand limit as x approaches 3. lim (x→3⁺) f(x) = lim (x→3⁺) (2x + 3) = 2(3) + 3 = 6 + 3 = 9. Step 4: Compare the values. f(3) = 9, left-hand limit = 0, right-hand limit = 9. Since the left-hand limit (0) is not equal to the function value (9), the function is not continuous at x = 3. The answer is No.

7. Sophia is analyzing the continuity of a function f(x) from its graph. The graph shows: f(7) = 9, lim(x→7⁻) f(x) = 9, lim(x→7⁺) f(x) = 9. Is f continuous at x = 7? Answer: A. yes Solution: Check if f(7) is defined. f(7) = 9, so yes, the function is defined at x = 7. Check if the limit exists as x approaches 7.
   Full step-by-step solution

   Step 1: Check if f(7) is defined. f(7) = 9, so yes, the function is defined at x = 7. Step 2: Check if the limit exists as x approaches 7. lim(x→7⁻) f(x) = 9 and lim(x→7⁺) f(x) = 9, so the left-hand and right-hand limits are equal. Therefore, lim(x→7) f(x) = 9. Step 3: Check if f(7) equals the limit. f(7) = 9 and lim(x→7) f(x) = 9, so yes, they are equal. Step 4: Since all three conditions are satisfied, the function is continuous at x = 7. The answer is yes.

8. Given the graph of function f(x) with points at x = -5, 0, 5, 10, determine if f is continuous at each point and classify any discontinuities. Answer: Continuous at x=-5, removable discontinuity at x=0, jump discontinuity at x=5, infinite discontinuity at x=10 Solution: At x = -5, the graph shows a solid point with no break, gap, or asymptote. The limit exists and equals the function value, so f is continuous at x = -5.
   Full step-by-step solution

   Step 1: At x = -5, the graph shows a solid point with no break, gap, or asymptote. The limit exists and equals the function value, so f is continuous at x = -5. Step 2: At x = 0, there is a hole in the graph (open circle) but the function approaches a finite limit from both sides. This indicates a removable discontinuity. Step 3: At x = 5, the graph shows a jump where the left-hand limit and right-hand limit both exist but are not equal. This is a jump discontinuity. Step 4: At x = 10, the graph has a vertical asymptote where the function approaches positive or negative infinity. This is an infinite discontinuity. Final classification: Continuous at x=-5, removable discontinuity at x=0, jump discontinuity at x=5, infinite discontinuity at x=10.