Function Continuity

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Answer Key & Explanations

Function Continuity · Grade 12 · Worksheet 2

1. 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

2. A pharmaceutical company is modeling the concentration of a new drug in the bloodstream over time. The concentration function is given by C(t) = (t² - 4)/(t - 2) for t ≠ 2, where t is measured in hours. The researchers need to determine if the concentration function is continuous at t = 2 hours to ensure predictable drug effects. If not continuous, they must identify the type of discontinuity present. Answer: The function has a removable discontinuity at t = 2. Solution: We are given the concentration function C(t) = (t² - 4)/(t - 2) for t ≠ 2. We need to check if C(t) is continuous at t = 2. If not, we must identify the type of discontinuity.
   Full step-by-step solution

   Step 1: Understand the problem We are given the concentration function C(t) = (t² - 4)/(t - 2) for t ≠ 2. We need to check if C(t) is continuous at t = 2. If not, we must identify the type of discontinuity. --- Step 2: Check if C(2) is defined C(2) would require substituting t = 2 into the function: C(2) = (2² - 4)/(2 - 2) = (4 - 4)/0 = 0/0. This is undefined. So C(t) is not defined at t = 2, meaning it cannot be continuous there unless the discontinuity is removable. --- Step 3: Simplify the function algebraically Factor the numerator: t² - 4 = (t - 2)(t + 2). So C(t) = [(t - 2)(t + 2)]/(t - 2) for t ≠ 2. Cancel (t - 2) (allowed since t ≠ 2): C(t) = t + 2 for t ≠ 2. --- Step 4: Find the limit as t approaches 2 lim (t→2) C(t) = lim (t→2) (t + 2) = 2 + 2 = 4. So the limit exists and equals 4. --- Step 5: Compare limit and function value We have: - C(2) is undefined - lim (t→2) C(t) = 4 exists Since the limit exists but the function is not defined at t = 2, this is a removable discontinuity. --- Step 6: Conclusion The function has a removable discontinuity at t = 2. If the researchers define C(2) = 4, the function becomes continuous.

3. 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).

4. Given the graph of Aroha's function f(x), identify all x-values where f(x) is discontinuous and classify each discontinuity type. Answer: x = -3 (jump), x = 1 (removable), x = 5 (infinite) Solution: Examine the graph at x = -3. The left-hand limit approaches 7, while the right-hand limit approaches 3. Since the left and right limits exist but are not equal, this is a jump discontinuity.
   Full step-by-step solution

   Step 1: Examine the graph at x = -3. The left-hand limit approaches 7, while the right-hand limit approaches 3. Since the left and right limits exist but are not equal, this is a jump discontinuity. Step 2: Examine the graph at x = 1. There is a hole at the point (1, 4), but the function is defined at f(1) = 2. Since the limit exists (approaches 4) but does not equal the function value (2), this is a removable discontinuity. Step 3: Examine the graph at x = 5. The function approaches positive infinity from both sides, creating a vertical asymptote. Since the limit does not exist (approaches infinity), this is an infinite discontinuity. Step 4: The function is continuous at all other x-values shown on the graph. The discontinuities are: x = -3 (jump), x = 1 (removable), x = 5 (infinite).

5. Olivia is a geologist studying a fault line. The graph of the function f(x) below models the elevation (in meters) of the terrain along a 10 km cross-section. Analyze the graph carefully and identify all points of discontinuity within the interval [-5, 5]. For each discontinuity, specify the x-coordinate and classify it as removable, jump, or infinite discontinuity. Answer: At x = -2: jump discontinuity; at x = 1: removable discontinuity; at x = 3: infinite discontinuity. Solution: Examine the graph from left to right. At x = -2, the left-hand limit is 5, the right-hand limit is 3, and f(-2) is defined as 5. Since left and right limits are not equal, this is a jump discontinuity.
   Full step-by-step solution

   Step 1: Examine the graph from left to right. At x = -2, the left-hand limit is 5, the right-hand limit is 3, and f(-2) is defined as 5. Since left and right limits are not equal, this is a jump discontinuity. Step 2: At x = 1, the function has a hole. The left-hand limit and right-hand limit both approach 7, but f(1) is undefined (hole). This is a removable discontinuity. Step 3: At x = 3, the function increases without bound as x approaches 3 from the left (to positive infinity) and decreases without bound from the right (to negative infinity). This is an infinite discontinuity. Step 4: No other breaks or gaps exist in the interval [-5, 5]. The answer is: at x = -2: jump discontinuity; at x = 1: removable discontinuity; at x = 3: infinite discontinuity.

6. Emma is analyzing the temperature distribution along a metal rod for a physics experiment. She models the temperature (in degrees Celsius) using the piecewise function f(x) = { 3x + 10 for x < 5, ax^2 + b for 5 ≤ x ≤ 10, 20x - 85 for x > 10 }, where x is the distance from the left end of the rod in centimeters. To ensure the model accurately represents a smooth temperature transition, the function must be continuous at both transition points x = 5 cm and x = 10 cm. What values must the parameters a and b have to ensure continuity throughout the rod? Answer: a = 0, b = 25 Solution: For continuity at x = 5, the left-hand limit and right-hand limit must be equal. Left-hand limit as x approaches 5 from the left: f(x) = 3x + 10, so limit = 3(5) + 10 = 15 + 10 = 25.
   Full step-by-step solution

   Step 1: For continuity at x = 5, the left-hand limit and right-hand limit must be equal. Left-hand limit as x approaches 5 from the left: f(x) = 3x + 10, so limit = 3(5) + 10 = 15 + 10 = 25. Right-hand limit as x approaches 5 from the right: f(x) = ax^2 + b, so limit = a(5)^2 + b = 25a + b. Set equal: 25a + b = 25. Step 2: For continuity at x = 10, the left-hand limit and right-hand limit must be equal. Left-hand limit as x approaches 10 from the left: f(x) = ax^2 + b, so limit = a(10)^2 + b = 100a + b. Right-hand limit as x approaches 10 from the right: f(x) = 20x - 85, so limit = 20(10) - 85 = 200 - 85 = 115. Set equal: 100a + b = 115. Step 3: Solve the system of equations: Equation 1: 25a + b = 25 Equation 2: 100a + b = 115 Step 4: Subtract Equation 1 from Equation 2 to eliminate b: (100a + b) - (25a + b) = 115 - 25 75a = 90 a = 90/75 = 6/5 = 1.2 Step 5: Substitute a = 1.2 into Equation 1: 25(1.2) + b = 25 30 + b = 25 b = 25 - 30 = -5 Step 6: Verify with Equation 2: 100(1.2) + (-5) = 120 - 5 = 115. This matches the right-hand limit at x = 10. The values are a = 1.2 and b = -5.

7. Consider the function f(x) = (x^3 - 8)/(x - 2) for x ≠ 2. Determine the value that f(2) should be assigned to make the function continuous at x = 2. Answer: 12 Solution: Factor the numerator using the difference of cubes formula: x^3 - 8 = (x - 2)(x^2 + 2x + 4) Simplify the function for x ≠ 2: f(x) = (x - 2)(x^2 + 2x + 4)/(x - 2) = x^2 + 2x + 4 To make the function continuous at x = 2, we need f(2) = lim(x→2) f(x) Evaluate the limit: lim(x→2) (x^2 + 2x + 4) =…
   Full step-by-step solution

   Step 1: Factor the numerator using the difference of cubes formula: x^3 - 8 = (x - 2)(x^2 + 2x + 4) Step 2: Simplify the function for x ≠ 2: f(x) = (x - 2)(x^2 + 2x + 4)/(x - 2) = x^2 + 2x + 4 Step 3: To make the function continuous at x = 2, we need f(2) = lim(x→2) f(x) Step 4: Evaluate the limit: lim(x→2) (x^2 + 2x + 4) = 2^2 + 2(2) + 4 = 4 + 4 + 4 = 12 Step 5: Therefore, f(2) should be assigned the value 12 to make the function continuous at x = 2. The answer is 12.