Continuity & Piecewise Functions - Business Calculus

1. The billing function is: $B(x) = \begin{cases} 0.12x & \text{if } 0 \le x \le 500 \\ 65 + 0.15(x-500) & \text{if } x > 500 \end{cases}$ What is the financial consequence for a customer at the 500 kWh threshold?
   Answer: The bill has a jump discontinuity, resulting in an immediate $5 surcharge for using just over 500 kWh. When you encounter piecewise functions in business calculus, always check for continuity at the boundary points—this reveals important economic implications like pricing jumps or rate changes. Let's examine what happens at exactly 500 kWh. Using the first piece: $B(500) = 0.12(500) = \$60$. Now check what happens just above 500 kWh, say at 500.01 kWh. Using the second piece: $B(500.01) = 65 + 0.15(500.01-500) = 65 + 0.15(0.01) = \$65.0015$. This creates a jump discontinuity—the bill instantly jumps from $60 to approximately $65 when usage exceeds 500 kWh, even by a tiny amount. That's a $5 surcharge for crossing the threshold. Answer A correctly identifies this jump discontinuity and the immediate $5 financial consequence. Answer B incorrectly claims the total bill is continuous—we just showed it's not. While the marginal rate does increase from $0.12 to $0.15 per kWh (a $0.03 increase, not $5), the total bill itself jumps. Answer C makes the same error, claiming continuity where none exists. Answer D mischaracterizes this as a "removable discontinuity," but this is intentional rate design, not a flaw—the jump is a deliberate feature of the two-tier system. Study tip: For piecewise functions, always test continuity at boundary points by calculating the function value from both pieces. In business contexts, discontinuities often represent real economic consequences like penalty fees or rate structure changes.
2. A company's revenue function has a discontinuity due to a pricing policy change. The function $R(x) = \begin{cases} 100x - x^2 & \text{if } x < 10 \\ ax + b & \text{if } x ≥ 10 \end{cases}$ represents revenue in thousands of dollars for $x$ thousand units sold. If $\lim_{x \to 10^+} R(x) = 1200$ and $\lim_{x \to 10^-} R(x) = 900$, what type of discontinuity exists at $x = 10$?
   Answer: A jump discontinuity because the one-sided limits exist but are unequal When analyzing discontinuities in piecewise functions, you need to examine the behavior of one-sided limits and determine whether the function "jumps" between different values. Let's verify the given limit information. For $x < 10$, we use $R(x) = 100x - x^2$. As $x$ approaches 10 from the left: $\lim_{x \to 10^-} R(x) = 100(10) - 10^2 = 1000 - 100 = 900$. This matches the given information. For the right-hand limit, we're told $\lim_{x \to 10^+} R(x) = 1200$, which means the linear piece $ax + b$ approaches 1200 as $x$ approaches 10 from the right. Since both one-sided limits exist but are finite and unequal (900 ≠ 1200), this creates a jump discontinuity. The function literally "jumps" from a revenue of $900,000 to $1,200,000 at the policy change point. Choice A incorrectly calls this removable—removable discontinuities occur when limits are equal but the function value differs, allowing you to "remove" the discontinuity by redefining the function at that point. Here, the unequal one-sided limits make removal impossible. Choice B misses the key point: having one-sided limits doesn't guarantee continuity. Continuity requires the limits to be equal and match the function value. Choice C confuses this with infinite discontinuities, where limits approach ±∞. Both our limits are finite values. Study tip: Jump discontinuities occur when one-sided limits exist, are finite, but unequal—visualize the graph literally "jumping" between two different y-values.
3. Consider the function $f(x) = \begin{cases} \frac{x^2 - 9}{x - 3} & \text{if } x ≠ 3 \\ c & \text{if } x = 3 \end{cases}$ At how many points in the interval $[0, 6]$ is this function discontinuous when $c = 5$?
   Answer: The function is discontinuous at exactly one point First, simplify $\frac{x^2 - 9}{x - 3} = \frac{(x-3)(x+3)}{x - 3} = x + 3$ for $x ≠ 3$. So $\lim_{x \to 3} f(x) = \lim_{x \to 3} (x + 3) = 6$. For continuity at $x = 3$, we need $f(3) = \lim_{x \to 3} f(x)$, which means $c = 6$. Since $c = 5 ≠ 6$, the function is discontinuous at $x = 3$. The function is continuous everywhere else in $[0, 6]$ since it equals $x + 3$ for all $x ≠ 3$, which is a continuous function. Therefore, there is exactly one point of discontinuity.
4. The function $g(t) = \begin{cases} t^2 + 4t - 5 & \text{if } t ≤ 2 \\ \frac{at + b}{t - 1} & \text{if } t > 2 \end{cases}$ represents the growth rate of bacteria colonies over time $t$ (in hours). For the growth rate to be continuous at $t = 2$, which condition must the constants $a$ and $b$ satisfy?
   Answer: $\frac{2a + b}{1} = 7$, which gives $2a + b = 7$ For continuity at $t = 2$: $\lim_{t \to 2^-} g(t) = g(2) = (2)^2 + 4(2) - 5 = 4 + 8 - 5 = 7$. $\lim_{t \to 2^+} g(t) = \frac{a(2) + b}{2 - 1} = \frac{2a + b}{1} = 2a + b$. For continuity: $7 = 2a + b$, so $2a + b = 7$. Choice C correctly states this condition and shows the substitution $\frac{2a + b}{1} = 7$. Choice A gives $a + b = 7$, which is insufficient. Choice B gives $a + b = 11$, which is incorrect. Choice D gives $2a + b = 11$, which doesn't match our calculation of 7.
5. A logistics company's shipping cost function transitions between two different rate structures at 50 pounds: $S(w) = \begin{cases} 2w + 15 & \text{if } 0 ≤ w < 50 \\ 1.5w + c & \text{if } w ≥ 50 \end{cases}$ where $w$ is weight in pounds. If the function is continuous at $w = 50$, but a customer ships exactly 50 pounds, which statement correctly describes the situation?
   Answer: The cost is $115 and the function is continuous but not differentiable First, find $c$ for continuity. $\lim_{w \to 50^-} S(w) = 2(50) + 15 = 115$. For continuity: $S(50) = 1.5(50) + c = 75 + c$. Setting $115 = 75 + c$ gives $c = 40$. So $S(50) = 1.5(50) + 40 = 115$. The cost for 50 pounds is $115. Check differentiability: Left derivative: $S'(50^-) = 2$. Right derivative: $S'(50^+) = 1.5$. Since $2 ≠ 1.5$, the function is not differentiable at $w = 50$. The function is continuous (we made it so) but not differentiable due to the change in slope. Choice A is wrong about discontinuity. Choice B has the wrong cost. Choice D claims differentiability, which is false.
6. The cost function is given by: $C(q) = \begin{cases} k + 5q & \text{if } 0 < q \le 50 \\ 2q + 200 & \text{if } q > 50 \end{cases}$ What must be the value of the special storage fee $k$ to ensure $C(q)$ is continuous at $q=50$?
   Answer: $k = 50$ For the function to be continuous at $q=50$, the left-hand limit and the right-hand limit must be equal. Left-hand limit: $\lim_{q\to50^-} C(q) = k + 5(50) = k + 250$. Right-hand limit: $\lim_{q\to50^+} C(q) = 2(50) + 200 = 100 + 200 = 300$. Set them equal: $k + 250 = 300$. Solving for $k$ gives $k = 300 - 250 = 50$.
7. The cost function is defined as: $C(w) = \begin{cases} 10 + 2.5w & \text{if } 0 < w \le 5 \\ 15 + 2w & \text{if } w > 5 \end{cases}$ Which of the following statements correctly analyzes the continuity of this function at $w=5$ and its business implication?
   Answer: The function has a jump discontinuity at $w=5$, creating a sudden price increase for packages just over 5 pounds. To analyze continuity at $w=5$, we check the value of the function and the one-sided limits. The value at $w=5$ is $C(5) = 10 + 2.5(5) = 22.5$. This is also the left-hand limit. The right-hand limit is $\lim_{w\to5^+} C(w) = 15 + 2(5) = 25$. Since the left-hand limit ($22.5) does not equal the right-hand limit ($25), there is a jump discontinuity. This implies that a package weighing slightly more than 5 pounds costs $25, a sudden increase from the $22.50 cost for a package weighing exactly 5 pounds.
8. The firm's cost function is: $C(h) = \begin{cases} 500h & \text{if } 0 < h \le 10 \\ k + 300h & \text{if } h > 10 \end{cases}$ To ensure a seamless transition in billing, what value of the fixed fee $k$ makes the cost function continuous at $h=10$?
   Answer: $k = 2000$ To ensure continuity at $h=10$, we must equate the left-hand and right-hand limits of $C(h)$. Left-hand limit: $\lim_{h\to10^-} C(h) = 500(10) = 5000$. Right-hand limit: $\lim_{h\to10^+} C(h) = k + 300(10) = k + 3000$. Setting the limits equal: $5000 = k + 3000$. Solving for $k$, we get $k = 5000 - 3000 = 2000$.
9. The earnings function is given by: $E(s) = \begin{cases} 0.04s & \text{if } 0 \le s \le 100,000 \\ B + 0.06s & \text{if } s > 100,000 \end{cases}$ To avoid a sudden drop or jump in pay, the earnings function must be continuous at the $100,000 threshold. What must the value of the bonus adjustment $B$ be?
   Answer: $B = -2000$ For continuity at $s=100,000$, the limits from both sides must be equal. Left-hand limit: $\lim_{s\to100000^-} E(s) = 0.04(100,000) = 4000$. Right-hand limit: $\lim_{s\to100000^+} E(s) = B + 0.06(100,000) = B + 6000$. Setting them equal: $4000 = B + 6000$. Solving for $B$ yields $B = 4000 - 6000 = -2000$. A negative value for $B$ is necessary to offset the jump caused by applying the higher 6% rate to the entire sales amount.
10. The current cost function is: $C(q) = \begin{cases} 10q & \text{if } 0 < q < 1000 \\ 8q & \text{if } q \ge 1000 \end{cases}$ To eliminate this illogical price drop, the company decides to add a fixed processing fee $F$ to all orders of 1,000 pounds or more, changing the second rule to $C(q) = 8q + F$. What value of $F$ makes the cost function continuous at $q=1,000$?
    Answer: $F = 2000$ When you encounter piecewise functions with discontinuities, the key to making them continuous is ensuring that the function values from both pieces match at the boundary point. For continuity at a point, the left-hand limit must equal the right-hand limit. Let's find what value of $F$ makes $C(q)$ continuous at $q = 1000$. First, calculate the cost just before the breakpoint using the first piece: $C(999) = 10(999) = 9990$. As $q$ approaches 1000 from the left, the cost approaches $10(1000) = 10000$. For continuity, the cost at $q = 1000$ using the second piece must equal this limit. With the processing fee, the second piece becomes $C(q) = 8q + F$. At $q = 1000$: $C(1000) = 8(1000) + F = 8000 + F$. Setting these equal: $8000 + F = 10000$, so $F = 2000$. Looking at the wrong answers: Choice A ($F = 2$) gives $C(1000) = 8002$, which is far too low. Choice B ($F = -2000$) actually makes the discontinuity worse by giving $C(1000) = 6000$, creating an even bigger price drop. Choice C ($F = 8000$) gives $C(1000) = 16000$, making it more expensive than the smaller quantity pricing. Choice D ($F = 2000$) correctly eliminates the discontinuity by making $C(1000) = 10000$. Study tip: For piecewise continuity problems, always evaluate the function at the boundary using both pieces and set them equal. The processing fee compensates for the lower per-unit rate at higher quantities.
11. The marginal cost is: $C'(q) = \begin{cases} 30 - 0.1q & \text{if } 0 \le q \le 100 \\ 15 & \text{if } q > 100 \end{cases}$ Given that the total cost function $C(q)$ is continuous and $C(0) = 1000$, which of the following is the correct total cost function?
    Answer: $C(q) = \begin{cases} 30q - 0.05q^2 + 1000 & \text{if } 0 \le q \le 100 \\ 15q + 2000 & \text{if } q > 100 \end{cases}$ First, find the total cost function for the first piece by integrating $C'(q)$: $C_1(q) = \int (30 - 0.1q) dq = 30q - 0.05q^2 + K$. Using $C(0)=1000$, we find $K=1000$. So, $C_1(q) = 30q - 0.05q^2 + 1000$. Next, find the cost at the boundary $q=100$: $C_1(100) = 30(100) - 0.05(100)^2 + 1000 = 3000 - 500 + 1000 = 3500$. Now, integrate the second piece: $C_2(q) = \int 15 dq = 15q + K_2$. For continuity, $C_2(100)$ must equal $C_1(100)$. So, $15(100) + K_2 = 3500$, which simplifies to $1500 + K_2 = 3500$. Solving gives $K_2 = 2000$. Therefore, the second piece is $15q + 2000$.
12. The profit model is given by: $P(t) = \begin{cases} 10\sqrt{t+4} & \text{if } 0 \le t < 12 \\ -0.1t^2 + 5t + 10.4 & \text{if } t \ge 12 \end{cases}$ Which statement provides the correct analysis of the profit model at $t=12$?
    Answer: The model has a jump discontinuity, implying the campaign caused an immediate, discrete increase in weekly profit. First, find the left-hand limit at $t=12$: $\lim_{t\to12^-} P(t) = 10\sqrt{12+4} = 10\sqrt{16} = 10(4) = 40$. Next, find the function value at $t=12$, which is also the right-hand limit: $P(12) = -0.1(12)^2 + 5(12) + 10.4 = -0.1(144) + 60 + 10.4 = -14.4 + 70.4 = 56$. Since the left-hand limit (40) is not equal to the right-hand limit (56), there is a jump discontinuity. The value jumps up, indicating an immediate increase in profit from $40,000 to $56,000 per week.
13. The cost function is given by: $C(x) = \begin{cases} 20x + 1000 & \text{if } 0 < x \le 100 \\ ax + b & \text{if } 100 < x \le 500 \\ 30x - 5000 & \text{if } x > 500 \end{cases}$ If the cost function is continuous for all $x > 0$, what is the value of the marginal cost $a$ for the intermediate batch size?
    Answer: $a = 17.5$ For continuity, the pieces must meet at the boundaries. At $x=100$: $20(100) + 1000 = a(100) + b \implies 3000 = 100a + b$. At $x=500$: $a(500) + b = 30(500) - 5000 \implies 500a + b = 15000 - 5000 = 10000$. We now have a system of two linear equations: 1) $100a + b = 3000$ 2) $500a + b = 10000$ Subtract equation (1) from equation (2): $(500a - 100a) + (b - b) = 10000 - 3000$, which gives $400a = 7000$. Solving for $a$: $a = \frac{7000}{400} = \frac{70}{4} = 17.5$.
14. The total cost for a one-day rental, $C(m)$, after driving $m$ miles is: $C(m) = \begin{cases} 50 + 0.40m & \text{if } 0 \le m \le 150 \\ 50 + 0.25m + S & \text{if } m > 150 \end{cases}$ To ensure the cost function is continuous, a 'high-mileage' surcharge $S$ is added to the second tier. What must be the value of $S$?
    Answer: $S = 22.50$ When you encounter a piecewise function in business calculus, continuity is crucial for real-world applications. A discontinuous pricing function would create unfair jumps in cost, so you need to ensure the function's value is the same at the boundary point regardless of which piece you use. To find the required surcharge $S$, you must set the two pieces equal at the boundary point $m = 150$. Using the first piece: $C(150) = 50 + 0.40(150) = 50 + 60 = 110$. Using the second piece: $C(150) = 50 + 0.25(150) + S = 50 + 37.5 + S = 87.5 + S$. For continuity, these must be equal: $110 = 87.5 + S$, which gives us $S = 110 - 87.5 = 22.50$. Looking at the wrong answers: Choice A ($S = -22.50$) represents the negative of the correct value, likely from incorrectly subtracting in the opposite direction. Choice B ($S = 0.15$) might come from confusing the difference in per-mile rates (0.40 - 0.25 = 0.15) with the surcharge needed. Choice C ($S = 37.50$) is simply the mileage cost from the second piece at $m = 150$, ignoring the flat fee entirely. Remember: for piecewise functions to be continuous, evaluate both pieces at the boundary point and set them equal. This ensures no sudden jumps in pricing, which is essential for fair business practices and realistic mathematical models.
15. The pricing model is given by the function: $C(x) = \begin{cases} 20x & \text{if } 1 \le x \le 50 \\ k + 15x & \text{if } x > 50 \end{cases}$ What value must the enterprise fee $k$ be for the cost function $C(x)$ to be continuous at $x=50$?
    Answer: $k = 250$ For $C(x)$ to be continuous at $x=50$, the left-hand limit must equal the right-hand limit. Left-hand limit: $\lim_{x\to50^-} C(x) = 20(50) = 1000$. Right-hand limit: $\lim_{x\to50^+} C(x) = k + 15(50) = k + 750$. Set the limits equal to each other: $1000 = k + 750$. Solving for $k$ gives $k = 1000 - 750 = 250$.
16. Consider the piecewise function $h(x) = \begin{cases} \frac{2x^2 - 8}{x - 2} & \text{if } x ≠ 2 \\ m & \text{if } x = 2 \end{cases}$ Which statement about the continuity of $h(x)$ at $x = 2$ is correct?
    Answer: $h(x)$ is continuous at $x = 2$ only when $m = 8$ When you encounter a piecewise function with a potential discontinuity, you need to check three conditions for continuity: the function must be defined at the point, the limit must exist, and the limit must equal the function value. First, let's find the limit as $x$ approaches 2. The expression $\frac{2x^2 - 8}{x - 2}$ appears to give $\frac{0}{0}$ when $x = 2$, so we need to simplify. Factor the numerator: $2x^2 - 8 = 2(x^2 - 4) = 2(x - 2)(x + 2)$. Therefore: $\frac{2x^2 - 8}{x - 2} = \frac{2(x - 2)(x + 2)}{x - 2} = 2(x + 2)$ for $x ≠ 2$. Now $\lim_{x \to 2} h(x) = \lim_{x \to 2} 2(x + 2) = 2(2 + 2) = 8$. For continuity at $x = 2$, we need $h(2) = \lim_{x \to 2} h(x)$, which means $m = 8$. Looking at the choices: A is correct because continuity requires $m = 8$. B is wrong because continuity depends on a specific value of $m$, not any value. C incorrectly claims $m = 4$, but our limit calculation shows the limit equals 8, not 4. D is wrong because we can indeed make the function continuous by choosing the right value of $m$. Study tip: For removable discontinuities (holes in graphs), always factor and cancel common terms to find the limit, then set the function value equal to that limit for continuity.
17. The tax function is: $T(i) = \begin{cases} 0.15i & \text{if } 0 \le i \le 40,000 \\ A + 0.25(i - 40,000) & \text{if } i > 40,000 \end{cases}$ In this function, $A$ represents the total tax paid on income from all lower brackets. What must be the value of $A$ for the tax function to be continuous at $i=40,000$?
    Answer: $A = 6000$ When you encounter a piecewise function that needs to be continuous, you're looking for the point where both pieces give the same output value. Continuity means there's no sudden jump or gap in the function. To find the value of $A$, you need to ensure that both pieces of the tax function give the same result at the boundary point $i = 40,000$. From the left side (using the first piece), the tax at $i = 40,000$ is: $T(40,000) = 0.15 \times 40,000 = 6,000$ From the right side (using the second piece), the tax approaches: $T(40,000) = A + 0.25(40,000 - 40,000) = A + 0$ For continuity, these must be equal: $6,000 = A$. Therefore, $A = 6,000$. Looking at the wrong answers: Choice (A) suggests $A = 0$, which would create a discontinuous drop from $6,000 to $0 at the boundary—clearly violating continuity. Choice (B) gives $A = 4,000$, which would mean the tax suddenly drops from $6,000 to $4,000 when crossing into the higher bracket, creating a "tax cliff" that defeats the system's purpose. Choice (C) proposes $A = 10,000$, which would cause an immediate jump from $6,000 to $10,000—again breaking continuity. Remember this pattern: for piecewise functions to be continuous, evaluate both pieces at the boundary point and set them equal. This technique appears frequently in business calculus when modeling real-world scenarios like tax systems, pricing structures, or cost functions that change at certain thresholds.

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