Business Calculus Quiz: Intermediate Value Theorem
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Intermediate Value TheoremQuestion 1 of 16

A company's weekly profit from selling xx units, P(x)P(x), is a continuous function. An analysis shows that $P(500) = -$2,000 and $P(1000) = $3,500. What does the Intermediate Value Theorem guarantee regarding the company's break-even point (where profit is $0) on the production interval from 500 to 1000 units?

There is exactly one break-even point between 500 and 1000 units.
There is at least one break-even point between 500 and 1000 units.
The break-even point must occur at a production level greater than 750 units.
A break-even point cannot be guaranteed because profit might not be a monotonic function.
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Business Calculus Quiz

Business Calculus Quiz: Intermediate Value Theorem

Practice Intermediate Value Theorem in Business Calculus with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Intermediate Value Theorem, giving you a quick way to practice the rules, question types, and explanations that matter most for Business Calculus.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

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Question 1

A company's weekly profit from selling xx units, P(x)P(x), is a continuous function. An analysis shows that $P(500) = -$2,000 and $P(1000) = $3,500. What does the Intermediate Value Theorem guarantee regarding the company's break-even point (where profit is $0) on the production interval from 500 to 1000 units?

  1. There is exactly one break-even point between 500 and 1000 units.
  2. There is at least one break-even point between 500 and 1000 units. (correct answer)
  3. The break-even point must occur at a production level greater than 750 units.
  4. A break-even point cannot be guaranteed because profit might not be a monotonic function.
Explanation: The Intermediate Value Theorem states that if a function P(x)P(x) is continuous on a closed interval [a,b][a, b], and NN is any number between P(a)P(a) and P(b)P(b), then there is at least one number cc in (a,b)(a, b) such that P(c)=NP(c) = N. Here, the function P(x)P(x) is continuous on [500,1000][500, 1000], P(500)=−$2,000,and$P(1000)=$3,500.Sincethebreak−evenvalue$N=0P(500) = -$2,000, and $P(1000) = $3,500. Since the break-even value $N=0 is between -$2,000 and $3,500, the theorem guarantees there is at least one production level $cinin(500, 1000)wherewhereP(c) = 0$.

Question 2

The total cost C(q)C(q) in dollars to produce qq widgets is a continuous function. The company has a production budget of $50,000. For the current month, it is known that $C(1000) = $48,000 and $C(1500) = $53,000. Which of the following conclusions is a valid application of the Intermediate Value Theorem?

  1. There must be a production level cc between 1000 and 1500 widgets for which the total cost is exactly $50,000. (correct answer)
  2. The marginal cost of production must be positive for all qq between 1000 and 1500.
  3. There must be a production level cc between 1000 and 1500 widgets for which the total cost is exactly $47,000.
  4. The cost function must be linear on the interval [1000,1500][1000, 1500].
Explanation: The cost function C(q)C(q) is continuous on the interval [1000,1500][1000, 1500]. The cost at the endpoints are C(1000)=$48,000and$C(1500)=$53,000.Thetargetbudgetvalue$N=$50,000isbetweenthesetwovalues.Therefore,theIntermediateValueTheoremguaranteesthatthereisatleastoneproductionlevel$cC(1000) = $48,000 and $C(1500) = $53,000. The target budget value $N = $50,000 is between these two values. Therefore, the Intermediate Value Theorem guarantees that there is at least one production level $c in the open interval (1000,1500)(1000, 1500) such that $C(c) = $50,000.

Question 3

A company's marginal profit function, MP(x)=P′(x)MP(x) = P'(x), is continuous, where xx is the number of units produced. It is found that MP(1000)=$5(profitisincreasing)and$MP(2000)=−$2(profitisdecreasing).WhatisthemostdirectconclusionfromtheIntermediateValueTheoremasappliedtothefunction$MP(x)MP(1000) = $5 (profit is increasing) and $MP(2000) = -$2 (profit is decreasing). What is the most direct conclusion from the Intermediate Value Theorem as applied to the function $MP(x)?

  1. The total profit P(x)P(x) must be $0 for some production level $cinin(1000, 2000)$.
  2. The total profit is maximized at some production level cc outside the interval (1000,2000)(1000, 2000).
  3. There exists a production level cc in (1000,2000)(1000, 2000) where MP(c)=0MP(c) = 0. (correct answer)
  4. The marginal profit must be $0 at either $x=1000ororx=2000$.
Explanation: When you encounter a marginal profit function question involving the Intermediate Value Theorem, focus on what the theorem actually tells us about continuous functions crossing specific values. The Intermediate Value Theorem states that if a function is continuous on an interval [a,b][a,b] and takes on values f(a)f(a) and f(b)f(b) with opposite signs, then there must exist some point cc between aa and bb where f(c)=0f(c) = 0. Here, MP(x)MP(x) is continuous, MP(1000)=5MP(1000) = 5 (positive), and MP(2000)=−2MP(2000) = -2 (negative). Since the marginal profit changes from positive to negative, it must cross zero at some production level cc in the interval (1000,2000)(1000, 2000). This is exactly what answer C states. Answer A incorrectly confuses marginal profit MP(x)=P′(x)MP(x) = P'(x) with total profit P(x)P(x). The theorem applies to the marginal profit function, not the total profit itself. Answer B makes an unsupported claim about profit maximization outside the given interval—the theorem doesn't tell us anything about points outside (1000,2000)(1000, 2000). Answer D misapplies the theorem by suggesting it requires MP(x)=0MP(x) = 0 at the endpoints, when we're actually given that MP(1000)=5MP(1000) = 5 and MP(2000)=−2MP(2000) = -2. Remember: The Intermediate Value Theorem is your tool for proving that continuous functions must hit specific values when they change from one side to another. Look for sign changes to identify where functions cross zero.

Question 4

The average cost per unit, AC(q)AC(q), for producing qq items is a continuous function for q>0q > 0. A factory manager observes that $AC(200) = $35 and $AC(600) = $25. The company has set a target to achieve an average cost of $30 per unit. What does the Intermediate Value Theorem imply about this target?

  1. The target average cost of $30 must be achieved at some production level $c$ between 200 and 600 units. (correct answer)
  2. The total cost, C(q)=q⋅AC(q)C(q) = q \cdot AC(q), must equal $30 for some production level $c$ in the interval.
  3. The average cost must be a strictly decreasing function on the interval (200,600)(200, 600).
  4. The target average cost of $30 can only be guaranteed if total cost is also between $25 and $35.
Explanation: The function under consideration is the average cost, AC(q)AC(q), which is continuous on [200,600][200, 600]. The values at the endpoints are AC(200)=35AC(200) = 35 and AC(600)=25AC(600) = 25. The target value N=30N=30 is between 2525 and 3535. By the Intermediate Value Theorem, there must be some production level cc in (200,600)(200, 600) where AC(c)=30AC(c) = 30.

Question 5

Let V(t)V(t) represent the value of a commercial real estate property, in millions of dollars, tt years after its purchase. Assume V(t)V(t) is a continuous function. The property was purchased for 2.5million(2.5 million (V(0)=2.5)andwasappraised10yearslaterfor) and was appraised 10 years later for 4.5 million (V(10)=4.5V(10)=4.5). Which of the following statements describes a situation that is impossible, given the conditions for the Intermediate Value Theorem?

  1. The property's value dipped to $2.4 million in year 3 before increasing.
  2. The property's value reached exactly $3.5 million in year 4 and again in year 8.
  3. The property's value never reached exactly $4.0 million at any time $tintheintervalin the interval(0, 10)$. (correct answer)
  4. The property was valued at $5.0 million in year 6 before its value decreased by year 10.
Explanation: The Intermediate Value Theorem guarantees that a continuous function on [a,b][a, b] must take on every value between f(a)f(a) and f(b)f(b). Here, the function V(t)V(t) is continuous on [0,10][0, 10], with V(0)=2.5V(0)=2.5 and V(10)=4.5V(10)=4.5. The value 4.04.0 million is between 2.52.5 and 4.54.5 million. Therefore, it is guaranteed that the property's value was exactly $4.0 million at some point. The statement that it never reached this value is impossible. The other options are possible because the IVT does not prevent the function from going outside the range of its endpoints or from hitting a value multiple times.

Question 6

The cost C(x)C(x) to produce xx units of a good is C(x)=10xC(x) = 10x for x≤1000x \le 1000. If production exceeds 1000 units, the company receives a bulk discount on materials, and the cost function becomes C(x)=5000+8xC(x) = 5000 + 8x for x>1000x > 1000. We know C(800)=8000C(800) = 8000 and C(1200)=14600C(1200) = 14600. Can we use the Intermediate Value Theorem to conclude that there is a production level c∈(800,1200)c \in (800, 1200) for which the cost is exactly $12,000?

  1. Yes, because C(x)C(x) is defined for all xx in [800,1200][800, 1200] and $12,000 is between $8,000 and $14,600.
  2. Yes, the IVT guarantees a value cc exists, although it does not provide a method to find it.
  3. No, because the value $12,000 is never actually achieved; the cost jumps from $10,000 to $13,000.
  4. No, because the function C(x)C(x) has a jump discontinuity at x=1000x=1000, so the theorem's conditions are not met. (correct answer)
Explanation: The Intermediate Value Theorem is a powerful tool in calculus, but it requires a crucial condition: the function must be continuous on the closed interval you're examining. When you encounter piecewise functions like this cost function, always check for continuity at the boundary points. Let's examine what happens at x=1000x = 1000. From the left, using C(x)=10xC(x) = 10x, we get C(1000)=10(1000)=10,000C(1000) = 10(1000) = 10,000. From the right, using C(x)=5000+8xC(x) = 5000 + 8x, we get C(1000)=5000+8(1000)=13,000C(1000) = 5000 + 8(1000) = 13,000. Since the left-hand limit (10,000) doesn't equal the right-hand limit (13,000), there's a jump discontinuity at x=1000x = 1000. This breaks the continuity requirement for the Intermediate Value Theorem. Choice A incorrectly assumes that being defined everywhere is sufficient—continuity requires more than just being defined. Choice B makes the same error, ignoring the discontinuity entirely. Choice C correctly identifies that 12,000 isn't achieved and mentions the jump from 10,000 to 13,000, but focuses on the conclusion rather than the fundamental reason the theorem doesn't apply. Choice D correctly identifies that the jump discontinuity violates the theorem's conditions. Study tip: Whenever applying the Intermediate Value Theorem to piecewise functions, first verify continuity at all boundary points. A function can be defined everywhere but still fail the theorem due to jump discontinuities. Always check that left-hand and right-hand limits match at transition points.

Question 7

A software license costs $50 per user for up to 20 users. If a company buys more than 20 licenses, the price for all licenses drops to $40 per user. The total cost function $C(n)forfornusersisthereforediscontinuousatusers is therefore discontinuous atn=20.Acompanyobservesthat. A company observes that C(15) = $750 and $C(25) = $1000. Can the Intermediate Value Theorem be used to guarantee a number of licenses $n$ between 15 and 25 for which the total cost is exactly $900?

  1. Yes, because $900 is between $750 and $1000, there must be a number of licenses $n$ that costs exactly $900.
  2. No, because the cost function C(n)C(n) is not continuous on the interval [15,25][15, 25]. (correct answer)
  3. No, because the number of licenses nn must be an integer, which violates a condition of the theorem.
  4. Yes, but only if the average cost per license is also continuous, which is not stated.
Explanation: The Intermediate Value Theorem requires the function to be continuous on the entire closed interval. The cost function C(n)C(n) has a jump discontinuity at n=20n=20, which is inside the interval [15,25][15, 25]. Because a key condition of the theorem is not met, it cannot be used to guarantee that any specific cost value is achieved within the interval.

Question 8

The number of daily users U(p)U(p) of a mobile app is a continuous function of the daily subscription price pp in cents. Market research shows that U(25)=10,000U(25) = 10,000 users and U(50)=4,000U(50) = 4,000 users. Applying the Intermediate Value Theorem to the interval p∈[25,50]p \in [25, 50], what is the correct interpretation of its conclusion?

  1. For any price cc between 25 and 50 cents, the number of users U(c)U(c) will be between 4,000 and 10,000.
  2. For any target number of users NN between 4,000 and 10,000, there exists a price cc between 25 and 50 cents that results in NN users. (correct answer)
  3. The daily revenue, R(p)=p⋅U(p)R(p) = p \cdot U(p), must take on every value between R(25)R(25) and R(50)R(50).
  4. The number of users must decrease linearly as the price increases from 25 to 50 cents.
Explanation: The Intermediate Value Theorem is applied to the continuous function U(p)U(p) on the interval [25,50][25, 50]. The theorem states that for any value NN between U(25)U(25) and U(50)U(50), there must exist a cc in the domain interval (25,50)(25, 50) such that U(c)=NU(c)=N. Here, U(25)=10,000U(25)=10,000 and U(50)=4,000U(50)=4,000. So, for any target user number NN between 4,000 and 10,000, there is a price cc between 25 and 50 cents that will achieve it. Choice A is a common misinterpretation; the theorem does not bound the function's range.

Question 9

A consulting firm models a project's cumulative profit P(t)P(t) (in thousands of dollars) over tt months as a continuous polynomial function. They calculate the following values: P(0)=−50P(0) = -50, P(6)=20P(6) = 20, and P(12)=−10P(12) = -10. What is the minimum number of times the project must have broken even (i.e., P(t)=0P(t) = 0) during the first 12 months, according to the Intermediate Value Theorem?

  1. Exactly once.
  2. Exactly twice.
  3. At least once, but it is impossible to determine if there was more than one.
  4. At least twice. (correct answer)
Explanation: When you encounter a continuous function with given values at specific points, the Intermediate Value Theorem (IVT) becomes your key tool for determining where the function must cross certain values—like zero for break-even points. The IVT states that if a continuous function takes on two different values, it must take on every value in between at least once. Here, you have P(0)=−50P(0) = -50, P(6)=20P(6) = 20, and P(12)=−10P(12) = -10. Since P(t)P(t) is continuous and goes from −50-50 at t=0t=0 to +20+20 at t=6t=6, it must cross zero at least once in the interval [0,6][0,6]. Similarly, since P(t)P(t) goes from +20+20 at t=6t=6 to −10-10 at t=12t=12, it must cross zero at least once in the interval [6,12][6,12]. This guarantees at least two break-even points during the 12 months. Answer A is wrong because there must be more than one crossing—the function changes sign twice. Answer B is incorrect because while there are at least two crossings, there could be more (the polynomial could oscillate and cross zero additional times between the given points). Answer C misses that you can definitively conclude there are at least two crossings, not just one. Answer D correctly identifies that there must be at least two break-even points based on the sign changes. Study tip: When applying IVT, count the sign changes between consecutive points—each sign change guarantees at least one zero crossing in that interval.

Question 10

The monthly revenue of a startup, R(t)R(t), is a continuous function of time tt (in months since launch). At t=3t=3 months, the revenue was $150,000. At $t=6months,therevenuewas$120,000.BasedontheIntermediateValueTheorem,whichofthefollowingrevenuegoalsisguaranteedtohavebeenmetatsometime$t months, the revenue was $120,000. Based on the Intermediate Value Theorem, which of the following revenue goals is guaranteed to have been met at some time $t within the interval (3,6)(3, 6)?

  1. A monthly revenue of $160,000.
  2. A monthly revenue of $115,000.
  3. A monthly revenue of $135,000. (correct answer)
  4. A peak monthly revenue greater than $150,000.
Explanation: When you encounter questions about continuous functions with given values at specific points, think about the Intermediate Value Theorem (IVT). This powerful theorem states that if a function is continuous on a closed interval, it must take on every value between its endpoint values at least once within that interval. Here, you know R(t)R(t) is continuous and that R(3)=150,000R(3) = 150,000 and R(6)=120,000R(6) = 120,000. Since the revenue decreased from $150,000 to $120,000 over the interval $(3,6)(3,6) ,theIVTguaranteesthatthefunctionmustpassthrougheveryvaluebetweentheseendpoints.Anyrevenuebetween, the IVT guarantees that the function must pass through every value between these endpoints. Any revenue between 120,000 and $150,000 must occur at some point during months 3 through 6. Looking at choice C, 135,000fallsdirectlybetween135,000 falls directly between 120,000 and $150,000, so the IVT guarantees this revenue level was achieved at some time tt in the interval. Choice A ($160,000) is incorrect because this value is above the starting point of 150,000.TheIVTonlyguaranteesvaluesbetweentheendpoints,notbeyondthem.ChoiceB(150,000. The IVT only guarantees values between the endpoints, not beyond them. Choice B (115,000) fails because it's below the ending value of $120,000 – again, outside the guaranteed range. Choice D is wrong because the IVT makes no promises about peaks or maximum values, only that intermediate values are achieved. Remember: the Intermediate Value Theorem is your tool for "guarantee" questions involving continuous functions. It only applies to values strictly between the given endpoints – anything outside that range requires additional information about the function's behavior.

Question 11

The net asset value of a company, V(t)V(t) in millions of dollars, is a continuous function of time tt in years. It is known that V(0)=10V(0) = 10 and V(4)=8V(4) = 8. A financial report states that at some time t0t_0 in the interval (0,4)(0, 4), the net asset value was exactly 99 million. Which of the following conditions is sufficient to guarantee this statement is true, but is not strictly necessary?

  1. V(t)V(t) is continuous on the interval [0,4][0, 4].
  2. The average rate of change of V(t)V(t) on [0,4][0, 4] is negative.
  3. V(t)V(t) is differentiable on the interval (0,4)(0, 4).
  4. V(t)V(t) is strictly decreasing on the interval [0,4][0, 4]. (correct answer)
Explanation: This question tests your understanding of the Intermediate Value Theorem and what conditions are sufficient versus necessary to guarantee certain outcomes. The financial report claims that V(t0)=9V(t_0) = 9 for some t0t_0 in (0,4)(0,4). Since we know V(0)=10V(0) = 10 and V(4)=8V(4) = 8, and 9 lies between these values, the Intermediate Value Theorem tells us this claim is true if V(t)V(t) is continuous. But the question asks for a condition that's sufficient but not strictly necessary. Answer D is correct because if V(t)V(t) is strictly decreasing on [0,4][0,4], then it must pass through every value between 10 and 8 exactly once. This guarantees there's a unique t0t_0 where V(t0)=9V(t_0) = 9. However, strict monotonicity is stronger than needed—continuity alone would suffice. Answer A is incorrect because continuity is necessary, not just sufficient but unnecessary. The Intermediate Value Theorem requires continuity as a minimum condition. Answer B is wrong because knowing the average rate of change is 8−104−0=−0.5\frac{8-10}{4-0} = -0.5 tells us nothing about whether the function actually hits 9. The function could jump discontinuously from above 9 to below 9. Answer C is incorrect because differentiability, while implying continuity, is still a necessary condition in this context. It's not stronger than what's required. Remember: when questions ask for "sufficient but not necessary" conditions, look for the strongest option that guarantees the result but goes beyond the minimum requirements.

Question 12

A city's water consumption model W(h)=50+30sin⁡(πh12)W(h) = 50 + 30\sin(\frac{\pi h}{12}) represents thousands of gallons per hour, where hh is hours after midnight. Environmental regulations require consumption to stay below 65,000 gallons per hour. If consumption at 6 AM was 50,000 gallons per hour and at noon was 80,000 gallons per hour, what does the Intermediate Value Theorem reveal about regulation compliance?

  1. Regulations were violated exactly once between 6 AM and noon since the sine function is strictly monotonic
  2. Regulations may have been violated, but the theorem cannot determine when without derivative analysis
  3. Regulations were violated at some point between 6 AM and noon, but timing cannot be determined (correct answer)
  4. Regulations were not violated because average consumption equals the regulation limit exactly
Explanation: W(h)W(h) is continuous everywhere. At 6 AM (h=6h=6): W(6)=50<65W(6) = 50 < 65. At noon (h=12h=12): W(12)=80>65W(12) = 80 > 65. Since 6565 lies between 5050 and 8080, IVT guarantees at least one time c∈(6,12)c \in (6,12) where W(c)=65W(c) = 65, meaning regulations were violated. Choice A incorrectly assumes exactly once and strict monotonicity. Choice B incorrectly suggests the theorem can't determine violation occurrence. Choice D confuses average values with the IVT application.

Question 13

A manufacturing cost function C(q)=2000+50q−0.1q2+0.001q3C(q) = 2000 + 50q - 0.1q^2 + 0.001q^3 represents total cost (in dollars) for producing qq units. Quality control requires identifying if production costs ever equal exactly $4,500 during the range of 20 to 40 units. Given $C(20)=3,980C(20) = 3,980 ,, C(30)=4,700C(30) = 4,700 ,and, and C(40)=5,800C(40) = 5,800 $, what does the Intermediate Value Theorem indicate?

  1. Production costs equal $4,500 at exactly one production level between 20 and 30 units since the cost function is strictly increasing
  2. The theorem cannot be applied because manufacturing costs have inherent discontinuities due to fixed setup costs and batch processing
  3. Production costs equal $4,500 at some production level between 20 and 30 units and definitely not between 30 and 40 units since 4,500 < 4,700
  4. Production costs equal $4,500 at some production level between 20 and 30 units, but the theorem cannot determine if this occurs elsewhere (correct answer)
Explanation: When you encounter questions about the Intermediate Value Theorem (IVT) in business contexts, remember that this theorem tells us about the existence of solutions, not their uniqueness or exact locations. The IVT states that if a function is continuous on an interval and takes on two different values at the endpoints, then it must take on every value between those endpoints at least once. The cost function C(q)=2000+50q−0.1q2+0.001q3C(q) = 2000 + 50q - 0.1q^2 + 0.001q^3 is a polynomial, making it continuous everywhere. Since C(20)=3,980<4,500<4,700=C(30)C(20) = 3,980 < 4,500 < 4,700 = C(30), the IVT guarantees that C(q)=4,500C(q) = 4,500 for some value between 20 and 30 units. However, the theorem only tells us that at least one such point exists—it doesn't rule out additional points elsewhere in the domain. Answer A incorrectly assumes the cost function is strictly increasing. While costs might increase overall, this particular function could have local fluctuations due to its cubic term. Answer B misapplies mathematical concepts—polynomial functions are continuous regardless of real-world manufacturing considerations. Answer C makes a logical error by claiming the theorem rules out solutions between 30 and 40 units; since C(30)=4,700>4,500C(30) = 4,700 > 4,500, we'd need to check if the function dips back down to $4,500 somewhere beyond 30 units, which the IVT cannot determine from the given information. Study tip: The IVT guarantees existence but never uniqueness. Always distinguish between "at least one solution exists" versus "exactly one solution exists."

Question 14

A logistics company's delivery efficiency E(r)=r3−6r2+11r−6r−2E(r) = \frac{r^3 - 6r^2 + 11r - 6}{r - 2} depends on route optimization parameter rr. The function appears undefined at r=2r = 2, but analysis shows lim⁡r→2E(r)=3\lim_{r \to 2} E(r) = 3. If efficiency measurements show E(1.5)=0.75E(1.5) = 0.75 and E(2.5)=4.75E(2.5) = 4.75, and the company needs exactly 3.5 efficiency units, what can be concluded?

  1. The target efficiency is achieved at least once in the interval (1.5,2.5)(1.5, 2.5) if we define E(2)=3E(2) = 3 to make the function continuous (correct answer)
  2. The target efficiency cannot be achieved because the function is discontinuous at r=2r = 2, preventing application of the Intermediate Value Theorem
  3. The target efficiency is achieved exactly twice: once in (1.5,2)(1.5, 2) and once in (2,2.5)(2, 2.5) due to the removable discontinuity
  4. The Intermediate Value Theorem cannot be applied to rational functions with discontinuities, regardless of limit behavior
Explanation: When you encounter a rational function with an apparent discontinuity, the key is determining whether the discontinuity is removable and how this affects the Intermediate Value Theorem's application. First, let's analyze what we have: E(r)E(r) appears undefined at r=2r = 2, but lim⁡r→2E(r)=3\lim_{r \to 2} E(r) = 3. This suggests a removable discontinuity - essentially a "hole" in the graph that can be filled by defining E(2)=3E(2) = 3. When we do this, the function becomes continuous on the interval [1.5,2.5][1.5, 2.5]. With E(1.5)=0.75E(1.5) = 0.75, E(2)=3E(2) = 3, and E(2.5)=4.75E(2.5) = 4.75, we have a continuous function where the target value 3.53.5 lies between E(2)=3E(2) = 3 and E(2.5)=4.75E(2.5) = 4.75. By the Intermediate Value Theorem, there must be at least one point in (2,2.5)(2, 2.5) where E(r)=3.5E(r) = 3.5. Since we're looking at the entire interval (1.5,2.5)(1.5, 2.5), answer A is correct. B is wrong because removable discontinuities don't prevent IVT application once the function is made continuous. C incorrectly assumes the function achieves 3.53.5 twice without sufficient evidence - we only know it occurs at least once in (2,2.5)(2, 2.5), not in (1.5,2)(1.5, 2) since 3.5>33.5 > 3. D makes a false blanket statement about rational functions - the IVT can apply when discontinuities are removable. Study tip: Remember that removable discontinuities (where a limit exists but the function isn't defined) can often be "fixed" to apply continuity theorems. Always check if the target value falls within your function's range on the given interval.

Question 15

A startup's valuation function V(m)=m3−12m2+36m+100V(m) = m^3 - 12m^2 + 36m + 100 gives company value (in millions) after mm months of operation. Investors want to know if the company will ever be valued at exactly $180 million during months 2 through 6. Given $V(2)=148V(2) = 148 ,, V(4)=164V(4) = 164 ,and, and V(6)=196V(6) = 196 $, which analysis using the Intermediate Value Theorem is correct?

  1. The valuation reaches $180 million exactly once between months 4 and 6 since $164<180<196164 < 180 < 196 $ and the function is continuous
  2. The valuation reaches $180 million at least once between months 4 and 6, and potentially multiple times across the entire interval [2,6] (correct answer)
  3. The theorem cannot be applied because we need the function values at integer month points only, not continuous monitoring
  4. The valuation never reaches exactly $180 million because the given values show the function is always increasing on [2,6]
Explanation: Since V(m)V(m) is polynomial (continuous), and V(4)=164<180<196=V(6)V(4) = 164 < 180 < 196 = V(6), IVT guarantees at least one c∈(4,6)c \in (4,6) where V(c)=180V(c) = 180. However, this cubic function could have local extrema, potentially crossing y=180y = 180 elsewhere in [2,6][2,6]. Choice A incorrectly assumes uniqueness in the subinterval. Choice C misunderstands continuity requirements. Choice D incorrectly concludes the function is always increasing from three points.

Question 16

A pharmaceutical company models drug concentration in the bloodstream with C(t)=40tt2+4C(t) = \frac{40t}{t^2 + 4} mg/L, where tt is hours after injection. Clinical guidelines require determining if the concentration reaches exactly 6 mg/L during the critical first 8 hours. Given that C(0)=0C(0) = 0, C(2)=10C(2) = 10, and C(8)=4.7C(8) = 4.7, what can be concluded using the Intermediate Value Theorem?

  1. The concentration reaches exactly 6 mg/L twice during the 8-hour period since the function increases then decreases
  2. The concentration reaches exactly 6 mg/L at least once between hours 2 and 8, but frequency cannot be determined (correct answer)
  3. The theorem cannot determine if 6 mg/L is reached because the function has a vertical asymptote
  4. The concentration reaches exactly 6 mg/L at most once because rational functions cross horizontal lines once
Explanation: The function C(t)=40tt2+4C(t) = \frac{40t}{t^2 + 4} is continuous on [0,8][0,8] (no vertical asymptotes since t2+4>0t^2 + 4 > 0 for all real tt). Since C(2)=10>6C(2) = 10 > 6 and C(8)=4.7<6C(8) = 4.7 < 6, IVT guarantees at least one c∈(2,8)c \in (2,8) where C(c)=6C(c) = 6. Choice A incorrectly assumes exactly twice. Choice C incorrectly claims a discontinuity. Choice D makes a false generalization about rational functions. The theorem guarantees existence but not uniqueness.