Business Calculus Quiz: Functions Domain And Range
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Functions Domain And RangeQuestion 1 of 20

A logistics company models delivery time as T(d)=2+dd+5T(d) = 2 + \frac{d}{d+5} hours, where dd is distance in miles. If the company only accepts orders within a 3-hour delivery window, and d≥0d \geq 0, what values can T(d)T(d) take?

[2,3][2, 3] since T(0)=2T(0) = 2 and we're limited by the 3-hour window
(2,3](2, 3] because the function never actually reaches its horizontal asymptote
[2,3)[2, 3) since dd+5\frac{d}{d+5} approaches but never equals 1
[0,3][0, 3] representing all possible delivery times within the constraint
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Business Calculus Quiz

Business Calculus Quiz: Functions Domain And Range

Practice Functions Domain And Range 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 Functions Domain And Range, giving you a quick way to practice the rules, question types, and explanations that matter most for Business Calculus.

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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 logistics company models delivery time as T(d)=2+dd+5T(d) = 2 + \frac{d}{d+5} hours, where dd is distance in miles. If the company only accepts orders within a 3-hour delivery window, and d≥0d \geq 0, what values can T(d)T(d) take?

  1. [2,3][2, 3] since T(0)=2T(0) = 2 and we're limited by the 3-hour window
  2. (2,3](2, 3] because the function never actually reaches its horizontal asymptote
  3. [2,3)[2, 3) since dd+5\frac{d}{d+5} approaches but never equals 1 (correct answer)
  4. [0,3][0, 3] representing all possible delivery times within the constraint
Explanation: As d→∞d \to \infty, dd+5=11+5d→1\frac{d}{d+5} = \frac{1}{1+\frac{5}{d}} \to 1, so T(d)→3T(d) \to 3. At d=0d = 0, T(0)=2+0=2T(0) = 2 + 0 = 2. Since dd+5\frac{d}{d+5} is increasing and bounded above by 1 (never reaching it), T(d)T(d) increases from 2 toward 3 but never reaches 3. The 3-hour constraint is satisfied since T(d)<3T(d) < 3 always. Choice A incorrectly includes 3. Choice B uses wrong interval notation. Choice D ignores that T(d)T(d) starts at 2, not 0.

Question 2

A company's profit function is defined as P(x)=25−x2P(x) = \sqrt{25 - x^2} thousand dollars, where xx represents the number of units produced (in hundreds). If the company must produce at least 200 units to meet contractual obligations, what is the domain of P(x)P(x) given this business constraint?

  1. [2,5][2, 5] (correct answer)
  2. [−5,5][-5, 5] excluding values less than 2
  3. [2,5)[2, 5) due to the open interval at the boundary
  4. [−5,−2]∪[2,5][-5, -2] \cup [2, 5] accounting for all mathematical solutions
Explanation: The mathematical domain requires 25−x2≥025 - x^2 \geq 0, so x2≤25x^2 \leq 25, giving −5≤x≤5-5 \leq x \leq 5. However, the business constraint requires at least 200 units, meaning x≥2x \geq 2 (since xx is in hundreds). Combining these constraints: x∈[2,5]x \in [2, 5]. Choice B is incorrect notation. Choice C incorrectly uses an open interval. Choice D ignores the business context that negative production is meaningless.

Question 3

A model for the number of new subscribers to a service $t$ months after an advertising campaign is given by $N(t) = 500\sqrt{2t-6}$. The model is considered valid only for the first year after the campaign begins. Determine the practical domain for $t$.

  1. $[0, 12]$
  2. $[3, \infty)$
  3. $[3, 12]$ (correct answer)
  4. $(3, 12)$
Explanation: There are two constraints on the domain. First, the mathematical constraint from the square root function requires the radicand to be non-negative: $2t-6 \ge 0$, which simplifies to $2t \ge 6$, or $t \ge 3$. Second, the problem states a practical constraint that the model is only valid for the first year, which means $0 \le t \le 12$. To find the practical domain, we must satisfy both conditions. The intersection of the interval $[3, \infty)$ and $[0, 12]$ is $[3, 12]$.

Question 4

The function g(x)=∣x2−4x+3∣g(x) = |x^2 - 4x + 3| represents the absolute deviation of a company's quarterly profits from a baseline trend. What is the range of this function?

  1. [0,∞)[0, \infty) since absolute values are always non-negative (correct answer)
  2. [1,∞)[1, \infty) based on the minimum value of the quadratic expression
  3. (−∞,∞)(-\infty, \infty) because quadratic functions have unlimited range
  4. [0,1][0, 1] representing the bounded deviation from the baseline
Explanation: First, find the minimum of h(x)=x2−4x+3h(x) = x^2 - 4x + 3. Completing the square: h(x)=(x−2)2−1h(x) = (x-2)^2 - 1. The minimum value is −1-1 at x=2x = 2. Since h(x)h(x) ranges from [−1,∞)[-1, \infty), then g(x)=∣h(x)∣g(x) = |h(x)| ranges from [0,∞)[0, \infty) because ∣−1∣=1|-1| = 1 and all values ≥−1\geq -1 under absolute value give [0,∞)[0, \infty). Choice B incorrectly states the range starts at 1. Choice C ignores the absolute value. Choice D incorrectly assumes the function is bounded above.

Question 5

A company's profit $P(x)$ from manufacturing and selling $x$ units of a product is modeled by the function $P(x) = -0.1x^2 + 80x - 7000$. For which interval of production levels $x$ does the company realize a positive profit?

  1. $(0, 100) \cup (700, \infty)$
  2. $[100, 700]$
  3. $(100, 700)$ (correct answer)
  4. $[0, \infty)$
Explanation: To find where the profit is positive, we need to solve the inequality $P(x) > 0$. First, we find the roots of $P(x) = 0$, which are the break-even points. Using the quadratic formula for $-0.1x^2 + 80x - 7000 = 0$: x=−80±802−4(−0.1)(−7000)2(−0.1)=−80±6400−2800−0.2=−80±3600−0.2=−80±60−0.2x = \frac{-80 \pm \sqrt{80^2 - 4(-0.1)(-7000)}}{2(-0.1)} = \frac{-80 \pm \sqrt{6400 - 2800}}{-0.2} = \frac{-80 \pm \sqrt{3600}}{-0.2} = \frac{-80 \pm 60}{-0.2} The two roots are $x_1 = \frac{-20}{-0.2} = 100$ and $x_2 = \frac{-140}{-0.2} = 700$. Since the parabola opens downward (due to the negative leading coefficient), the function is positive between the roots. Therefore, the company makes a profit for production levels in the interval $(100, 700)$. The endpoints are not included because the profit is zero at those points.

Question 6

The total cost to produce $x$ items is given by $C(x) = 2x^2 - 10x + 800$. The factory has a maximum capacity of 200 items. The average cost per item is given by the function $A(x) = \frac{C(x)}{x}$. What is the practical domain of the average cost function $A(x)$?

  1. $(0, 200]$ (correct answer)
  2. $[1, 200]$
  3. $(-\infty, 0) \cup (0, 200]$
  4. $[0, 200]$
Explanation: The average cost function is $A(x) = \frac{2x^2 - 10x + 800}{x}$. There are three constraints on the domain. First, from a mathematical perspective, the denominator cannot be zero, so $x \neq 0$. Second, from a practical business perspective, the number of items produced, $x$, cannot be negative. Thus, $x > 0$. Third, the factory has a maximum capacity of 200 items, which means $x \le 200$. Combining these conditions ($x > 0$ and $x \le 200$), the practical domain is the interval $(0, 200]$.

Question 7

A product's price-demand function is $p(x) = 120 - 0.5x$, where $x$ is the number of units sold. The revenue function is $R(x) = x \cdot p(x)$. Considering all practical production levels, what is the range of the revenue function?

  1. $[0, 120]$
  2. $(-\infty, 7200]$
  3. $[0, 240]$
  4. $[0, 7200]$ (correct answer)
Explanation: First, establish the revenue function: $R(x) = x(120 - 0.5x) = 120x - 0.5x^2$. Next, determine the practical domain for $x$. We need $x \ge 0$ (cannot sell negative units) and $p(x) \ge 0$ (price cannot be negative). $120 - 0.5x \ge 0$ implies $x \le 240$. So the domain is $[0, 240]$. The revenue function is a downward-opening parabola. Its maximum value occurs at the vertex. The x-coordinate of the vertex is $x = -b/(2a) = -120/(2(-0.5)) = 120$. The maximum revenue is $R(120) = 120(120) - 0.5(120)^2 = 14400 - 7200 = 7200$. The minimum revenue occurs at the endpoints of the domain: $R(0) = 0$ and $R(240) = 0$. Therefore, the range of possible revenue is $[0, 7200]$.

Question 8

The daily productivity $P$ of a worker, in units assembled, is a function of hours worked, $h$, given by $P(h) = \frac{200h}{h^2+4}$. The worker must work at least 2 hours but no more than 10 hours. For the model to be realistic, productivity $P(h)$ must be at least 25 units. Determine the valid domain for $h$ under all these conditions.

  1. $[2, 10]$
  2. $[2, 4+2\sqrt{3}]$ (correct answer)
  3. $[4-2\sqrt{3}, 4+2\sqrt{3}]$
  4. $[4+2\sqrt{3}, 10]$
Explanation: When finding a domain with multiple constraints, you need to satisfy ALL conditions simultaneously. Here you have three restrictions: the work-hour constraint (2 ≤ h ≤ 10), and the productivity constraint (P(h) ≥ 25). Start by solving the productivity inequality: 200hh2+4≥25\frac{200h}{h^2+4} \geq 25. Cross-multiply to get 200h≥25(h2+4)200h \geq 25(h^2+4), which simplifies to 200h≥25h2+100200h \geq 25h^2 + 100. Rearranging gives 25h2−200h+100≤025h^2 - 200h + 100 \leq 0, or h2−8h+4≤0h^2 - 8h + 4 \leq 0. Using the quadratic formula: h=8±64−162=8±482=8±432=4±23h = \frac{8 \pm \sqrt{64-16}}{2} = \frac{8 \pm \sqrt{48}}{2} = \frac{8 \pm 4\sqrt{3}}{2} = 4 \pm 2\sqrt{3}. Since the parabola opens upward, the inequality is satisfied when 4−23≤h≤4+234 - 2\sqrt{3} \leq h \leq 4 + 2\sqrt{3}. Now find the intersection with [2, 10]. Since 4−23≈0.54<24 - 2\sqrt{3} \approx 0.54 < 2 and 4+23≈7.46<104 + 2\sqrt{3} \approx 7.46 < 10, the valid domain is [2,4+23][2, 4 + 2\sqrt{3}]. Choice A gives [2, 10] but ignores the productivity constraint entirely. Choice C gives the full solution to the productivity inequality without considering the work-hour minimum of 2. Choice D represents values where productivity actually decreases below 25 units again. Study tip: Domain problems with multiple constraints require finding the intersection of all individual constraint sets. Always verify your boundary points by substituting back into the original function.

Question 9

The number of users of a new software, in thousands, is modeled by $N(t) = 50 \ln(0.4t - 8)$, where $t$ is the number of months after its launch. For the model to be mathematically defined, what must be the domain of $t$?

  1. $(20, \infty)$ (correct answer)
  2. $[20, \infty)$
  3. $(0, 20)$
  4. $(-\infty, 20)$
Explanation: The natural logarithm function, $\ln(u)$, is only defined when its argument $u$ is strictly positive. In this model, the argument is $(0.4t - 8)$. Therefore, we must have the inequality $0.4t - 8 > 0$. To solve for $t$, we first add 8 to both sides: $0.4t > 8$. Then, we divide both sides by 0.4: $t > \frac{8}{0.4}$, which simplifies to $t > 20$. In interval notation, the domain is $(20, \infty)$.

Question 10

A subscription service uses a tiered pricing model based on usage levels. The monthly cost function is defined piecewise as follows:

20 & \text{if } 0 \leq x < 100 \\ 20 + 0.5(x-100) & \text{if } 100 \leq x < 500 \\ 220 + 0.8(x-500) & \text{if } x \geq 500 \end{cases}

where xx represents gigabytes of data used.

If the company limits individual accounts to a maximum monthly bill of $350, what is the effective domain of the cost function $f(x)f(x) $?

  1. [0,500)[0, 500) since costs exceed $350 in the third tier
  2. [0,662.5][0, 662.5] based on solving 220+0.8(x−500)=350220 + 0.8(x-500) = 350 (correct answer)
  3. [0,660][0, 660] using the constraint 0.8(x−500)≤1300.8(x-500) \leq 130
  4. [0,625][0, 625] from the inequality 220+0.8x≤350220 + 0.8x \leq 350
Explanation: We need to find where f(x)=350f(x) = 350. In the first tier (x<100x < 100), f(x)=20<350f(x) = 20 < 350. In the second tier (100≤x<500100 \leq x < 500), f(x)=20+0.5(x−100)≤20+0.5(400)=220<350f(x) = 20 + 0.5(x-100) \leq 20 + 0.5(400) = 220 < 350. In the third tier (x≥500x \geq 500), we solve 220+0.8(x−500)=350220 + 0.8(x-500) = 350, giving 0.8(x−500)=1300.8(x-500) = 130, so x−500=162.5x-500 = 162.5, thus x=662.5x = 662.5. Choice A stops too early. Choice C makes an algebraic error. Choice D incorrectly writes the third-tier formula.

Question 11

A company's efficiency rating is modeled by E(t)=t2+6t+8t+4E(t) = \frac{t^2 + 6t + 8}{t + 4} where t≥0t \geq 0 represents months of operation. If t=−4t = -4 would represent a theoretical state before company formation, what is the domain of E(t)E(t) in the business context, and what happens to the function near the excluded point?

  1. Domain: [0,∞)[0, \infty); the original function form must be preserved for business analysis
  2. Domain: (−∞,∞)∖{−4}(-\infty, \infty) \setminus \{-4\}; there is a vertical asymptote at t=−4t = -4
  3. Domain: [0,∞)[0, \infty); the function simplifies to t+2t + 2 after factoring
  4. Domain: [0,∞)[0, \infty); the function has a removable discontinuity at t=−4t = -4 (correct answer)
Explanation: When analyzing rational functions in business contexts, you need to consider both mathematical behavior and practical constraints. Here, the domain is restricted by business reality (t≥0t \geq 0 since negative time doesn't make sense) and potential mathematical issues. First, let's factor the numerator: t2+6t+8=(t+2)(t+4)t^2 + 6t + 8 = (t + 2)(t + 4). This gives us: E(t)=(t+2)(t+4)t+4E(t) = \frac{(t + 2)(t + 4)}{t + 4} For t≠−4t \neq -4, this simplifies to E(t)=t+2E(t) = t + 2. Since the business context requires t≥0t \geq 0, the domain is [0,∞)[0, \infty). At t=−4t = -4, the original function has the indeterminate form 00\frac{0}{0}, but the simplified form shows the function approaches −4+2=−2-4 + 2 = -2. This creates a removable discontinuity (a "hole") at the point (−4,−2)(-4, -2). Answer A is wrong because it ignores the mathematical simplification that's crucial for understanding the function's behavior. Answer B incorrectly uses the full real line as the domain, ignoring the business constraint that t≥0t \geq 0, and wrongly claims there's a vertical asymptote when the discontinuity is actually removable. Answer C correctly identifies the simplified form but fails to recognize that the original function still has a removable discontinuity at t=−4t = -4. Answer D is correct because it properly identifies both the business-appropriate domain [0,∞)[0, \infty) and recognizes the removable discontinuity. Study tip: Always factor rational functions completely to identify whether discontinuities are removable (common factors) or vertical asymptotes (factors only in denominator).

Question 12

A financial model uses F(x)=log⁡2(x2−6x+9)F(x) = \log_2(x^2 - 6x + 9) to represent investment risk, where xx is the investment amount in thousands. Given that log⁡2(1)=0\log_2(1) = 0 represents neutral risk, what is the domain of F(x)F(x), and what risk level does x=3x = 3 represent?

  1. Domain: x≠3x \neq 3; x=3x = 3 represents undefined risk due to division by zero
  2. Domain: (−∞,3)∪(3,∞)(-\infty, 3) \cup (3, \infty); F(3)F(3) is undefined in the logarithm
  3. Domain: (−∞,∞)∖{3}(-\infty, \infty) \setminus \{3\}; x=3x = 3 approaches minimum risk as F(x)→−∞F(x) \to -\infty (correct answer)
  4. Domain: (0,∞)(0, \infty); x=3x = 3 gives neutral risk since log⁡2(9)>0\log_2(9) > 0
Explanation: Note that x2−6x+9=(x−3)2≥0x^2 - 6x + 9 = (x-3)^2 \geq 0 for all real xx. For the logarithm to be defined, we need (x−3)2>0(x-3)^2 > 0, which means x≠3x \neq 3. So domain is (−∞,∞)∖{3}(-\infty, \infty) \setminus \{3\}. At x=3x = 3, (x−3)2=0(x-3)^2 = 0, and log⁡2(0)\log_2(0) is undefined, but as x→3x \to 3, (x−3)2→0+(x-3)^2 \to 0^+ and log⁡2((x−3)2)→−∞\log_2((x-3)^2) \to -\infty. Choice A gives incomplete domain. Choice B uses wrong notation. Choice D misunderstands the domain requirement and incorrectly evaluates the behavior at x=3x = 3.

Question 13

The cost $C(x)$ in dollars to ship a package weighing $x$ pounds is given by the function: $$\large C(x) = \begin{cases} 15 & \text{if } 0 < x \le 2 \ 15 + 4(x-2) & \text{if } 2 < x \le 10 \ 47 + 2(x-10) & \text{if } x > 10 \end{cases}

  1. $[27, 57]$ (correct answer)
  2. $[27, 47]$
  3. $[47, 57]$
  4. $[27, 67]$
Explanation: The domain of package weights is $[5, 15]$. This interval spans two pieces of the cost function. For weights $x$ in $[5, 10], the cost is $C(x) = 15 + 4(x-2)$. This is an increasing linear function. The costs for this part of the domain range from $C(5) = 15 + 4(3) = 27$ to $C(10) = 15 + 4(8) = 47$. For weights $x$ in $(10, 15], the cost is $C(x) = 47 + 2(x-10)$. This is also an increasing linear function. The costs for this part of the domain range from just above $C(10) = 47$ to $C(15) = 47 + 2(5) = 57$. Combining the ranges from both pieces, $[27, 47] and $(47, 57], gives a total range of $[27, 57]$.

Question 14

A company's daily cost to produce $x$ units is $C(x) = 0.2x^2 - 20x + 800$. Production capacity limits the output to be between 30 and 80 units per day, inclusive. What is the range of possible daily costs?

  1. $[300, 480]$ (correct answer)
  2. $[380, 480]$
  3. $[300, 380]$
  4. $[30, 80]$
Explanation: The domain for $x$ is $[30, 80]$. The cost function $C(x)$ is an upward-opening parabola. The minimum value of the function occurs at the vertex. The x-coordinate of the vertex is $x_v = -b/(2a) = -(-20)/(2 \cdot 0.2) = 20/0.4 = 50$. Since $50$ is in the domain $[30, 80]$, the minimum cost is $C(50) = 0.2(50)^2 - 20(50) + 800 = 500 - 1000 + 800 = 300$. The maximum cost will occur at the endpoint of the domain that is farthest from the vertex. The distances are |30 - 50| = 20 and |80 - 50| = 30. The maximum cost occurs at $x=80$. $C(80) = 0.2(80)^2 - 20(80) + 800 = 1280 - 1600 + 800 = 480$. For completeness, $C(30)=380$. The range is the interval from the minimum cost to the maximum cost, which is $[300, 480]$.

Question 15

A company finds that its weekly widget production, $W$, is a function of the number of labor hours, $h$, given by $W(h) = 50\sqrt{h}$. The number of available labor hours is a function of the number of employees, $n$, given by $h(n) = 30n$. The company operates with a staff of 10 to 15 employees, inclusive. What is the range of weekly widget production?

  1. $[300, 450]$
  2. $[50\sqrt{10}, 50\sqrt{15}]$
  3. $[500\sqrt{3}, 750\sqrt{2}]$ (correct answer)
  4. $[10, 15]$
Explanation: First, create the composite function for production in terms of employees, $W(h(n))$. Substitute $h(n) = 30n$ into $W(h)$: $W(n) = 50\sqrt{30n}$. The domain for the number of employees $n$ is $[10, 15]$. Since $W(n)$ is an increasing function, the minimum and maximum production will occur at the minimum and maximum number of employees. Minimum production (at $n=10$): $W(10) = 50\sqrt{30 \cdot 10} = 50\sqrt{300} = 50\sqrt{100 \cdot 3} = 50 \cdot 10\sqrt{3} = 500\sqrt{3}$. Maximum production (at $n=15$): $W(15) = 50\sqrt{30 \cdot 15} = 50\sqrt{450} = 50\sqrt{225 \cdot 2} = 50 \cdot 15\sqrt{2} = 750\sqrt{2}$. Thus, the range of widget production is $[500\sqrt{3}, 750\sqrt{2}]$.

Question 16

A consultant charges a fee based on the length of a project. The fee $F(d)$, in thousands of dollars, for a project lasting $d$ full months is given by $F(d) = 2d^2 + 10$. Due to her schedule, she only accepts projects that last 3, 4, 5, or 6 full months. What is the range of her fee?

  1. The set $\left\{3, 4, 5, 6\right\}$
  2. The set $\left\{28, 42, 60, 82\right\}$ (correct answer)
  3. The interval $[28, 82]$
  4. The interval $[18, \infty)$
Explanation: When you encounter a function with a restricted domain, you're finding the range by evaluating the function at each allowed input value, not by considering all possible outputs over an interval. Here, the consultant's fee function is F(d)=2d2+10F(d) = 2d^2 + 10, but she only accepts projects lasting 3, 4, 5, or 6 months. To find the range, you need to calculate F(d)F(d) at each of these specific values:
  • F(3)=2(3)2+10=2(9)+10=28F(3) = 2(3)^2 + 10 = 2(9) + 10 = 28
  • F(4)=2(4)2+10=2(16)+10=42F(4) = 2(4)^2 + 10 = 2(16) + 10 = 42
  • F(5)=2(5)2+10=2(25)+10=60F(5) = 2(5)^2 + 10 = 2(25) + 10 = 60
  • F(6)=2(6)2+10=2(36)+10=82F(6) = 2(6)^2 + 10 = 2(36) + 10 = 82
The range is the set of all possible output values: {28,42,60,82}\{28, 42, 60, 82\}, which is answer B. Answer A gives you the domain values (the allowed project lengths), not the range. Answer C represents the interval from the minimum to maximum fee, but this would only be correct if the consultant accepted projects of any length between 3 and 6 months, including fractional months. Since she only accepts whole-month projects, the range consists of discrete values, not a continuous interval. Answer D appears to consider the range of the function without the domain restriction, starting from F(3)=18F(3) = 18 if the minimum were 3, but this miscalculates F(3)F(3). Remember: when a function has a restricted domain with specific discrete values, the range consists of the function outputs at those exact points only.

Question 17

A company's market share percentage, $M$, after $t$ years of operation is modeled by $M(t) = \frac{100t}{t+5}$ for $t \ge 0$. What is the practical range for the company's market share according to this model?

  1. $[0, 100]$
  2. $[0, 100)$ (correct answer)
  3. $[0, \infty)$
  4. $(0, 100)$
Explanation: The domain of the function is $t \ge 0$. To find the range, we evaluate the function at the minimum value of the domain and analyze its long-term behavior. At $t=0$, the market share is $M(0) = \frac{100(0)}{0+5} = 0$. This is the minimum value in the range. To find the upper bound, we examine the limit as $t$ approaches infinity, which corresponds to the horizontal asymptote of the function: $ \lim_{t\to\infty} \frac{100t}{t+5} = \lim_{t\to\infty} \frac{100}{1+5/t} = 100$. Since the function is always increasing for $t \ge 0$ and approaches 100 but never reaches it, the range is $[0, 100)$.

Question 18

A company's weekly profit, PP, from selling nn units of a product is given by the function P(n)=−0.5n2+80n−2000P(n) = -0.5n^2 + 80n - 2000. The company has the capacity to produce at most 100 units per week, and company policy requires that they produce at least 20 units per week. Which of the following best describes the practical domain of the profit function P(n)P(n)?

  1. The interval of real numbers [20,100][20, 100].
  2. The set of all integers nn such that 32≤n≤10032 \le n \le 100.
  3. The set of all integers nn such that 0≤n≤1000 \le n \le 100.
  4. The set of all integers nn such that 20≤n≤10020 \le n \le 100. (correct answer)
Explanation: The domain of a function consists of all possible input values. In this context, the input variable is nn, the number of units. Since units of a product are discrete, nn must be an integer. The problem states two constraints on nn: a minimum production of 20 units (n≥20n \ge 20) and a maximum capacity of 100 units (n≤100n \le 100). Combining these constraints, the practical domain is the set of all integers nn satisfying 20≤n≤10020 \le n \le 100.

Question 19

A marketing firm models the number of weekly sales of a product, SS (in thousands), as a function of its price, pp (in dollars), with the function S(p)=1200−15pp−20S(p) = \frac{\sqrt{1200 - 15p}}{p-20}. For the model to be meaningful, both the price pp and the number of sales SS must be non-negative. What is the practical domain of the function S(p)S(p)?

  1. [0,20)∪(20,80][0, 20) \cup (20, 80] (correct answer)
  2. [0,80][0, 80]
  3. [0,20)∪(20,∞)[0, 20) \cup (20, \infty)
  4. (20,80](20, 80]
Explanation: There are three conditions for the practical domain. First, the price must be non-negative, so p≥0p \ge 0. Second, the term in the denominator cannot be zero, so p−20≠0p-20 \neq 0, which means p≠20p \neq 20. Third, the term under the square root must be non-negative, so 1200−15p≥01200 - 15p \ge 0, which simplifies to 1200≥15p1200 \ge 15p, or p≤80p \le 80. Combining these three conditions (p≥0p \ge 0, p≠20p \neq 20, and p≤80p \le 80), we get the domain [0,20)∪(20,80][0, 20) \cup (20, 80]. The condition that sales SS must be non-negative is satisfied on this domain since the numerator 1200−15p\sqrt{1200-15p} is always non-negative, but this distractor D considers only when the denominator is also positive, which is not required for S to be non-negative.

Question 20

The number of new users, NN (in hundreds), for a mobile app tt months after its launch is modeled by N(t)=50−900−36tN(t) = 50 - \sqrt{900 - 36t}. The model is considered valid only when the expression for N(t)N(t) is a real, non-negative number. What is the range of the function N(t)N(t) over its valid domain?

  1. [0,50][0, 50]
  2. [20,50][20, 50] (correct answer)
  3. [0,25][0, 25]
  4. [20,25][20, 25]
Explanation: First, find the valid domain for tt. The term under the square root must be non-negative: 900−36t≥0  ⟹  900≥36t  ⟹  25≥t900 - 36t \ge 0 \implies 900 \ge 36t \implies 25 \ge t. Also, time must be non-negative, so t≥0t \ge 0. The model also requires N(t)≥0N(t) \ge 0: 50−900−36t≥0  ⟹  50≥900−36t  ⟹  2500≥900−36t  ⟹  1600≥−36t  ⟹  −1600/36≤t50 - \sqrt{900-36t} \ge 0 \implies 50 \ge \sqrt{900-36t} \implies 2500 \ge 900-36t \implies 1600 \ge -36t \implies -1600/36 \le t, which is t≥−44.4...t \ge -44.4.... This does not add a new constraint to t≥0t \ge 0. So the valid domain is [0,25][0, 25]. Now, find the range on this domain. The function 900−36t\sqrt{900-36t} is a decreasing function of tt. Since it is being subtracted from 50, N(t)N(t) is an increasing function of tt. The minimum value will occur at the smallest value of tt in the domain, t=0t=0: N(0)=50−900−0=50−30=20N(0) = 50 - \sqrt{900 - 0} = 50 - 30 = 20. The maximum value will occur at the largest value of tt, t=25t=25: N(25)=50−900−36(25)=50−900−900=50−0=50N(25) = 50 - \sqrt{900 - 36(25)} = 50 - \sqrt{900-900} = 50 - 0 = 50. So, the range of N(t)N(t) is [20,50][20, 50].