Functions: Domain & Range - Business Calculus

1. 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?
   Answer: The set $\left\{28, 42, 60, 82\right\}$ 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 $F(d) = 2d^2 + 10$, but she only accepts projects lasting 3, 4, 5, or 6 months. To find the range, you need to calculate $F(d)$ at each of these specific values: - $F(3) = 2(3)^2 + 10 = 2(9) + 10 = 28$ - $F(4) = 2(4)^2 + 10 = 2(16) + 10 = 42$ - $F(5) = 2(5)^2 + 10 = 2(25) + 10 = 60$ - $F(6) = 2(6)^2 + 10 = 2(36) + 10 = 82$ The range is the set of all possible output values: $\{28, 42, 60, 82\}$, 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 $F(3) = 18$ if the minimum were 3, but this miscalculates $F(3)$. Remember: when a function has a restricted domain with specific discrete values, the range consists of the function outputs at those exact points only.
2. A delivery service charges for shipping based on the weight, $w$, of a package in pounds. The cost function $C(w)$ is a piecewise function. For packages weighing more than 5 pounds but no more than 15 pounds, the cost is given by $C(w) = 10 + 2(w-5)$. For packages weighing more than 15 pounds up to a maximum of 30 pounds, the cost is given by $C(w) = 30 + 1.5(w-15)$. What is the range of possible shipping costs for a package that weighs more than 10 pounds but no more than 20 pounds?
   Answer: $(20, 37.50]$ The domain of interest is $(10, 20]$. We need to evaluate the range of the cost function $C(w)$ on this domain. The domain is split across two pieces of the function. First piece for $w \in (10, 15]$: The relevant cost function is $C(w) = 10 + 2(w-5) = 2w$. The costs for this interval are $C(10) = 2(10) = 20$ and $C(15) = 2(15) = 30$. Since $w > 10$, the costs are in $(20, 30]$. Second piece for $w \in (15, 20]$: The relevant cost function is $C(w) = 30 + 1.5(w-15) = 30 + 1.5w - 22.5 = 1.5w + 7.5$. The costs for this interval are $C(15) = 1.5(15)+7.5 = 22.5+7.5 = 30$ and $C(20) = 1.5(20)+7.5 = 30+7.5=37.50$. Since $w>15$, the costs are in $(30, 37.50]$. Combining the two ranges, we have $(20, 30] \cup (30, 37.50]$. This union is the interval $(20, 37.50]$.
3. If the company limits individual accounts to a maximum monthly bill of $350, what is the effective domain of the cost function $f(x)$?
   Answer: $[0, 662.5]$ based on solving $220 + 0.8(x-500) = 350$ We need to find where $f(x) = 350$. In the first tier ($x < 100$), $f(x) = 20 < 350$. In the second tier ($100 \leq x < 500$), $f(x) = 20 + 0.5(x-100) \leq 20 + 0.5(400) = 220 < 350$. In the third tier ($x \geq 500$), we solve $220 + 0.8(x-500) = 350$, giving $0.8(x-500) = 130$, so $x-500 = 162.5$, thus $x = 662.5$. Choice A stops too early. Choice C makes an algebraic error. Choice D incorrectly writes the third-tier formula.
4. A financial model uses $F(x) = \log_2(x^2 - 6x + 9)$ to represent investment risk, where $x$ is the investment amount in thousands. Given that $\log_2(1) = 0$ represents neutral risk, what is the domain of $F(x)$, and what risk level does $x = 3$ represent?
   Answer: Domain: $(-\infty, \infty) \setminus \{3\}$; $x = 3$ approaches minimum risk as $F(x) \to -\infty$ Note that $x^2 - 6x + 9 = (x-3)^2 \geq 0$ for all real $x$. For the logarithm to be defined, we need $(x-3)^2 > 0$, which means $x \neq 3$. So domain is $(-\infty, \infty) \setminus \{3\}$. At $x = 3$, $(x-3)^2 = 0$, and $\log_2(0)$ is undefined, but as $x \to 3$, $(x-3)^2 \to 0^+$ and $\log_2((x-3)^2) \to -\infty$. Choice A gives incomplete domain. Choice B uses wrong notation. Choice D misunderstands the domain requirement and incorrectly evaluates the behavior at $x = 3$.
5. 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 $A(x)$?
   Answer: $(0, 200]$ 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 $(0, 200]$.
6. A manufacturer's profit function is $P(q) = -2q^2 + 160q - 2000$, where $q$ is the number of units produced and sold. Due to production capacity, the company can produce at most 50 units. Assuming at least one unit is produced, what is the range of the profit function for this company?
   Answer: $[-1842, 1200]$ The domain is for $q$ in the set of integers from 1 to 50, but for range of a continuous model we consider the real interval $[1, 50]$. The profit function is a downward-opening parabola. The maximum occurs at the vertex, where $q = -b/(2a) = -160/(2(-2)) = 40$. Since $q=40$ is in the domain $[1, 50]$, the maximum profit is $P(40) = -2(40)^2 + 160(40) - 2000 = 1200$. The minimum profit will occur at one of the endpoints of the domain, $q=1$ or $q=50$. We evaluate $P(1) = -2(1)^2 + 160(1) - 2000 = -1842$. We evaluate $P(50) = -2(50)^2 + 160(50) - 2000 = 1000$. The minimum value on the interval is $-1842$. Therefore, the range is $[-1842, 1200]$. Choice A uses $P(0)$ as the minimum, but the problem states at least one unit is produced.
7. The number of new users, $N$ (in hundreds), for a mobile app $t$ months after its launch is modeled by $N(t) = 50 - \sqrt{900 - 36t}$. The model is considered valid only when the expression for $N(t)$ is a real, non-negative number. What is the range of the function $N(t)$ over its valid domain?
   Answer: $[20, 50]$ First, find the valid domain for $t$. The term under the square root must be non-negative: $900 - 36t \ge 0 \implies 900 \ge 36t \implies 25 \ge t$. Also, time must be non-negative, so $t \ge 0$. The model also requires $N(t) \ge 0$: $50 - \sqrt{900-36t} \ge 0 \implies 50 \ge \sqrt{900-36t} \implies 2500 \ge 900-36t \implies 1600 \ge -36t \implies -1600/36 \le t$, which is $t \ge -44.4...$. This does not add a new constraint to $t \ge 0$. So the valid domain is $[0, 25]$. Now, find the range on this domain. The function $\sqrt{900-36t}$ is a decreasing function of $t$. Since it is being subtracted from 50, $N(t)$ is an increasing function of $t$. The minimum value will occur at the smallest value of $t$ in the domain, $t=0$: $N(0) = 50 - \sqrt{900 - 0} = 50 - 30 = 20$. The maximum value will occur at the largest value of $t$, $t=25$: $N(25) = 50 - \sqrt{900 - 36(25)} = 50 - \sqrt{900-900} = 50 - 0 = 50$. So, the range of $N(t)$ is $[20, 50]$.
8. 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 $t$?
   Answer: $(20, \infty)$ 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 $(20, \infty)$.
9. A company's efficiency rating is modeled by $E(t) = \frac{t^2 + 6t + 8}{t + 4}$ where $t \geq 0$ represents months of operation. If $t = -4$ would represent a theoretical state before company formation, what is the domain of $E(t)$ in the business context, and what happens to the function near the excluded point?
   Answer: Domain: $[0, \infty)$; the function has a removable discontinuity at $t = -4$ 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 ($t \geq 0$ since negative time doesn't make sense) and potential mathematical issues. First, let's factor the numerator: $t^2 + 6t + 8 = (t + 2)(t + 4)$. This gives us: $E(t) = \frac{(t + 2)(t + 4)}{t + 4}$ For $t \neq -4$, this simplifies to $E(t) = t + 2$. Since the business context requires $t \geq 0$, the domain is $[0, \infty)$. At $t = -4$, the original function has the indeterminate form $\frac{0}{0}$, but the simplified form shows the function approaches $-4 + 2 = -2$. This creates a removable discontinuity (a "hole") at the point $(-4, -2)$. 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 $t \geq 0$, 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 $t = -4$. Answer D is correct because it properly identifies both the business-appropriate domain $[0, \infty)$ 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).
10. 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?
    Answer: $[0, 100)$ 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 $[0, 100)$.
11. A company's profit function is defined as $P(x) = \sqrt{25 - x^2}$ thousand dollars, where $x$ 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 $P(x)$ given this business constraint?
    Answer: $[2, 5]$ The mathematical domain requires $25 - x^2 \geq 0$, so $x^2 \leq 25$, giving $-5 \leq x \leq 5$. However, the business constraint requires at least 200 units, meaning $x \geq 2$ (since $x$ is in hundreds). Combining these constraints: $x \in [2, 5]$. Choice B is incorrect notation. Choice C incorrectly uses an open interval. Choice D ignores the business context that negative production is meaningless.
12. A marketing firm models the number of weekly sales of a product, $S$ (in thousands), as a function of its price, $p$ (in dollars), with the function $S(p) = \frac{\sqrt{1200 - 15p}}{p-20}$. For the model to be meaningful, both the price $p$ and the number of sales $S$ must be non-negative. What is the practical domain of the function $S(p)$?
    Answer: $[0, 20) \cup (20, 80]$ There are three conditions for the practical domain. First, the price must be non-negative, so $p \ge 0$. Second, the term in the denominator cannot be zero, so $p-20 \neq 0$, which means $p \neq 20$. Third, the term under the square root must be non-negative, so $1200 - 15p \ge 0$, which simplifies to $1200 \ge 15p$, or $p \le 80$. Combining these three conditions ($p \ge 0$, $p \neq 20$, and $p \le 80$), we get the domain $[0, 20) \cup (20, 80]$. The condition that sales $S$ must be non-negative is satisfied on this domain since the numerator $\sqrt{1200-15p}$ 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.
13. The total revenue from selling $x$ units of a new tablet is given by the function $R(x)$. What is the correct business interpretation of the expression $\frac{R(a+h) - R(a)}{h}$?
    Answer: The average rate of change in revenue as sales increase from $a$ to $a+h$ tablets. The expression $\frac{R(a+h) - R(a)}{h}$ is the difference quotient. $R(a+h) - R(a)$ represents the change in revenue when the number of units sold changes from $a$ to $a+h$. The denominator, $h$, represents the change in the number of units sold. A change in a quantity divided by the change in another variable gives the average rate of change. Therefore, the expression represents the average rate of change in revenue with respect to the number of units sold over the interval $[a, a+h]$.
14. To produce a cylindrical container with a fixed volume of $50\pi$ cubic inches, the material cost, $C$, is a function of the container's radius, $r$. The cost function is $C(r) = 0.4\pi r^2 + \frac{10\pi}{r}$, where the first term is the cost of the top and bottom and the second term is the cost of the side. Which of the following represents the practical domain of this cost function?
    Answer: $(0, \infty)$ The input to the function is the radius, $r$. From a practical standpoint, a physical dimension like radius cannot be negative, so $r \ge 0$. The cost function is $C(r) = 0.4\pi r^2 + \frac{10\pi}{r}$. Looking at the function, we see that $r$ is in the denominator of the second term. Division by zero is undefined, so $r$ cannot be 0. Therefore, we must have $r > 0$. There are no other mathematical or practical constraints given that would place an upper limit on the radius (although a very large or very small radius would be impractical, the mathematical domain has no upper bound). Thus, the practical domain is all positive real numbers, which is the interval $(0, \infty)$.
15. 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 $t$.
    Answer: $[3, 12]$ 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 $[3, 12]$.
16. A company is building a rectangular warehouse. Three sides will be constructed from standard fencing that costs $20 per linear foot, and the fourth side, which faces the main road, will be constructed from premium fencing that costs $30 per linear foot. The company has a total budget of $12,000 for the fencing. If $x$ represents the length of the premium fencing side, the area of the warehouse is given by the function $A(x) = x\left( \frac{12000 - 30x}{40} \right)$. What is the practical domain of the area function $A(x)$?
    Answer: $(0, 400)$ Let $x$ be the length of the premium side and $y$ be the length of the adjacent sides. The cost equation is $30x + 20x + 20y + 20y = 12000$, which simplifies to $50x + 40y = 12000$. This doesn't match the function provided. Let's re-read. $x$ is one side (premium). The other side parallel to it and the two other sides are standard. Wait, the stem says 'Three sides will be constructed from standard fencing...and the fourth side...from premium fencing.' This implies one side is $x$ (premium) and the other three are standard. Let the side parallel to $x$ have length $x$ and the other two sides have length $y$. Cost: $30x + 20(x+2y) = 12000$. This is $50x+40y=12000$. Area is $A=xy$. From cost, $y = (12000-50x)/40$. So $A(x) = x((12000-50x)/40)$. The function in the stem must have come from a different setup: one premium side $x$, two adjacent standard sides $y$. The fourth side is not fenced (e.g., against a building). Cost: $30x + 2(20y) = 12000$. Area $A=xy$. From cost, $40y = 12000-30x$, so $y = (12000-30x)/40$. Thus $A(x) = x(\frac{12000-30x}{40})$. For the domain to be practical, the physical dimensions $x$ and $y$ must be positive. $x>0$. Also, $y>0$, which means $\frac{12000 - 30x}{40} > 0$. This implies $12000 - 30x > 0$, so $12000 > 30x$, which gives $x < 400$. Combining the conditions, the domain is $(0, 400)$.
17. The function $g(x) = |x^2 - 4x + 3|$ represents the absolute deviation of a company's quarterly profits from a baseline trend. What is the range of this function?
    Answer: $[0, \infty)$ since absolute values are always non-negative First, find the minimum of $h(x) = x^2 - 4x + 3$. Completing the square: $h(x) = (x-2)^2 - 1$. The minimum value is $-1$ at $x = 2$. Since $h(x)$ ranges from $[-1, \infty)$, then $g(x) = |h(x)|$ ranges from $[0, \infty)$ because $|-1| = 1$ and all values $\geq -1$ under absolute value give $[0, \infty)$. Choice B incorrectly states the range starts at 1. Choice C ignores the absolute value. Choice D incorrectly assumes the function is bounded above.
18. A retailer's daily profit function is $P(n) = -2n^2 + 60n - 200$ dollars, where $n$ is the number of items sold. The retailer can only sell whole items and must sell at least 5 items to cover fixed costs. If the retailer stops selling when profit becomes negative, what is the practical domain of $P(n)$, and what is the maximum profit?
    Answer: Domain: $\{5, 6, 7, \ldots, 25\}$; Maximum profit: $250 When analyzing profit functions in business contexts, you need to determine both the practical domain (realistic input values) and optimize the function within those constraints. To find the practical domain, start with the given constraints: the retailer must sell whole items (integers only), needs at least 5 items to cover fixed costs, and stops when profit becomes negative. So you need to find where $P(n) = -2n^2 + 60n - 200 = 0$: $-2n^2 + 60n - 200 = 0$ $n^2 - 30n + 100 = 0$ $(n - 20)(n - 10) = 0$ This gives $n = 10$ and $n = 20$. Since the parabola opens downward (negative coefficient of $n^2$), profit is positive between these values and negative outside them. Combined with the minimum of 5 items, the domain is $\{5, 6, 7, \ldots, 20\}$. Wait—let me check $n = 25$: $P(25) = -2(625) + 60(25) - 200 = -1250 + 1500 - 200 = 50 > 0$. Let me solve this more carefully. Setting $P(n) = 0$ and using the quadratic formula: $n = \frac{60 \pm \sqrt{3600 - 1600}}{4} = \frac{60 \pm \sqrt{2000}}{4} \approx \frac{60 \pm 44.72}{4}$. This gives approximately $n = 3.82$ and $n = 26.18$. So profit is positive for $4 \leq n \leq 26$, making the practical domain $\{5, 6, 7, \ldots, 25\}$. For maximum profit, use $n = -\frac{b}{2a} = -\frac{60}{2(-2)} = 15$: $P(15) = -450 + 900 - 200 = 250$. Choice A has an unbounded domain. Choice B stops at 24 instead of 25 and gives the wrong maximum. Choice C includes 26, where profit would be negative. Study tip: Always check endpoint values when determining practical domains, and remember that quadratic profit functions typically have both lower and upper bounds.
19. 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?
    Answer: $[0, 7200]$ 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 $[0, 7200]$.
20. 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.
    Answer: $[2, 4+2\sqrt{3}]$ 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: $\frac{200h}{h^2+4} \geq 25$. Cross-multiply to get $200h \geq 25(h^2+4)$, which simplifies to $200h \geq 25h^2 + 100$. Rearranging gives $25h^2 - 200h + 100 \leq 0$, or $h^2 - 8h + 4 \leq 0$. Using the quadratic formula: $h = \frac{8 \pm \sqrt{64-16}}{2} = \frac{8 \pm \sqrt{48}}{2} = \frac{8 \pm 4\sqrt{3}}{2} = 4 \pm 2\sqrt{3}$. Since the parabola opens upward, the inequality is satisfied when $4 - 2\sqrt{3} \leq h \leq 4 + 2\sqrt{3}$. Now find the intersection with [2, 10]. Since $4 - 2\sqrt{3} \approx 0.54 < 2$ and $4 + 2\sqrt{3} \approx 7.46 < 10$, the valid domain is $[2, 4 + 2\sqrt{3}]$. 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.

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