Function Notation & Interpretation - IB Mathematics Applications and Interpretation

1. The profit, $P$, in euros (EUR), from selling $x$ handmade scarves is modelled by the function $P(x) = 25x - 300$. The model is considered valid for the first 100 scarves made each month. Which statement best describes the domain of this function in this specific context?
   Answer: $\{x \in \mathbb{Z} \mid 0 \le x \le 100\}$ The domain represents the set of all possible input values. In this context, the input $x$ is the number of handmade scarves. Since scarves are discrete, countable items, $x$ must be an integer ($\mathbb{Z}$). The context states the model is valid for the first 100 scarves, and the number of scarves cannot be negative, so the domain is all integers from 0 to 100, inclusive. This is represented by $\{x \in \mathbb{Z} \mid 0 \le x \le 100\}$.
2. A rectangular garden has a fixed perimeter of 100 metres. The area of the garden, $A$, can be modelled as a function of its width, $w$, by $A(w) = w(50-w)$. What is the practical domain for the width, $w$?
   Answer: $0 < w < 50$ When you encounter optimization problems involving real-world constraints, identifying the practical domain requires considering both mathematical and physical limitations. The function $A(w) = w(50-w)$ models area as a function of width, but not all mathematical values of $w$ make sense in reality. Since this is a rectangular garden, the width must be positive, so $w > 0$. To find the upper bound, consider what the function tells us: if the area is $A(w) = w(50-w)$, then the length must be $(50-w)$ since area equals width times length. This makes sense because with a perimeter of 100 meters, we have $2w + 2l = 100$, which gives us $l = 50 - w$. For a rectangle to exist, both dimensions must be positive. While $w > 0$ is obvious, we also need the length to be positive: $50 - w > 0$, which means $w < 50$. Therefore, the practical domain is $0 < w < 50$. Looking at the wrong answers: Choice A ($w > 0$) ignores the upper constraint that length must also be positive. Choice B ($0 < w < 25$) incorrectly assumes the garden must be a square or that width cannot exceed length, but the problem doesn't specify which dimension is called "width." Choice C ($w \in \mathbb{R}$) ignores all physical constraints entirely. Remember: In optimization problems with real-world contexts, always check that all physical quantities remain positive and meaningful throughout your domain.
3. The velocity, $v$, in metres per second, of a wave in shallow water is modelled by $v(d) = \sqrt{9.8d}$, where $d$ is the depth of the water in metres. The model is used for a coastal area where the depth ranges from 0.5 metres to 10 metres. What is the domain of the function in this specific context?
   Answer: $0.5 \le d \le 10$ The domain is the set of valid inputs, $d$. Mathematically, the square root requires $d \ge 0$. However, the problem specifies that the model is being applied to a context where the depth, $d$, ranges from 0.5 metres to 10 metres. The practical domain is therefore restricted by this context. The correct domain is $0.5 \le d \le 10$.
4. The function $C(x) = 0.75x + 3500$ models the total cost for a factory to produce $x$ units of a product. The function $R(x) = 2x$ models the total revenue from selling $x$ units. What is the meaning of the solution to the equation $R(x) = C(x)$?
   Answer: The number of units that must be sold for revenue to equal cost, also known as the break-even point. When you encounter cost and revenue functions in business mathematics, you're typically dealing with break-even analysis. The key insight is understanding what happens when these two functions are equal. The equation $R(x) = C(x)$ means $2x = 0.75x + 3500$. Solving this: $2x - 0.75x = 3500$, so $1.25x = 3500$, giving us $x = 2800$ units. At this point, the company brings in exactly as much money as it spends—this is the break-even point where profit equals zero. Option A is incorrect because profit maximization occurs where the difference between revenue and cost is greatest, not where they're equal. For linear functions like these, profit actually increases indefinitely as you sell more units beyond the break-even point. Option B confuses break-even with profitability. To make a profit, the company needs to sell more than the break-even quantity, not exactly the break-even amount. Option C misunderstands what's being covered. The fixed costs ($3500$ in the cost function) represent overhead expenses, but break-even analysis considers total costs versus total revenue, not just fixed costs. Option D correctly identifies that when $R(x) = C(x)$, revenue equals cost—the definition of break-even point. Study tip: Remember that break-even occurs at the intersection of cost and revenue functions. Below this point, you lose money; above it, you make profit. This concept appears frequently in IB Math Applications, so practice identifying these relationships in word problems.
5. The concentration of a certain medication in a patient's bloodstream, $C$, in milligrams per litre (mg/L), is modelled by the function $C(t) = \frac{100t}{t^2 + 4}$, where $t$ is the number of hours after the medication was taken, for $t \ge 0$. What does the value $C(2)$ represent?
   Answer: The concentration of the medication in the bloodstream after 2 hours. In the function notation $C(t)$, $t$ is the input (time in hours) and $C(t)$ is the output (concentration in mg/L). Therefore, $C(2)$ represents the output of the function when the input is $t=2$. This means it is the concentration of the medication in the bloodstream after 2 hours. While $t=2$ happens to be where the maximum concentration occurs for this function, the expression $C(2)$ itself simply means the value of C at t=2.
6. The height, $h$, in metres, of a ball thrown upwards from the top of a 50-metre building is given by $h(t) = -4.9t^2 + 20t + 50$, where $t$ is the time in seconds. The ball hits the ground after approximately 5.88 seconds. What is the most appropriate range for this function in this context?
   Answer: $\{h \in \mathbb{R} \mid 0 \le h \le 70.4\}$ The range represents the set of all possible output values (heights). The minimum height is 0 metres, when the ball hits the ground. The maximum height occurs at the vertex of the parabola. The time at the vertex is $t = -b/(2a) = -20/(2 \times -4.9) \approx 2.04$ seconds. The maximum height is $h(2.04) = -4.9(2.04)^2 + 20(2.04) + 50 \approx 70.4$ metres. Therefore, the practical range of heights is from 0 to 70.4 metres, inclusive.
7. The temperature of a cup of coffee, $T$, in degrees Celsius, is modelled by the function $T(m) = 20 + 70(0.95)^m$, where $m$ is the number of minutes after it is poured. What is the practical meaning of the expression $T(5) - T(10)$?
   Answer: The decrease in the coffee's temperature between the 5th minute and the 10th minute. $T(5)$ is the temperature at 5 minutes, and $T(10)$ is the temperature at 10 minutes. The difference, $T(5) - T(10)$, represents the total change in temperature between these two points in time. Since the coffee is cooling, $T(5)$ will be greater than $T(10)$, so the result is a positive value representing the total temperature drop, or decrease, over that 5-minute interval.
8. The fuel efficiency, $E$, of a car in kilometres per litre (km/L) is modelled by the function $E(s) = -0.01s^2 + 1.2s + 10$, where $s$ is the speed in kilometres per hour (km/h). The car's speed can realistically vary from 20 km/h to 120 km/h. What is the practical range of the car's fuel efficiency based on this model?
   Answer: $10 \le E \le 46$ The range is the set of output values $E$ corresponding to the given domain for $s$, which is $[20, 120]$. We must find the minimum and maximum values of $E(s)$ on this interval. The vertex of the parabola is at $s = -1.2/(2 \times -0.01) = 60$ km/h. Since 60 is in the domain, the maximum efficiency is $E(60) = -0.01(60)^2 + 1.2(60) + 10 = 46$ km/L. To find the minimum on the interval, we check the endpoints: $E(20) = -0.01(400) + 24 + 10 = 30$ km/L and $E(120) = -0.01(14400) + 144 + 10 = 10$ km/L. The minimum value on the interval is 10. Thus the range is $10 \le E \le 46$.
9. An open-top box is made by cutting congruent squares of side length $x$ cm from the corners of a rectangular piece of cardboard measuring 30 cm by 40 cm. The volume of the box, $V$, in cm³, is modelled by $V(x) = x(30-2x)(40-2x)$. Which of the following represents the practical domain of this function?
   Answer: $0 < x < 15$ The domain is the set of possible values for $x$. For the volume to be physically possible, all dimensions must be positive. The side length of the cut square, $x$, must be greater than 0. The side length of the box's base, $30-2x$, must be greater than 0, which implies $30 > 2x$, or $x < 15$. The other side length, $40-2x$, must also be greater than 0, which implies $40 > 2x$, or $x < 20$. For all three conditions to be met ($x>0$, $x<15$, and $x<20$), $x$ must be between 0 and 15. Thus, the practical domain is $0 < x < 15$.
10. A psychologist models a person's proficiency score, $S$, on a new video game by the function $S(h) = \frac{100h}{h+5}$, where $h$ is the number of hours practiced, for $h \ge 0$. Which statement best describes the range of this function in this context?
    Answer: The score is always non-negative and approaches a maximum of 100. The range is the set of all possible output values, $S$. When $h=0$, $S(0) = 0$. As the number of hours $h$ becomes very large, the value of $S(h)$ approaches the horizontal asymptote of the function, which is $y = \frac{100}{1} = 100$. The score will get closer and closer to 100 but never reach it. Therefore, the range of possible scores is $[0, 100)$, which means the score is always non-negative and approaches a maximum of 100.
11. The time, $T$, in minutes, required for a robot to assemble $n$ products is modelled by the function $T(n) = 20 + \frac{500}{n}$. The robot can assemble between 10 and 100 products per shift. Which statement best describes the domain of this function for one shift?
    Answer: $\{n \in \mathbb{Z} \mid 10 \le n \le 100\}$ The domain is the set of valid inputs, $n$. The variable $n$ represents the number of products, which must be a whole number (an integer). The problem states the robot assembles between 10 and 100 products, so $10 \le n \le 100$. Combining these, the domain is the set of integers from 10 to 100, inclusive, which is written as $\{n \in \mathbb{Z} \mid 10 \le n \le 100\}$.
12. The depth of water, $D$, in metres, at a harbour entrance is modelled by the function $D(t) = 12 + 4\cos(0.5t)$, where $t$ is the number of hours after midnight. What is the range of water depths predicted by this model?
    Answer: $8 \le D \le 16$ When you encounter a trigonometric function modeling a real-world situation, you need to identify the range by analyzing how the trigonometric component behaves. This function represents a sinusoidal model where the cosine function creates periodic variation around a central value. The function $D(t) = 12 + 4\cos(0.5t)$ has the form of a vertical shift plus an amplitude variation. Since cosine oscillates between -1 and 1, you can find the extreme values by substituting these bounds. When $\cos(0.5t) = 1$, the depth reaches its maximum: $D = 12 + 4(1) = 16$ metres. When $\cos(0.5t) = -1$, the depth reaches its minimum: $D = 12 + 4(-1) = 8$ metres. Therefore, the range is $8 \le D \le 16$. Option A correctly identifies this range. Option B gives $0 \le t \le 24$, which describes the domain (input values) for a 24-hour period, not the range of water depths. This is a common confusion between domain and range. Option C states $12 \le D \le 16$, which only captures the upper half of the oscillation—this would be correct if the cosine term could only be positive, but cosine takes both positive and negative values. Option D suggests $D \ge 0$, which is far too broad and doesn't reflect the specific constraints of this model. Remember: for sinusoidal functions in the form $a + b\cos(ct)$, the range is always $[a-|b|, a+|b|]$. The coefficient $a$ is the midline, and $|b|$ is the amplitude of oscillation.
13. The perceived brightness, $B$, of a light source is modelled by $B(d) = \frac{1200}{d^2}$, where $d$ is the distance from the source in metres, for $d > 0$. Which statement correctly interprets the function's behaviour as $d$ increases?
    Answer: As distance increases, the brightness decreases and approaches zero. When you encounter functions modeling real-world phenomena like light brightness, focus on analyzing the function's behavior by examining its mathematical form and considering what happens to the output as the input changes. The function $B(d) = \frac{1200}{d^2}$ is an inverse square relationship, which is fundamental in physics for phenomena like light intensity, gravitational force, and sound intensity. As distance $d$ increases, you're dividing 1200 by increasingly larger values of $d^2$, making $B(d)$ smaller and smaller. To see this clearly, test some values: at $d = 1$, $B = 1200$; at $d = 10$, $B = 12$; at $d = 100$, $B = 0.12$. As $d$ approaches infinity, $B$ approaches zero but never quite reaches it. This confirms that answer A is correct. Looking at the incorrect options: B suggests brightness increases, which contradicts the inverse relationship—larger denominators create smaller fractions. C claims brightness is constant, but this function clearly depends on distance; a constant function would have no variable in it. D describes linear decrease, but this function decreases much more rapidly than a linear relationship—it follows an inverse square pattern where doubling the distance quarters the brightness. Remember that inverse square relationships are everywhere in physics and always follow this pattern: as one quantity increases, the other decreases rapidly and approaches (but never reaches) zero. Recognizing $\frac{\text{constant}}{x^2}$ immediately tells you the behavior.
14. Using the piecewise function for parking cost, what is the interpretation of $C(3.5)$?
    Answer: It is the cost for parking for 3.5 hours, calculated as $8 + 4(3.5-1)$, which is $18. To find $C(3.5)$, we must first determine which piece of the function to use. Since the input $t=3.5$ satisfies the condition $1 < t \le 5$, we use the second formula: $C(t) = 8 + 4(t-1)$. Substituting $t=3.5$ gives $C(3.5) = 8 + 4(3.5-1) = 8 + 4(2.5) = 8 + 10 = 18$. This value represents the total cost in dollars for parking for 3.5 hours.
15. The number of users of a social media app, $N$, in millions, is given by $N(t) = 2t^2 + 10t + 50$, where $t$ is the number of months since its launch. What does the expression $\frac{N(6) - N(3)}{6-3}$ represent?
    Answer: The average rate of change in the number of users, in millions per month, from month 3 to month 6. The expression $\frac{f(b) - f(a)}{b-a}$ is the formula for the average rate of change of a function $f$ over the interval $[a, b]$. In this context, $N(6) - N(3)$ is the total change in the number of users (in millions) from month 3 to month 6. Dividing by the time interval, $6-3$ months, gives the average rate of change over that period, with units of millions of users per month.
16. The population of a species of owl in a forest is modelled by $P(t) = 120 + 20t - 0.5t^2$, where $t$ is the number of years since 2020. The model is considered valid for $0 \le t \le 30$. What is the practical meaning of solving the inequality $P(t) > 250$?
    Answer: Finding the time period during which the owl population is above 250. The function $P(t)$ gives the population at a given time $t$. The inequality $P(t) > 250$ means 'the population is greater than 250'. Solving for $t$ in this inequality means finding the values of $t$ (the time in years) for which this condition is true. Therefore, the solution represents the time period during which the owl population exceeds 250.
17. The sound intensity level, $L$, in decibels (dB), is given by the function $L(I) = 10 \log_{10}\left(\frac{I}{I_0}\right)$, where $I$ is the sound intensity and $I_0$ is a reference intensity. What is the correct interpretation of $L(100I_0)$?
    Answer: The decibel level of a sound that is 100 times more intense than the reference. When you encounter logarithmic functions like this sound intensity formula, focus on understanding what the input variable represents and how the function transforms it. The function $L(I) = 10 \log_{10}\left(\frac{I}{I_0}\right)$ converts sound intensity ratios into decibel levels. To find $L(100I_0)$, substitute $I = 100I_0$ into the formula: $L(100I_0) = 10 \log_{10}\left(\frac{100I_0}{I_0}\right) = 10 \log_{10}(100) = 10 \times 2 = 20 \text{ dB}$ This represents the decibel level when the sound intensity is 100 times the reference intensity $I_0$. The key insight is that $L(100I_0)$ is asking for the output (decibel level) when the input intensity is $100I_0$. Option A incorrectly reverses the relationship—it describes finding intensity from a given decibel level, which would require solving $L(I) = 100$ for $I$. Option B contains two errors: it suggests the sound is less intense (when $100I_0 > I_0$) and misinterprets what the function calculates. Option C describes the change in decibels rather than the absolute decibel level, which isn't what $L(100I_0)$ represents. Option D correctly identifies that we're finding the decibel level (the function's output) for a sound with intensity $100I_0$, which is indeed 100 times more intense than the reference $I_0$. Study tip: In function notation $f(x)$, always identify what goes in (the input) and what comes out (the output). Don't confuse finding $f(a)$ with solving $f(x) = a$.
18. The value of an investment, $V$, in thousands of dollars, is modelled by $V(t) = 10(1.04)^t$, where $t$ is the number of years since the initial investment. Let $V^{-1}(x)$ be the inverse function of $V(t)$. What is the practical interpretation of $V^{-1}(20)$?
    Answer: The number of years it takes for the investment's value to reach $20,000. An inverse function reverses the input and output. If $V(t)$ takes a time $t$ and gives a value $V$, then $V^{-1}(x)$ takes a value $x$ and gives the time $t$ at which that value was reached. The input to $V^{-1}$ is 20, which represents a value of $20,000 (since V is in thousands). The output will be the time $t$. Therefore, $V^{-1}(20)$ is the number of years it takes for the investment's value to reach $20,000.
19. A company's net profit, $P$, in thousands of dollars, is modelled by $P(t) = -2t^2 + 16t - 24$, where $t$ is the number of years since 2010. The model is valid for $0 \le t \le 10$. What is the practical interpretation of the solutions to $P(t) = 0$?
    Answer: The years when the company's net profit was zero, known as the break-even points. The equation $P(t) = 0$ sets the net profit function equal to zero. The solutions for $t$ are the times when the profit is zero. In a business context, this is known as the break-even point, where revenue equals costs. Solving $-2t^2 + 16t - 24 = 0$ gives $t=2$ and $t=6$, which are the years (after 2010) when the company broke even.
20. The number of bacteria in a petri dish, $N$, is modelled by the function $N(t) = 500e^{0.1t}$, where $t$ is the time in hours. What does the value 500 represent in the context of this model?
    Answer: The initial number of bacteria in the dish. The function is of the form $N(t) = N_0 e^{kt}$, where $N_0$ is the initial amount. To find the initial number of bacteria, we evaluate the function at time $t=0$. $N(0) = 500e^{0.1 \times 0} = 500e^0 = 500 \times 1 = 500$. Thus, 500 represents the initial number of bacteria.

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