FSP Training 06

Preparation for Mock Exam 07

Instructions

This training page covers all of Section 06 and prepares you for Mock Exam 07. Set A trains setting up: you translate a text into mathematics and check your translation. Set B is a practice mock in the format of Mock Exam 07: 180 minutes plus 5 minutes reading time, 100 points, and all 3 problems must be completed. Write down every step as you would in the exam. The points show what each part is worth.

In the real FSP, decimals are written with a comma (1,25 means 1.25).

Set A: Set Up & Check (50 minutes)

From Text to Mathematics (xx)

Translate each statement into mathematics. Then answer the check question: it tells you whether your translation works. (QU = quantity units, MU = monetary units)

1. An FSP task says: “Find the antiderivative of \(f(x) = \frac{2x^4 - 9}{x^3}\) and show your work.” Here \(x > 0\). Rewrite \(f(x)\) as a sum of powers of \(x\) and give all antiderivatives of \(f\).

   Check: Differentiate your antiderivative. Do you get \(f\) back? A classmate integrates the numerator and the denominator separately and gets \(G(x) = \frac{0.4x^5 - 9x}{0.25x^4}\). Compute \(G'(1)\) and compare it with \(f(1)\).

2. A company forges kitchen knives: “At an output of \(x\) QU, the marginal costs are \(C'(x) = 0.3x^2 - 2.8x + 15\) MU per QU. The total costs of 10 QU are 200 MU.” Write down the condition that fixes the constant of integration and determine \(C(x)\) and the fixed costs.

   Check: Compute \(C(10)\) with your function. How much of the 200 MU are fixed costs, and how much are variable costs?

3. A new tennis club: “\(t\) weeks after the opening, new members sign up at a rate of \(n(t) = 45 + 4t\) members per week.” The club wants to know when it has 650 members. Write the condition “650 members have signed up” as an equation with a definite integral and solve it.

   Check: Compute \(\int_0^T n(t) \, dt\) for your \(T\). A classmate solves \(n(T) = 650\) instead and gets \(T = 151.25\). What does the classmate’s number measure?

4. A workshop makes leather belts to order: “At an output of \(x\) belts per day (\(x \geq 10\)), the marginal costs are \(C'(x) = 4 + \frac{80}{x}\) EUR per belt.” Write the rise of the daily costs when the output rises from 20 to 40 belts as a definite integral and compute it. Give the exact value and a decimal rounded to two decimal places.

   Check: A classmate uses the power rule for \(\frac{80}{x} = 80x^{-1}\) and writes \(80 \cdot \frac{x^0}{0}\). Explain what goes wrong. A second classmate writes \(80\ln(40 - 20)\) for the logarithm part. Compare it with your result.

5. An FSP-style task says: “The graph of \(f(x) = e^{0.2x}\), the \(x\)-axis and the lines \(x = 0\) and \(x = 10\) enclose a region. The vertical line \(x = c\) divides this region into two parts of equal size. Determine \(c\).” Write down the equation for \(c\) with integrals and solve it. Round to two decimal places.

   Check: A classmate takes the middle of the interval, \(c = 5\). Compute the area to the left of \(x = 5\) and compare it with half of the whole area. Why must \(c\) lie to the right of the middle?

6. A laser tag arena in a city: “Nobody pays more than 150 EUR for a group booking (maximum price), and at a price of 0 EUR, 2,000 bookings per month would be made (market saturation). The demand is linear. The market price is 90 EUR per booking.” Determine the demand function \(D(q)\) for \(q\) bookings per month and the number of bookings at the market price. Write the consumer surplus as an integral and compute it.

   Check: The region of the consumer surplus is a triangle. Compute its area with \(\frac{1}{2} \cdot \text{base} \cdot \text{height}\). Do both results agree?

7. The market for honey in a region, with \(q\) in tonnes per year and prices in EUR per kg: “Demand \(D(q) = 19 - 0.25q\), supply \(S(q) = 4 + 0.125q\). The state sets a minimum price of 13 EUR per kg to support the beekeepers.” Determine the equilibrium and the quantity traded at the minimum price. Write the new consumer surplus and the new producer surplus as integrals and compute them in EUR.

   Check: Another proposal is a minimum price of 7 EUR per kg. Explain why this minimum price would change nothing on the market.

8. A company starts producing a new kitchen blender: “\(t\) months after the start, the company produces \(s(t) = 50 + 7t - 0.25t^2\) hundred blenders per month.” An FSP task says: “Compute the average monthly output volume within the first 12 months.” Write the average as a formula with an integral and compute it.

   Check: A classmate takes the mean of the first and the last value, \(\frac{s(0) + s(12)}{2}\). Compute it. Why does it differ from your result, and for which kind of function would it give the average value?

9. An FSP task says: “Find the antiderivative of \(f(x) = (2x - 3) \cdot e^{0.5x}\) and show your work.” Write down \(u\), \(du\), \(dv\) and \(v\) and evaluate the integral with integration by parts.

   Check: Differentiate your result. A classmate writes \(v = 0.5e^{0.5x}\). Find the classmate’s result and show by differentiating that it is wrong.

10. An FSP-style task says: “Determine the second derivative of \(f(x) = -0.25x^4 + 1.5x^3 + 6x^2 - 5x\). Draw the graph of \(f''\) and use it to describe the curvature of \(G_f\).” Compute \(f''(x)\) and its zeros, sketch the graph of \(f''\) and explain where \(G_f\) is concave up, where it is concave down and where its inflection points lie.

    Check: Compute \(f'(0)\) and \(f''(0)\). A classmate says: “\(f'(0) < 0\), so \(G_f\) is concave down at \(x = 0\).” Is the classmate right?

CautionSolution

1. Rewrite: \(f(x) = \frac{2x^4}{x^3} - \frac{9}{x^3} = 2x - 9x^{-3}\). The power rule term by term (the exponent grows by 1, then divide by the new exponent) gives

   \[F(x) = x^2 - 9 \cdot \frac{x^{-2}}{-2} + C = x^2 + \frac{9}{2x^2} + C\]

   Check: \(F'(x) = 2x + \frac{9}{2} \cdot (-2) x^{-3} = 2x - 9x^{-3} = f(x)\) ✓. The classmate’s function is \(G(x) = \frac{0.4x^5}{0.25x^4} - \frac{9x}{0.25x^4} = 1.6x - 36x^{-3}\), so \(G'(x) = 1.6 + 108x^{-4}\) and \(G'(1) = 109.6\). But \(f(1) = 2 - 9 = -7\), so \(G\) is not an antiderivative of \(f\).

   Typical mistake: the classmate’s way. There is no rule that integrates a quotient part by part: rewrite the quotient as a sum of powers first. A second mistake is a lost sign, \(-9 \cdot \frac{x^{-2}}{-2} = -4.5x^{-2}\).

2. The sentence about the 10 QU fixes the constant: \(C(10) = 200\). Reverse the power rule term by term:

   \[C(x) = 0.1x^3 - 1.4x^2 + 15x + K, \qquad C(10) = 100 - 140 + 150 + K = 110 + K = 200 \iff K = 90\]

   \[C(x) = 0.1x^3 - 1.4x^2 + 15x + 90\]

   The fixed costs are \(C(0) = 90\) MU.

   Check: \(C(10) = 100 - 140 + 150 + 90 = 200\) ✓. Of the 200 MU, 90 MU are fixed costs and \(200 - 90 = 110\) MU are variable costs, \(\int_0^{10} C'(x) \, dx = 110\).

   Typical mistake: \(C(x) = 0.1x^3 - 1.4x^2 + 15x + 200\). The constant of integration equals the value at \(x = 0\), the fixed costs. This function gives \(C(10) = 310 \neq 200\). A condition at another output has to be substituted into \(C(x)\) and solved for the constant.

3. The members who have signed up after \(T\) weeks are the integral of the rate from 0 to \(T\):

   \[\int_0^T (45 + 4t) \, dt = \Big[45t + 2t^2\Big]_0^T = 45T + 2T^2 = 650\]

   Zero form: \(2T^2 + 45T - 650 = 0\). Quadratic formula: \(T = \frac{-45 \pm \sqrt{2025 + 5200}}{4} = \frac{-45 \pm 85}{4}\), so \(T = 10\) or \(T = -32.5\). The negative solution lies before the opening, so the club has 650 members after 10 weeks.

   Check: \(\int_0^{10} (45 + 4t) \, dt = 450 + 200 = 650\) ✓. The classmate’s \(T = 151.25\) solves \(45 + 4T = 650\): after 151.25 weeks, the rate would reach 650 new members per week. It says nothing about the total.

   Typical mistake: the classmate’s. \(n(t)\) is a rate (members per week), and the total number of members is the area under its graph.

4. The rise of the costs is the definite integral of the marginal costs. The term \(\frac{80}{x}\) needs \(\int \frac{1}{x} \, dx = \ln|x| + C\):

   \[\int_{20}^{40} \left(4 + \frac{80}{x}\right) dx = \Big[4x + 80\ln|x|\Big]_{20}^{40} = (160 + 80\ln 40) - (80 + 80\ln 20) = 80 + 80\ln 2 \approx 135.45 \text{ EUR}\]

   (With the logarithm rule \(\ln 40 - \ln 20 = \ln\frac{40}{20} = \ln 2\).)

   Check: the power rule \(\frac{x^{n+1}}{n+1}\) only works for \(n \neq -1\). For \(n = -1\) it divides by \(n + 1 = 0\), which is not defined: this is the case that needs \(\ln|x|\). The second classmate gets \(80\ln 20 \approx 239.66\) instead of \(80\ln 2 \approx 55.45\). The evaluation bar gives \(\ln 40 - \ln 20 = \ln 2\), not \(\ln(40 - 20)\).

   Typical mistake: both classmates’ steps.

5. \(f(x) = e^{0.2x} > 0\), so the whole area is

   \[\int_0^{10} e^{0.2x} \, dx = \Big[5e^{0.2x}\Big]_0^{10} = 5(e^2 - 1) \approx 31.95\]

   Equation: the part from 0 to \(c\) is half of it:

   \[\int_0^c e^{0.2x} \, dx = \frac{1}{2}\int_0^{10} e^{0.2x} \, dx \iff 5\left(e^{0.2c} - 1\right) = 2.5\left(e^2 - 1\right)\]

   \[e^{0.2c} = \frac{e^2 + 1}{2} \approx 4.19 \iff 0.2c = \ln\left(\frac{e^2 + 1}{2}\right) \iff c = 5\ln\left(\frac{e^2 + 1}{2}\right) \approx 7.17\]

   Check: \(\int_0^{5} e^{0.2x} \, dx = 5(e - 1) \approx 8.59\), much less than half of the area, \(15.97\). The graph rises, so the region is higher on the right. A part of equal size on the left must be wider: \(c\) lies to the right of the middle ✓.

   Typical mistake: the middle of the interval. It halves the area only if the graph is symmetric on the interval. A second mistake is \(e^{0.2c} = \frac{e^2}{2}\): the lower limit gives \(e^0 = 1\), which must be subtracted on both sides.

6. Demand: the line through \((0, 150)\) and \((2000, 0)\) has the slope \(\frac{0 - 150}{2000 - 0} = -0.075\), so \(D(q) = 150 - 0.075q\).

   Bookings at the market price: \(150 - 0.075q = 90 \iff q = 800\) bookings per month.

   \[CS = \int_0^{800} \left[(150 - 0.075q) - 90\right] dq = \Big[60q - 0.0375q^2\Big]_0^{800} = 48000 - 24000 = 24000 \text{ EUR}\]

   Check: the triangle has the base 800 bookings and the height \(150 - 90 = 60\) EUR, so \(\frac{1}{2} \cdot 800 \cdot 60 = 24{,}000\) EUR ✓.

   Typical mistake: the upper limit 2,000 (the market saturation). At the market price only 800 bookings are made, so the consumer surplus ends at \(q = 800\). A second mistake is the slope \(-\frac{2000}{150}\): the slope of \(D\) is the change of the price divided by the change of the quantity.

7. Equilibrium: \(19 - 0.25q = 4 + 0.125q \iff 15 = 0.375q \iff q^* = 40\) t, and \(p^* = 19 - 10 = 9\) EUR per kg.

   At the minimum price of 13 EUR per kg (above \(p^*\)), the buyers take only \(19 - 0.25q = 13 \iff q = 24\) t. The beekeepers would like to sell \(4 + 0.125q = 13 \iff q = 72\) t, but only 24 t are traded.

   Units: EUR per kg times tonnes gives thousand EUR, because 1 t = 1,000 kg.

   \[CS_{new} = \int_0^{24} \left[(19 - 0.25q) - 13\right] dq = \Big[6q - 0.125q^2\Big]_0^{24} = 144 - 72 = 72\]

   \[PS_{new} = \int_0^{24} \left[13 - (4 + 0.125q)\right] dq = \Big[9q - 0.0625q^2\Big]_0^{24} = 216 - 36 = 180\]

   The new consumer surplus is 72,000 EUR and the new producer surplus 180,000 EUR per year.

   Check: 7 EUR per kg lies below the equilibrium price of 9 EUR per kg. The market price of 9 EUR is allowed anyway, so buyers and beekeepers keep trading 40 t at 9 EUR per kg. Only a minimum price above \(p^*\) changes the market.

   Typical mistake: the upper limit 72 (what the beekeepers would like to sell) or 40 (the old equilibrium quantity). Only the quantity that the buyers take at the minimum price is traded.

8. The average value on \([0; 12]\):

   \[\bar{s} = \frac{1}{12 - 0}\int_0^{12} \left(50 + 7t - 0.25t^2\right) dt = \frac{1}{12}\left[50t + 3.5t^2 - \frac{t^3}{12}\right]_0^{12} = \frac{1}{12}(600 + 504 - 144) = \frac{960}{12} = 80\]

   On average, the company produces 80 hundred, i.e. 8,000 blenders per month in the first 12 months.

   Check: \(s(0) = 50\) and \(s(12) = 50 + 84 - 36 = 98\), so the classmate gets \(\frac{50 + 98}{2} = 74 \neq 80\). The graph of \(s\) is concave down (\(s''(t) = -0.5 < 0\)): between the two ends it lies above the straight line through \((0, 50)\) and \((12, 98)\), e.g. \(s(6) = 83 > 74\). The mean of the two end values gives the average value only for a linear function, whose graph is that straight line.

   Typical mistake: forgetting the factor \(\frac{1}{12}\). The integral 960 is the total output of the 12 months (96,000 blenders), not the monthly average.

9. LIATE: \(2x - 3\) is algebraic, so \(u = 2x - 3\) and \(dv = e^{0.5x} \, dx\). Then \(du = 2 \, dx\) and \(v = \frac{1}{0.5}e^{0.5x} = 2e^{0.5x}\).

   \[\int (2x - 3) \cdot e^{0.5x} \, dx = (2x - 3) \cdot 2e^{0.5x} - \int 2e^{0.5x} \cdot 2 \, dx = (4x - 6)e^{0.5x} - 8e^{0.5x} + C = (4x - 14) \cdot e^{0.5x} + C\]

   Check: \(\left((4x - 14)e^{0.5x}\right)' = 4e^{0.5x} + (4x - 14) \cdot 0.5e^{0.5x} = e^{0.5x}(4 + 2x - 7) = (2x - 3) \cdot e^{0.5x}\) ✓. With \(v = 0.5e^{0.5x}\), the classmate gets \((2x - 3) \cdot 0.5e^{0.5x} - \int 0.5e^{0.5x} \cdot 2 \, dx = (x - 3.5)e^{0.5x} + C\). Its derivative is \(e^{0.5x} + (x - 3.5) \cdot 0.5e^{0.5x} = (0.5x - 0.75)e^{0.5x}\), only a quarter of \(f(x)\).

   Typical mistake: the classmate’s \(v\). \(0.5e^{0.5x}\) is the derivative of \(e^{0.5x}\), but \(v\) must be an antiderivative of \(dv\): \(\int e^{0.5x} \, dx = 2e^{0.5x}\).

10. \(f'(x) = -x^3 + 4.5x^2 + 12x - 5\) and

    \[f''(x) = -3x^2 + 9x + 12 = -3\left(x^2 - 3x - 4\right) = -3(x + 1)(x - 4)\]

    The zeros of \(f''\) are \(x = -1\) and \(x = 4\). The graph of \(f''\) is a parabola that opens downward (\(-3 < 0\)), with the \(y\)-intercept 12 and the vertex at \(x = 1.5\), \(f''(1.5) = 18.75\).

    

    Figure 1: Graph of f″ and the curvature of the graph of f

    Curvature of \(G_f\):

    • \(f''(x) < 0\) for \(x < -1\) and for \(x > 4\): \(G_f\) is concave down there
    • \(f''(x) > 0\) for \(-1 < x < 4\): \(G_f\) is concave up there
    • \(f''\) changes its sign at \(x = -1\) and at \(x = 4\): inflection points \((-1, 9.25)\) and \((4, 108)\), since \(f(-1) = -0.25 - 1.5 + 6 + 5 = 9.25\) and \(f(4) = -64 + 96 + 96 - 20 = 108\)

    Check: \(f'(0) = -5\) and \(f''(0) = 12 > 0\). At \(x = 0\), \(G_f\) falls (\(f' < 0\)), but it is concave up (\(f'' > 0\)). The classmate is wrong.

    Typical mistake: the classmate’s. The sign of \(f'\) only tells whether \(G_f\) rises or falls, and the sign of \(f\) whether it lies above or below the \(x\)-axis. The curvature comes from the sign of \(f''\). A second mistake is to take the zeros of \(f'\) (the stationary points) as inflection points.

One Paragraph, All Conditions (xx)

A company produces smoke detectors and sells them to DIY stores. Its marginal costs are a quadratic function of the output: they are lowest at an output of 4,000 smoke detectors per month, with 20 EUR per smoke detector. The variable costs of the first 6,000 smoke detectors per month are 138,000 EUR. The rent for the hall and the salaries of the permanent staff come to 80,000 EUR per month, even without production. The DIY stores pay 47 EUR per smoke detector, however many they buy. The company can produce at most 12,000 smoke detectors per month. The owner wants to know the output with the highest profit, a proof that it really is a maximum, and the maximum profit.

(QU = quantity units, MU = monetary units. 1 QU = 1,000 smoke detectors, 1 MU = 1,000 EUR, so a price of \(p\) MU per QU is a price of \(p\) EUR per smoke detector)

List every condition and every function in this paragraph as a mathematical statement, with \(x\) in QU and all money in MU. Do not solve anything.

Check: An assistant writes the sentence about the first 6,000 smoke detectors as \(C(6) = 138\). Explain what is wrong and give the correct condition.

CautionSolution

1. Marginal costs, a quadratic function with its lowest point (vertex) at 4 QU and 20 MU per QU: \(C'(x) = a(x - 4)^2 + 20\) with \(a > 0\) (the parabola opens upward, because 20 is the lowest value)
2. Variable costs of the first 6 QU, 138,000 EUR \(= 138\) MU: \(\int_0^6 C'(x) \, dx = C(6) - C(0) = 138\)
3. Fixed costs, 80,000 EUR \(= 80\) MU: \(C(0) = 80\)
4. Cost function: \(C(x) = \int C'(x) \, dx\), with the constant of integration from \(C(0) = 80\)
5. Revenue at 47 EUR per smoke detector \(= 47\) MU per QU: \(R(x) = 47x\)
6. Profit: \(P(x) = R(x) - C(x)\), so \(P'(x) = 47 - C'(x)\)
7. Domain: \(0 \leq x \leq 12\) (capacity 12,000 smoke detectors)
8. Highest profit on \([0; 12]\): \(P'(x) = 0\), i.e. \(C'(x) = 47\), with the proof \(P''(x) = -C''(x) < 0\), and a comparison with the values at the ends \(x = 0\) and \(x = 12\)

Check: \(C(6)\) are the total costs of 6 QU, and they contain the fixed costs: \(C(6) = 80 + 138 = 218\). The variable costs are the costs without the fixed costs, so the correct condition is \(C(6) - C(0) = \int_0^6 C'(x) \, dx = 138\).

Typical mistake: \(C'(4) = 0\) for the first sentence. At 4 QU the marginal costs are 20 MU per QU, not 0. The sentence gives the vertex of \(C'\): \(C'(4) = 20\) and \(C''(4) = 0\), and the vertex form already contains both.

FSP Words So Far

The FSP problems use these words from Sections 01 to 06. Learn what each one means and how to write it in mathematics.

Word	Meaning	Mathematics
quantity unit (QU)	a fixed number of items, e.g. 1 QU = 100 items	the output \(x\) is counted in QU
monetary unit (MU)	a fixed amount of money, e.g. 1 MU = 100 EUR	results in MU, convert to EUR at the end
fixed costs	costs that arise even if nothing is produced	the constant term of \(C(x)\)
variable costs	costs that grow with the output	\(C(x) - C(0)\) (for linear costs: output times costs per QU)
total costs	fixed costs + variable costs	\(C(x)\)
revenue	quantity times price	\(R(x) = x \cdot \text{price}\)
profit	revenue − total costs	\(P(x) = R(x) - C(x)\)
break-even point	the output at which the profit changes from negative (a loss) to positive	\(P(x) = 0\), the smaller zero
profit limit (upper break-even point)	the larger output with zero profit: above it, the profit turns negative again	\(P(x) = 0\), the larger zero
profit zone	all outputs with a positive profit	\(P(x) > 0\): from the break-even point to the profit limit (or the capacity)
capacity	the maximum possible output	an upper bound: \(x \leq\) capacity
growth factor	the factor by which a quantity is multiplied each period	\(q\) in (starting value) \(\cdot\, q^t\)
growth rate	the change per period in percent	\(q = 1 + r\) with \(r\) as a decimal, a loss of 4 % gives \(q = 0.96\)
doubling time	the time after which a quantity has doubled	\(q^T = 2\), continuous: \(e^{kT} = 2\), so \(T = \frac{\ln 2}{k}\)
half-life (halving time)	the time after which a quantity has halved	\(q^T = 0.5\), continuous: \(e^{-kT} = 0.5\), so \(T = \frac{\ln 2}{k}\)
“deviates from \(a\) by at most \(d\)” (tolerance)	the distance from \(a\) is at most \(d\)	\(\lvert x - a \rvert \leq d \iff a - d \leq x \leq a + d\)
“share of \(A\) among \(B\)”	the part of \(B\) that also belongs to \(A\): the base is \(B\)	\(\dfrac{\text{number in } A \cap B}{\text{number in } B}\)
“neither \(A\) nor \(B\)”	not in \(A\) and not in \(B\), i.e. outside both circles	\(\overline{A \cup B}\)
contingency table	a table of counts for two properties: the inner fields are the regions of the Venn diagram, the margins are the totals	inner fields \(A \cap B\), \(A \cap \overline{B}\), \(\overline{A} \cap B\), \(\overline{A} \cap \overline{B}\)
gross price, net price	gross price = net price + VAT	with 19 % VAT: gross \(= 1.19 \cdot\) net
system of linear equations	several linear equations that must hold at the same time	solve by substitution or elimination, check the solution in every equation
no solution, infinitely many solutions	the conditions contradict each other, or one condition only repeats another	elimination leaves \(0 = d\) with \(d \neq 0\), or \(0 = 0\)
demand, supply	the quantity buyers want to buy and the quantity sellers offer at a given price	as terms in the price \(p\), e.g. \(Q_d(p)\), \(Q_s(p)\), or as prices \(p_D(x)\), \(p_S(x)\) for the quantity \(x\)
market equilibrium	the quantity demanded equals the quantity supplied	demand = supply: \(Q_d(p) = Q_s(p)\), or \(p_D(x) = p_S(x)\). With demand and supply as prices: \(D(q) = S(q)\) gives \(q^*\), and \(p^* = D(q^*)\)
price-demand function	gives the price at which \(x\) QU can be sold	\(p(x)\), revenue \(R(x) = x \cdot p(x)\)
average cost	total costs divided by the number of units	\(\dfrac{C(x)}{x}\) with \(x \neq 0\)
zero form	all terms on one side, 0 on the other side	\(ax^2 + bx + c = 0\), then factor or use the quadratic formula
discriminant	decides how many real solutions a quadratic equation has	\(\Delta = b^2 - 4ac\): \(> 0\) two, \(= 0\) one, \(< 0\) none
biquadratic equation	an equation with only \(x^4\), \(x^2\) and a constant	substitute \(u = x^2\) in \(ax^4 + bx^2 + c\), solve for \(u\), then \(x = \pm\sqrt{u}\) for \(u \geq 0\)
domain restrictions (of an equation)	the conditions the unknown must meet so that every term is defined	denominators \(\neq 0\), under a root \(\geq 0\), inside a logarithm \(> 0\)
extraneous solution	a candidate that solves the squared equation but not the original one	check every candidate in the original equation
function	assigns to every input exactly one output	\(f(x)\), its graph passes the vertical line test
domain (of a function)	all inputs for which the function is defined, in a context only the sensible ones	e.g. \([0; \text{capacity}]\)
range	all outputs that the function takes on its domain	if the function grows or falls: between the values at the two ends, for a parabola also check the vertex
maximum price	the price at which nobody buys any more	\(Q_d(p) = 0\), or \(c\) in \(p = c - dQ\)
market saturation	the quantity demanded at a price of 0	\(a\) in \(Q_d = a - bp\), or the zero of \(p(x)\)
shortage, surplus	at a price below the equilibrium price, demand exceeds supply (shortage), above it, supply exceeds demand (surplus)	\(Q_d(p) > Q_s(p)\), or \(Q_d(p) < Q_s(p)\)
contribution margin	price minus variable costs per unit: the part of the price that covers the fixed costs	break-even quantity \(= \dfrac{\text{fixed costs}}{\text{contribution margin}}\)
vertex	the highest or lowest point of a parabola	\(x_v = -\dfrac{b}{2a}\), or \((h, k)\) in \(a(x - h)^2 + k\)
profit-maximizing output	the output with the highest profit	\(x_v\) of \(P(x)\), a maximum since \(a < 0\), and the price \(p(x_v)\). If the capacity lies below \(x_v\): the capacity and its price
shift, stretch, compression	moving or scaling a graph	\(a \cdot f(b(x - h)) + k\): first \(b\), then \(h\), then \(a\), then \(k\)
mirror image at the \(y\)-axis	the graph flipped from right to left	\(f(-x)\)
mirror image at the \(x\)-axis	the graph flipped upside down	\(-f(x)\)
composite function	first the inner function, then the outer function	\((f \circ g)(x) = f(g(x))\)
inverse function	reverses a function: gives the input that belongs to an output	solve \(y = f(x)\) for \(x\), e.g. \(x(p)\) from \(p(x)\)
polynomial of third degree	a function with powers of \(x\) up to \(x^3\)	\(ax^3 + bx^2 + cx + d\) with \(a \neq 0\): four unknowns, so four conditions
degree, leading coefficient	the highest power of \(x\), and the number in front of it	together they decide the end behavior
zero of multiplicity \(m\)	the factor \((x - c)\) appears \(m\) times in the factored form	odd \(m\): the graph crosses the \(x\)-axis at \(c\), even \(m\): it touches it
polynomial long division	divides a polynomial by \((x - c)\) once \(c\) is a known zero	\(f(x) = (x - c) \cdot q(x)\), the remainder must be 0
asymptotic behavior (end behavior)	what happens to \(f(x)\) for \(x \to \infty\) and \(x \to -\infty\)	polynomial: degree and sign of the leading coefficient. Rational function: its asymptotes
intercepts	the points where the graph meets the axes	\(y\)-intercept \((0, f(0))\), \(x\)-intercepts from \(f(x) = 0\)
rational function	a quotient of two polynomials	\(\dfrac{p(x)}{q(x)}\) with the domain \(q(x) \neq 0\)
vertical asymptote	a vertical line near which the values grow without bound	a zero of the denominator that does not cancel
hole	a single missing point of the graph	a factor that cancels, its \(y\)-value comes from the simplified term
horizontal asymptote	a horizontal line that the graph approaches for \(x \to \pm\infty\)	\(y = c\) with \(f(x) \to c\)
oblique asymptote	a slanted line that the graph approaches for \(x \to \pm\infty\)	the linear part of the long division, if the numerator’s degree is one higher
grows less than proportionally	grows, but twice the input gives less than twice the output	\(a \cdot x^b\) with \(0 < b < 1\)
continuous growth rate	the rate \(k\) of a model with the base \(e\)	\(N_0 \cdot e^{kt}\), growth factor per period \(e^k\)
saturation level	a level that a quantity approaches but never exceeds	a horizontal asymptote \(y = L\), e.g. of \(L - a \cdot e^{-kt}\) or \(\dfrac{L}{1 + A e^{-kt}}\)
behavior for \(t \to \infty\)	the value a quantity approaches in the long run	\(\lim\limits_{t \to \infty} N(t)\), often with \(e^{-kt} \to 0\)
radian	an angle measure in which a full turn is \(2\pi\)	\(180° = \pi\), arc length \(s = r\theta\)
amplitude	half the distance between the highest and the lowest value	\(\lvert A \rvert = \dfrac{\max - \min}{2}\)
midline	the horizontal line halfway between the highest and the lowest value	\(y = D = \dfrac{\max + \min}{2}\)
period	the length of one full cycle	\(B = \dfrac{2\pi}{\text{period}}\)
phase shift	the horizontal shift of a sine or cosine curve	\(C\) in \(A\sin(B(t - C)) + D\) or \(A\cos(B(t - C)) + D\)
mirror image at the midline	the graph flipped upside down about its midline \(y = D\)	\(2D - f(t)\)
limit	the value that \(f(x)\) approaches when \(x\) approaches a number \(a\), or \(\pm\infty\)	\(\lim\limits_{x \to a} f(x)\), one-sided: \(x \to a^-\) (from the left), \(x \to a^+\) (from the right)
continuous (at \(a\))	the graph has no jump and no gap at \(a\)	\(f(a)\) is defined, \(\lim\limits_{x \to a} f(x)\) exists, and both are equal
removable discontinuity	a gap that a single value could close, e.g. a hole	the limit exists, but \(f(a)\) is not defined or differs from it
derivative	the instantaneous rate of change: the slope of the tangent	\(f'(x) = \lim\limits_{h \to 0} \dfrac{f(x + h) - f(x)}{h}\)
secant, slope of the secant	the line through two points of the graph, and its slope (the average rate of change)	\(\dfrac{f(b) - f(a)}{b - a}\) on \([a; b]\)
tangent, tangential gradient	the line that touches the graph at a point, and its slope	\(y = f'(a)(x - a) + f(a)\) with the slope \(f'(a)\)
angle of intersection (with the \(x\)-axis)	the angle at which a line with slope \(m\) crosses the \(x\)-axis	\(\alpha = \arctan(m)\), with the indicator D for degrees
marginal cost, marginal revenue, marginal profit	about the change of the costs, the revenue or the profit caused by one more unit	\(C'(x)\), \(R'(x)\), \(P'(x) = R'(x) - C'(x)\)
stationary point (critical point)	a point of the graph with a horizontal tangent	\(f'(x_0) = 0\)
nature of a stationary point	whether it is a local maximum, a local minimum or a saddle point	sign change of \(f'\) at \(x_0\), or the sign of \(f''(x_0)\)
local maximum, local minimum	a point that is higher (lower) than all points nearby	\(f'(x_0) = 0\) and \(f''(x_0) < 0\) (maximum) or \(f''(x_0) > 0\) (minimum)
saddle point	a stationary point at which \(f'\) does not change its sign: the graph levels off and keeps its direction	\(f'(x_0) = 0\) without a sign change, e.g. \(f''(x_0) = 0\) and \(f'''(x_0) \neq 0\)
necessary condition, sufficient condition	a condition every candidate must meet, and a condition that proves the kind of point (“prove it really is a maximum”)	necessary: \(f'(x_0) = 0\), sufficient: e.g. \(f''(x_0) < 0\) for a maximum
global maximum, global minimum (on an interval)	the highest or lowest value on the whole interval	compare the local extrema with the values at the ends of the interval
concave up, concave down	the graph curves upward like a cup, or downward like a cap	\(f''(x) > 0\), or \(f''(x) < 0\)
inflection point	the point at which the curvature changes sign: the graph is steepest or flattest there	\(f''(x_0) = 0\) and \(f'''(x_0) \neq 0\), or a sign change of \(f''\)
“the curvature changes sign”	the graph changes from concave down to concave up, or back	an inflection point: \(f''(x_0) = 0\). For a cubic cost function, the marginal costs are lowest there
inflectional tangent	the tangent at an inflection point: it shows the fastest growth or decline	\(y = f'(a)(x - a) + f(a)\) with \(f''(a) = 0\)
unit costs, lowest unit costs	the total costs per unit, and the output at which they are lowest	\(k(x) = \dfrac{C(x)}{x}\), minimum from \(k'(x) = 0\) and \(k''(x) > 0\)
long-term price floor	the lowest unit costs: below this price, the company cannot cover its total costs in the long run	the minimum value of \(k(x) = \dfrac{C(x)}{x}\)
variable costs per unit	the variable costs divided by the output	\(k_v(x) = \dfrac{C(x) - C(0)}{x}\)
short-term lower limit price	the lowest variable costs per unit: below this price, the company cannot even cover its variable costs	the minimum value of \(k_v(x)\)
function family (German: Funktionsschar)	functions with the same term that differ only in a parameter	\(f_a(x)\), e.g. \(f_a(x) = x^2 - ax + 1\)
parameter	a letter that is fixed within one function of the family but changes from one function to the next	solve a condition for it, e.g. \(f_a'(2) = 0\)
antiderivative (German: Stammfunktion)	a function whose derivative is \(f\)	\(F'(x) = f(x)\)
constant of integration	the free constant: two antiderivatives of \(f\) differ only by a constant	\(F(x) + C\), fixed by one condition, e.g. \(C(0) =\) fixed costs or \(P(0) = -\)fixed costs
indefinite integral	the family of all antiderivatives	\(\int f(x) \, dx = F(x) + C\)
definite integral	a number: the signed area between the graph and the \(x\)-axis from \(a\) to \(b\)	\(\int_a^b f(x) \, dx = F(b) - F(a)\) (Fundamental Theorem of Calculus)
lower limit, upper limit	the start \(a\) and the end \(b\) of the interval in \(\int_a^b f(x) \, dx\)	“in the second month”: from \(t = 1\) to \(t = 2\)
signed area	area above the \(x\)-axis counts positive, area below it negative	\(\int_a^b f(x) \, dx\), it can be negative or 0
area (between a graph and the \(x\)-axis)	always a positive number	split at the zeros and add the absolute values of the integrals of the pieces
net change	the total change of a quantity, computed from its rate of change	\(F(b) - F(a) = \int_a^b F'(x) \, dx\), e.g. extra costs \(\int_a^b C'(x) \, dx\)
“total … within the first … months”	everything that has come together from the start up to month \(b\)	\(\int_0^b (\text{rate}) \, dt\)
area enclosed by two graphs	the region between the graphs from one intersection point to the next	\(\int_a^b [f(x) - g(x)] \, dx\) with \(f\) on top, limits from \(f(x) = g(x)\), split where the graphs cross
“both regions have equal size”	two regions have the same area (not: the integral over both is 0)	compute both areas separately and compare them
consumer surplus	what the buyers gain: they would have paid more than the market price	\(CS = \int_0^{q^*} [D(q) - p^*] \, dq\)
producer surplus	what the sellers gain: they would have sold for less than the market price	\(PS = \int_0^{q^*} [p^* - S(q)] \, dq\)
total surplus	consumer surplus + producer surplus	\(\int_0^{q^*} [D(q) - S(q)] \, dq\), the area between demand and supply
minimum price	a price set by the state: no unit may be sold for less	above \(p^*\): only the quantity demanded at this price is traded, and the surpluses end there. Below \(p^*\): no effect
average value	the constant height that gives the same area (the same total) over the interval	\(f_{avg} = \frac{1}{b - a}\int_a^b f(x) \, dx\), “average monthly … within the first \(b\) months”: \(\frac{1}{b}\int_0^b f(t) \, dt\)
“prove that \(F\) is an antiderivative of \(f\)”	show it by differentiating, no integrating needed	compute \(F'(x)\), e.g. with the product and chain rule, and show \(F'(x) = f(x)\)
integration by parts	the integration rule that reverses the product rule	\(\int u \, dv = uv - \int v \, du\), choose \(u\) with LIATE

In set-builder notation, \(\{x \in \mathbb{R} \mid \dots\}\) and \(\{x \in \mathbb{R} : \dots\}\) mean the same.

Set B: Practice Mock (180 minutes, 100 points)

Problem 1: A Cubic Read from Its Graph [25 pts. total]

The figure shows the graph \(G_f\) of a function \(f\) on the interval \([-5.5; 2.5]\). The graph touches the \(x\)-axis at \(x = 1\). The three marked points lie on grid points. Round to two decimal places.

Note: Print the page or copy the figure onto squared paper. In the exam, you draw directly into the printed figure.

Figure 2: Graph of f

1. Graph the derivative \(f'\) directly into the figure. Give the zeros of \(f'\) and the \(x\)-value at which \(f'\) has its lowest value, and explain each of them with the graph of \(f\). [5 pts.]

2. Determine the \(x\)-values of the local extrema and of the inflection points of an antiderivative \(F\) of \(f\), with the type of each extremum. Give the intervals on which \(F\) is concave up and concave down. Then sketch the antiderivative \(F\) whose graph passes through the point \((1, 0)\) in a new coordinate system.

Hint: Because \(F' = f\), the rules from Session 05-05 apply to \(F\): \(F\) increases where \(f\) is positive, \(F\) has a local extremum where \(f\) changes sign, and \(F\) has an inflection point where \(f\) has a local extremum.

[7 pts.]

3. \(G_f\) is the graph of a polynomial function of third degree, \(f(x) = ax^3 + bx^2 + cx + d\). Read four conditions from the figure and determine \(a\), \(b\), \(c\) and \(d\) computationally. [6 pts.]

(for verification: \(f(x) = 0.125x^3 + 0.375x^2 - 1.125x + 0.625\))

4. Determine the equation of the tangent \(t\) to \(G_f\) at \(x = -4\). Determine the other point of \(G_f\) at which the tangent is parallel to \(t\). [5 pts.]

5. A line with slope \(m\) meets the \(x\)-axis at the angle \(\alpha = \arctan(m)\). Compute the angle of intersection \(\alpha\) between \(t\) and the \(x\)-axis.

Hint: The indicator D must show. SHIFT + tan gives \(\arctan\), just as SHIFT + sin gives \(\sin^{-1}\).

[2 pts.]

CautionSolutions

a) Reading the graph of \(f\):

• \(G_f\) has horizontal tangents at the highest point \((-3, 4)\) and at \(x = 1\), where it touches the \(x\)-axis. So \(f'(-3) = 0\) and \(f'(1) = 0\): the zeros of \(f'\) are \(x = -3\) and \(x = 1\).
• \(f\) increases for \(x < -3\) and for \(x > 1\), so \(f' > 0\) there. \(f\) decreases between \(-3\) and \(1\), so \(f' < 0\) there.
• Between the two extrema, \(G_f\) falls most steeply at its inflection point, where the curvature changes from concave down to concave up. In the figure this happens at about \(x = -1\). There \(f'\) has its lowest value, about \(-1.5\) (on the grid, the graph falls by about 1.5 per unit near \(x = -1\)).

\(f\) is a polynomial of third degree, so \(f'\) is a parabola that opens upward, with the zeros \(-3\) and \(1\) and the vertex at about \((-1, -1.5)\). With c): \(f'(x) = 0.375x^2 + 0.75x - 1.125 = 0.375(x + 3)(x - 1)\), \(f''(-1) = 0.75 \cdot (-1) + 0.75 = 0\) and \(f'(-1) = -1.5\) ✓.

Figure 3: Graph of f (light) and its derivative f′

Grading: 1 Pt. zeros of \(f'\) at the extrema of \(f\), 1 Pt. sign of \(f'\) on the three intervals, 1 Pt. lowest value of \(f'\) at the inflection point \(x \approx -1\), 2 Pts. graph of \(f'\) (parabola through the zeros, vertex near \((-1, -1.5)\)).

b) Sign of \(f = F'\): \(f(x) < 0\) for \(x < -5\), and \(f(x) > 0\) for \(-5 < x < 1\) and for \(x > 1\). At \(x = 1\) the graph only touches the \(x\)-axis.

• \(x = -5\): \(f\) changes from \(-\) to \(+\), so \(F\) has a local minimum there.
• \(x = 1\): \(f(1) = 0\), but \(f\) does not change its sign. \(F\) has a horizontal tangent there and keeps increasing: no extremum, a saddle point of \(F\).

Inflection points of \(F\) at the local extrema of \(f\): at \(x = -3\) (maximum of \(f\)) and at \(x = 1\) (minimum of \(f\)). The saddle point at \(x = 1\) is an inflection point with a horizontal tangent.

Concavity: \(F'' = f'\). \(f\) increases on \((-5.5; -3)\) and on \((1; 2.5)\), so \(F\) is concave up there. \(f\) decreases on \((-3; 1)\), so \(F\) is concave down there.

Sketch: \(F\) falls to its minimum at \(x = -5\), then rises. It rises most steeply at \(x = -3\), levels off to a horizontal tangent at \((1, 0)\) and rises again. Your sketch only needs this shape. The figure shows the exact antiderivative \(F(x) = \frac{1}{32}(x - 1)^3(x + 7)\) for comparison, with \(F(-5) = -13.5\) and \(F(-3) = -8\).

Figure 4: The antiderivative F with F(1) = 0

Grading: 1 Pt. minimum of \(F\) at \(x = -5\), 1 Pt. no extremum at \(x = 1\) with the reason (no sign change of \(f\)), 1 Pt. inflection points at \(x = -3\) and \(x = 1\), 1 Pt. concavity intervals, 3 Pts. sketch (minimum, horizontal tangent at \((1, 0)\), correct curvature).

c) Conditions from the figure (with \(f'(x) = 3ax^2 + 2bx + c\)):

1. \(G_f\) touches the \(x\)-axis at \(x = 1\): \(f(1) = 0\), so \(a + b + c + d = 0\)
2. The tangent at \(x = 1\) is horizontal: \(f'(1) = 0\), so \(3a + 2b + c = 0\)
3. Point \((-3, 4)\): \(f(-3) = 4\), so \(-27a + 9b - 3c + d = 4\)
4. Zero at \(x = -5\): \(f(-5) = 0\), so \(-125a + 25b - 5c + d = 0\)

Solving: subtract equation 1 from equations 3 and 4 to remove \(d\):

• 3 − 1: \(-28a + 8b - 4c = 4\), so \(-7a + 2b - c = 1\)
• 4 − 1: \(-126a + 24b - 6c = 0\), so \(-21a + 4b - c = 0\)
• From 2: \(c = -3a - 2b\). Substituted: \(-4a + 4b = 1\) and \(-18a + 6b = 0\), so \(b = 3a\)
• Then \(-4a + 12a = 1\), so \(a = \frac{1}{8}\), \(b = \frac{3}{8}\), \(c = -\frac{3}{8} - \frac{6}{8} = -\frac{9}{8}\), \(d = -a - b - c = \frac{5}{8}\)

\[f(x) = 0.125x^3 + 0.375x^2 - 1.125x + 0.625\]

Check: \(f'(-3) = 3.375 - 2.25 - 1.125 = 0\) ✓, so \((-3, 4)\) is the highest point, as the figure shows. A shorter way uses the factored form: the double zero at \(x = 1\) and the zero at \(x = -5\) give \(f(x) = a(x - 1)^2(x + 5)\), and \(f(-3) = 32a = 4\) gives \(a = \frac{1}{8}\) ✓.

Grading: 2 Pts. four correct conditions, 2 Pts. equations in \(a\), \(b\), \(c\), \(d\), 1 Pt. solving the system, 1 Pt. \(f(x)\).

d) Point and slope: \(f(-4) = \frac{1}{8} \cdot 25 \cdot 1 = 3.125\) and \(f'(-4) = 6 - 3 - 1.125 = 1.875\).

\[t: \; y = 1.875(x + 4) + 3.125 = 1.875x + 10.625\]

Parallel tangent: parallel lines have the same slope, so \(f'(x) = 1.875\):

\[0.375x^2 + 0.75x - 1.125 = 1.875 \iff x^2 + 2x - 8 = 0 \iff (x + 4)(x - 2) = 0\]

\(x = -4\) is the point of contact of \(t\). The other point is \(x = 2\) with \(f(2) = \frac{1}{8} \cdot 1 \cdot 7 = 0.875\), so \((2, 0.875)\).

Grading: 1 Pt. \(f(-4)\) and \(f'(-4)\), 1 Pt. equation of \(t\), 1 Pt. condition \(f'(x) = 1.875\), 1 Pt. solutions of the quadratic equation, 1 Pt. point \((2, 0.875)\).

e)

\[\alpha = \arctan(1.875) \approx 61.93°\]

Grading: 1 Pt. \(\alpha = \arctan(1.875)\) with the slope of \(t\), 1 Pt. \(61.93°\) (indicator D).

Problem 2: An Exponential Function, Its Antiderivatives and an Area [40 pts. total]

Let the function \(f\) be given by

\[f(x) = \left(4 - x^2\right) \cdot e^{0.5x}, \quad x \in \mathbb{R}\]

Round all results to two decimal places.

1. Determine the \(x\)- and \(y\)-intercepts of \(G_f\). Investigate the asymptotic behavior of \(f\).

Hint: For \(x \to -\infty\), put \(t = -x\). Then \(e^{0.5x} = e^{-0.5t}\) with \(t \to \infty\).

[4 pts.]

2. Determine the first and the second derivative of \(f\). Factor out \(e^{0.5x}\) in both. [5 pts.]

(for verification: \(f'(x) = -0.5\left(x^2 + 4x - 4\right) \cdot e^{0.5x}\) and \(f''(x) = -0.25\left(x^2 + 8x + 4\right) \cdot e^{0.5x}\))

3. Determine the nature and the coordinates of all stationary points of \(f\). [5 pts.]

4. Determine the inflection points of \(f\). [4 pts.]

5. Sketch \(G_f\) in the interval \([-8; 2.5]\). Mark the intercepts, the stationary points and the inflection points. [5 pts.]

6. Show that \(G_f\) is not point-symmetric to the origin (that would mean \(f(-x) = -f(x)\) for every \(x\)). Give a function \(f_1(x)\) which is a mirror image of \(f(x)\) mirrored at the \(y\)-axis and give a reason for your choice. [3 pts.]

7. Prove that \(F\) given by \(F(x) = -2(x - 2)^2 \cdot e^{0.5x}\) is an antiderivative of \(f\). [4 pts.]

8. Give all antiderivatives of \(f\). Determine the antiderivative \(F_0\) of \(f\) whose graph passes through the point \((-2, 0)\). [3 pts.]

(for verification: \(F_0(x) = -2(x - 2)^2 \cdot e^{0.5x} + 32e^{-1}\))

1. Between its two zeros, \(G_f\) and the \(x\)-axis enclose a region. Compute its area. [3 pts.]

10. The vertical line \(x = c\) with \(-2 < c < 2\) cuts the region from i) into a left and a right part with the same area. Show that \(c\) satisfies \((c - 2)^2 \cdot e^{0.5c} = 8e^{-1}\), and solve this equation. Explain with f) why \(c\) lies to the right of the \(y\)-axis.

Hint: Solve the last equation with SOLVE (SHIFT + CALC) and the start value \(0\).

[4 pts.]

CautionSolutions

a) \(y\)-intercept: \(f(0) = 4 \cdot e^0 = 4\), i.e. \((0, 4)\).

\(x\)-intercepts: \(e^{0.5x} > 0\) for every \(x\), so only the bracket can become zero: \(4 - x^2 = 0 \iff x = -2\) or \(x = 2\), i.e. \((-2, 0)\) and \((2, 0)\).

\(x \to -\infty\): write \(f(x) = 4e^{0.5x} - x^2 \cdot e^{0.5x}\). Here \(e^{0.5x} \to 0\), and the exponential beats the power: with \(t = -x\), \(x^2 \cdot e^{0.5x} = t^2 \cdot e^{-0.5t} \to 0\) for \(t \to \infty\). So

\[\lim_{x \to -\infty} f(x) = 0\]

\(x \to \infty\): \(4 - x^2 \to -\infty\) and \(e^{0.5x} \to +\infty\):

\[\lim_{x \to \infty} f(x) = -\infty\]

Horizontal asymptote: \(y = 0\) (the \(x\)-axis), only for \(x \to -\infty\). For \(x < -2\), \(4 - x^2 < 0\), so the graph approaches the \(x\)-axis from below.

Grading: 1 Pt. \(y\)-intercept, 1 Pt. \(x\)-intercepts with the reason \(e^{0.5x} > 0\), 1 Pt. limit for \(x \to -\infty\) with the reason, 1 Pt. limit for \(x \to \infty\) and the asymptote \(y = 0\) only on the left.

b) First derivative (product rule, and the chain rule for \(e^{0.5x}\)): \(u = 4 - x^2\), \(v = e^{0.5x}\), \(u' = -2x\), \(v' = 0.5e^{0.5x}\).

\[f'(x) = -2x \cdot e^{0.5x} + \left(4 - x^2\right) \cdot 0.5e^{0.5x} = e^{0.5x}\left(-0.5x^2 - 2x + 2\right) = -0.5\left(x^2 + 4x - 4\right) \cdot e^{0.5x}\]

Second derivative (product rule on \(f'(x) = \left(-0.5x^2 - 2x + 2\right) \cdot e^{0.5x}\), with \(u' = -x - 2\)):

\[f''(x) = (-x - 2) \cdot e^{0.5x} + \left(-0.5x^2 - 2x + 2\right) \cdot 0.5e^{0.5x} = e^{0.5x}\left(-0.25x^2 - 2x - 1\right) = -0.25\left(x^2 + 8x + 4\right) \cdot e^{0.5x}\]

Grading: 1 Pt. \(u\), \(v\), \(u'\), \(v'\) with the chain rule, 1 Pt. product rule for \(f'\), 1 Pt. \(f'(x)\) factored, 1 Pt. product rule for \(f''\), 1 Pt. \(f''(x)\) factored.

c) Necessary condition: \(f'(x) = 0\). Since \(-0.5e^{0.5x} \neq 0\):

\[x^2 + 4x - 4 = 0 \iff x = \frac{-4 \pm \sqrt{16 + 16}}{2} = -2 \pm 2\sqrt{2}, \qquad x_1 \approx -4.83, \quad x_2 \approx 0.83\]

Sign change: \(x^2 + 4x - 4\) is a parabola that opens upward: negative between its zeros, positive outside. The factor \(-0.5e^{0.5x}\) is negative, so \(f' < 0\) for \(x < x_1\), \(f' > 0\) for \(x_1 < x < x_2\) and \(f' < 0\) for \(x > x_2\).

• \(x_1 \approx -4.83\): \(f'\) changes from \(-\) to \(+\), local minimum at \((-4.83, -1.73)\)
• \(x_2 \approx 0.83\): \(f'\) changes from \(+\) to \(-\), local maximum at \((0.83, 5.01)\)

(With the second derivative: \(f''(-4.83) \approx 0.25 > 0\) and \(f''(0.83) \approx -4.28 < 0\).)

Grading: 1 Pt. necessary condition with \(e^{0.5x} \neq 0\), 1 Pt. both \(x\)-values, 2 Pts. nature of both points, 1 Pt. coordinates.

d) Necessary condition: \(f''(x) = 0\). Since \(-0.25e^{0.5x} \neq 0\):

\[x^2 + 8x + 4 = 0 \iff x = \frac{-8 \pm \sqrt{64 - 16}}{2} = -4 \pm 2\sqrt{3}, \qquad x_3 \approx -7.46, \quad x_4 \approx -0.54\]

Sign change: \(x^2 + 8x + 4\) has two simple zeros, so it changes its sign at both, and so does \(f''\).

Coordinates: inflection points at \((-7.46, -1.24)\) and \((-0.54, 2.84)\). At \(x_4 \approx -0.54\), between the minimum and the maximum, \(G_f\) rises most steeply, with the slope \(f'(-0.54) \approx 2.24\).

Grading: 1 Pt. equation \(x^2 + 8x + 4 = 0\), 1 Pt. both \(x\)-values, 1 Pt. sign change, 1 Pt. coordinates.

e) Table of values (rounded to two decimal places):

\(x\)	\(-8\)	\(-7\)	\(-6\)	\(-5\)	\(-4\)	\(-3\)	\(-2\)	\(-1\)	\(0\)	\(1\)	\(2\)	\(2.5\)
\(f(x)\)	\(-1.10\)	\(-1.36\)	\(-1.59\)	\(-1.72\)	\(-1.62\)	\(-1.12\)	\(0\)	\(1.82\)	\(4\)	\(4.95\)	\(0\)	\(-7.85\)

Figure 5: Graph of f(x) = (4 − x²)e^(0.5x)

Grading: 1 Pt. labelled axes and a suitable scale on \([-8; 2.5]\), 1 Pt. intercepts, 1 Pt. stationary points, 1 Pt. inflection points with the correct curvature, 1 Pt. behavior at both ends (approach to the \(x\)-axis from below on the left, steep fall on the right).

f) Point symmetry to the origin would need \(f(-x) = -f(x)\) for every \(x\). At \(x = 0\) this gives \(f(0) = -f(0)\), i.e. \(f(0) = 0\). But \(f(0) = 4\), so \(G_f\) is not point-symmetric to the origin. (Another counterexample: \(f(-1) = 3e^{-0.5} \approx 1.82\), but \(-f(1) = -3e^{0.5} \approx -4.95\).)

Mirror image at the \(y\)-axis:

\[f_1(x) = f(-x) = \left(4 - (-x)^2\right) \cdot e^{0.5 \cdot (-x)} = \left(4 - x^2\right) \cdot e^{-0.5x}\]

Reason: mirroring at the \(y\)-axis moves every point \((x, y)\) of \(G_f\) to \((-x, y)\). So \(f_1\) takes at \(x\) the value that \(f\) takes at \(-x\): \(f_1(x) = f(-x)\).

Grading: 1 Pt. counterexample \(f(0) = 4 \neq 0\), 1 Pt. \(f_1(x)\), 1 Pt. reason.

g) \(F\) is an antiderivative of \(f\) if \(F'(x) = f(x)\). Product rule on \(F(x) = -2(x - 2)^2 \cdot e^{0.5x}\): \(u = -2(x - 2)^2\) with \(u' = -4(x - 2)\) (chain rule), \(v = e^{0.5x}\) with \(v' = 0.5e^{0.5x}\).

\[F'(x) = -4(x - 2) \cdot e^{0.5x} - 2(x - 2)^2 \cdot 0.5e^{0.5x} = -(x - 2) \cdot e^{0.5x}\left[4 + (x - 2)\right] = -(x - 2)(x + 2) \cdot e^{0.5x}\]

\[= \left(4 - x^2\right) \cdot e^{0.5x} = f(x) \quad \checkmark\]

Grading: 1 Pt. \(u'\) with the chain rule, 1 Pt. \(v'\), 1 Pt. product rule, 1 Pt. simplification to \(f(x)\).

h) All antiderivatives: \(F(x) + C = -2(x - 2)^2 \cdot e^{0.5x} + C\) with \(C \in \mathbb{R}\).

Through \((-2, 0)\): \(F(-2) + C = -2 \cdot 16 \cdot e^{-1} + C = -32e^{-1} + C = 0 \iff C = 32e^{-1} \approx 11.77\).

\[F_0(x) = -2(x - 2)^2 \cdot e^{0.5x} + 32e^{-1}\]

Grading: 1 Pt. \(F(x) + C\), 1 Pt. condition \(F_0(-2) = 0\), 1 Pt. \(F_0(x)\).

i) Between the zeros \(-2\) and \(2\), \(4 - x^2 \geq 0\), so \(f(x) \geq 0\) and the area is the integral:

\[A = \int_{-2}^{2} f(x) \, dx = F_0(2) - F_0(-2) = \left(0 + 32e^{-1}\right) - 0 = 32e^{-1} \approx 11.77\]

Grading: 1 Pt. limits \(-2\) and \(2\) with \(f \geq 0\), 1 Pt. evaluation with an antiderivative, 1 Pt. exact value and decimal.

j) Equation: the part from \(-2\) to \(c\) has half the area, with \(-2 < c < 2\):

\[\int_{-2}^{c} f(x) \, dx = \frac{1}{2} \cdot 32e^{-1} = 16e^{-1}\]

With \(F_0(-2) = 0\), the left side is \(F_0(c)\):

\[-2(c - 2)^2 \cdot e^{0.5c} + 32e^{-1} = 16e^{-1} \iff -2(c - 2)^2 \cdot e^{0.5c} = -16e^{-1} \iff (c - 2)^2 \cdot e^{0.5c} = 8e^{-1} \quad \checkmark\]

SOLVE with the start value \(0\) gives \(c \approx 0.48\), and this value lies in \((-2; 2)\).

Why \(c > 0\): by f), \(f(-x) = f_1(x) = \left(4 - x^2\right) \cdot e^{-0.5x}\). For \(0 < x < 2\), \(e^{0.5x} > 1 > e^{-0.5x}\), so \(f(x) > f(-x)\): at every \(x\) of the right half, the graph is higher than at the mirrored point of the left half, e.g. \(f(1) \approx 4.95\) and \(f(-1) \approx 1.82\). The right half \([0; 2]\) holds more than half of the area, so the dividing line lies to the right of the \(y\)-axis.

Figure 6: The region from i) and the line x = c

Grading: 1 Pt. equation with the integral and half the area, 1 Pt. transformation to \((c - 2)^2 \cdot e^{0.5c} = 8e^{-1}\), 1 Pt. \(c \approx 0.48\) with SOLVE, 1 Pt. explanation with the mirror image.

Problem 3: Grapes and Wine from a Wine Region [35 pts. total]

In a wine region, winegrowers sell their grapes to the wineries of the region. The market for grapes is modelled with \(q\) in thousand tonnes of grapes per year and prices in EUR per tonne. The winegrowers would sell the \(q\)-th thousand tonnes for at least

\[S(q) = 290 + 7q + 0.5q^2\]

EUR per tonne (supply function). Round to two decimal places.

1. Determine the linear demand function assuming a maximum price of 1,700 EUR per tonne and a market saturation at 68,000 tonnes of grapes per year. Give \(D(q)\) with \(q\) in thousand tonnes. [3 pts.]

(for verification: \(D(q) = 1700 - 25q\))

2. Determine the market equilibrium. [4 pts.]

3. Compute the consumer surplus and the producer surplus in EUR. Explain what the producer surplus means for the winegrowers. [7 pts.]

4. After a year with low prices, the state guarantees a minimum price of 1,050 EUR per tonne: no grapes may be sold for less. Determine the quantity that the wineries buy and the quantity that the winegrowers want to sell at this price. The state buys the remaining grapes at the minimum price. Compute the costs for the state and the new consumer surplus of the wineries. [6 pts.]

One winery of the region makes wine from its own grapes and sells it in bottles. At an output of \(x\) thousand bottles per year, its marginal profit is

\[P'(x) = -0.06x^2 + 1.56x + 7.2\]

EUR per bottle. The fixed costs of the winery are 250,000 EUR per year.

5. Determine the output with the highest profit and prove that no other output brings a higher profit. [4 pts.]

6. Determine the profit function \(P(x)\) in thousand EUR. Compute the maximum profit in EUR. [4 pts.]

A sparkling wine cellar of the region launches a new sparkling wine. \(t\) months after the launch, it sells

\[s(t) = 8 - 6e^{-t/3}\]

thousand bottles per month.

7. Compute the average monthly sales volume within the first 12 months. Determine the time at which the monthly sales reach this average. [7 pts.]

CautionSolutions

a) Points: maximum price 1,700 EUR per tonne, so \(D(0) = 1700\). Market saturation 68,000 t \(= 68\) thousand tonnes, so \(D(68) = 0\).

Slope:

\[m = \frac{0 - 1700}{68 - 0} = -25, \qquad D(q) = 1700 - 25q\]

Grading: 1 Pt. both points with \(q\) in thousand tonnes, 1 Pt. slope, 1 Pt. \(D(q)\).

b) Condition: \(D(q) = S(q)\):

\[1700 - 25q = 290 + 7q + 0.5q^2 \iff 0.5q^2 + 32q - 1410 = 0 \iff q^2 + 64q - 2820 = 0\]

\[q = \frac{-64 \pm \sqrt{4096 + 11280}}{2} = \frac{-64 \pm 124}{2}, \qquad q = 30 \text{ or } q = -94\]

The negative quantity makes no sense, so \(q^* = 30\) and \(p^* = D(30) = 1700 - 750 = 950\). Check: \(S(30) = 290 + 210 + 450 = 950\) ✓.

Market equilibrium: \((30, 950)\). The wineries buy 30,000 tonnes of grapes per year at a price of 950 EUR per tonne.

Grading: 1 Pt. condition \(D(q) = S(q)\), 1 Pt. zero form, 1 Pt. \(q^* = 30\) (negative solution excluded), 1 Pt. \(p^* = 950\) with the interpretation.

c) Units: EUR per tonne times thousand tonnes gives thousand EUR.

\[CS = \int_0^{30} \left[(1700 - 25q) - 950\right] dq = \Big[750q - 12.5q^2\Big]_0^{30} = 22500 - 11250 = 11250\]

\[PS = \int_0^{30} \left[950 - \left(290 + 7q + 0.5q^2\right)\right] dq = \left[660q - 3.5q^2 - \frac{q^3}{6}\right]_0^{30} = 19800 - 3150 - 4500 = 12150\]

The consumer surplus is 11.25 million EUR and the producer surplus 12.15 million EUR per year.

Meaning: many winegrowers would also sell their grapes for less than 950 EUR per tonne, but all of them receive the market price. Together, they receive 12.15 million EUR per year more than the lowest prices at which they would sell.

Grading: 1 Pt. units, 1 Pt. integral for \(CS\), 1 Pt. value of \(CS\), 1 Pt. integral for \(PS\), 2 Pts. antiderivative and value of \(PS\), 1 Pt. meaning.

d) Quantity bought (the wineries buy only what they demand): \(1700 - 25q = 1050 \iff q = 26\), i.e. 26,000 tonnes.

Quantity offered:

\[290 + 7q + 0.5q^2 = 1050 \iff q^2 + 14q - 1520 = 0 \iff q = \frac{-14 + \sqrt{196 + 6080}}{2} \approx \frac{-14 + 79.22}{2} \approx 32.61\]

(the negative solution is excluded). The winegrowers want to sell about 32,610 tonnes.

Costs for the state: it buys \(32.61 - 26 = 6.61\) thousand tonnes at 1,050 EUR per tonne: \(6.61 \cdot 1050 \approx 6941\) thousand EUR (with the unrounded quantity), i.e. about 6.94 million EUR per year.

New consumer surplus (only 26 thousand tonnes are bought by the wineries):

\[CS_{new} = \int_0^{26} \left[(1700 - 25q) - 1050\right] dq = \Big[650q - 12.5q^2\Big]_0^{26} = 16900 - 8450 = 8450\]

The wineries’ consumer surplus falls from 11.25 to 8.45 million EUR.

Grading: 1 Pt. 26,000 t, 2 Pts. quantity offered (equation and solution), 1 Pt. costs for the state, 2 Pts. new consumer surplus (integral with the upper limit 26 and value).

e) Necessary condition:

\[P'(x) = 0 \iff -0.06x^2 + 1.56x + 7.2 = 0 \iff x^2 - 26x - 120 = 0 \iff (x - 30)(x + 4) = 0\]

So \(x = 30\) (\(x = -4\) lies outside the domain).

Sufficient condition: \(P''(x) = -0.12x + 1.56\), so \(P''(30) = -2.04 < 0\): maximum.

No other output: \(P'(x) = -0.06(x - 30)(x + 4)\) is positive for \(0 \leq x < 30\) and negative for \(x > 30\). The profit rises up to 30 thousand bottles and falls after that, so the maximum at 30,000 bottles per year is the highest profit of all outputs.

Grading: 1 Pt. equation \(P'(x) = 0\), 1 Pt. \(x = 30\), 1 Pt. \(P''(30) < 0\), 1 Pt. sign of \(P'\) on both sides (no other output).

f) Antiderivative: EUR per bottle times thousand bottles gives thousand EUR.

\[P(x) = \int \left(-0.06x^2 + 1.56x + 7.2\right) dx = -0.02x^3 + 0.78x^2 + 7.2x + C\]

Constant of integration: without sales, the winery has no revenue but still pays its fixed costs, so \(P(0) = -250\) and \(C = -250\):

\[P(x) = -0.02x^3 + 0.78x^2 + 7.2x - 250\]

Maximum profit: \(P(30) = -540 + 702 + 216 - 250 = 128\) thousand EUR, i.e. 128,000 EUR per year.

Grading: 1 Pt. antiderivative, 1 Pt. \(P(0) = -250\) with the reason, 1 Pt. \(P(x)\), 1 Pt. maximum profit in EUR.

g) Average value on \([0; 12]\), with \(\int e^{-t/3} \, dt = \frac{1}{-1/3}e^{-t/3} = -3e^{-t/3}\):

\[\bar{s} = \frac{1}{12 - 0}\int_0^{12} \left(8 - 6e^{-t/3}\right) dt = \frac{1}{12}\Big[8t + 18e^{-t/3}\Big]_0^{12} = \frac{1}{12}\left[\left(96 + 18e^{-4}\right) - 18\right] = 6.5 + 1.5e^{-4} \approx 6.53\]

In the first 12 months, the cellar sells on average about 6,530 bottles per month.

Time: \(s(t) = 6.5 + 1.5e^{-4}\):

\[8 - 6e^{-t/3} = 6.5 + 1.5e^{-4} \iff e^{-t/3} = \frac{1 - e^{-4}}{4} \approx 0.25 \iff t = -3\ln\left(\frac{1 - e^{-4}}{4}\right) \approx 4.21\]

\(s'(t) = 2e^{-t/3} > 0\), so the monthly sales grow all the time and reach the average only once: about 4.2 months after the launch, in the fifth month. Before, the sales lie below the average, and after, above it.

Grading: 1 Pt. formula with \(\frac{1}{12}\), 1 Pt. antiderivative, 1 Pt. evaluation, 1 Pt. average with the unit, 1 Pt. equation \(s(t) = \bar{s}\), 1 Pt. logarithm, 1 Pt. \(t \approx 4.21\) with the meaning.

Exam Preparation Checklist

Before the exam, ensure you can:

• Compute limits from a graph, by factoring and cancelling and for \(x \to \pm\infty\), and check continuity with the three conditions
• Tell a hole (removable discontinuity) from a vertical asymptote with one-sided limits
• Compute the slope of a secant and the tangential gradient, and interpret marginal cost, marginal revenue and marginal profit
• Differentiate with the power, product, quotient and chain rules, including \(e^x\), \(e^{g(x)}\) and \(\ln(x)\), and check a derivative value with the d/dx template
• Investigate the asymptotic behavior of functions like \((x^2 + bx + c) \cdot e^{-kx}\) with \(x^n \cdot e^{-kx} \to 0\) for \(x \to \infty\)
• Determine the nature and the coordinates of all stationary points with the sign change of \(f'\) or the second derivative test, and recognize a saddle point
• Determine inflection points with \(f''\) (sign change or \(f'''\)), assess how many inflection points a function has to possess at least, and explain their meaning, e.g. the fastest growth or the lowest marginal costs
• Determine a tangent or an inflectional tangent and the angle of intersection \(\alpha = \arctan(m)\) with the indicator D
• Find global extrema on a closed interval with the values at the ends, e.g. the highest profit within the capacity
• Sketch a graph with the 6-step algorithm and argue whether statements about \(f\), \(f'\) and \(f''\) are true or false
• Determine a cubic cost, revenue or sales function from conditions: an extremum or a tangent gives two equations, an inflectional tangent three
• Get the cost function from the marginal costs and the fixed costs
• Compute the maximum profit, prove that it is a maximum and give the price, and determine the break-even point and the profit limit
• Compute the lowest unit costs, the long-term price floor and the short-term lower limit price, and decide with them
• Work with a function family: stationary points in terms of the parameter, and the parameter for a given location or value
• Graph \(f'\) from the graph of \(f\), and sketch a possible antiderivative \(F\) from the graph of \(f\): \(F\) increases where \(f\) is positive, has a local extremum where \(f\) changes sign and an inflection point where \(f\) has a local extremum
• Find antiderivatives with the power, constant multiple and sum rules, \(\int e^{ax} \, dx\) and \(\int \frac{1}{x} \, dx\), and fix the constant of integration with a condition such as \(C(0) =\) fixed costs or \(P(0) = -\)fixed costs
• Prove that \(F\) is an antiderivative of \(f\) by differentiating, and find antiderivatives of products such as \((ax + b) \cdot e^{kx}\) with integration by parts
• Compute definite integrals with the Fundamental Theorem of Calculus, and tell the signed area from the area by splitting at the zeros
• Compute totals and net changes from rates, e.g. “the total … within the first … months” or the extra costs \(\int_a^b C'(x) \, dx\), and solve for an unknown limit
• Compute the area enclosed by two graphs, set up the equation for a line that divides a region into two parts of equal size, and show computationally that two regions have equal size
• Determine the market equilibrium, the consumer surplus and the producer surplus, and what a minimum price changes
• Compute the average value of a function, e.g. the average monthly sales volume within the first … months

Remember: Show all work, label units, and interpret your results in the business context.