Tasks 04-02 - Power, Root & Rational Functions

Section 04: Advanced Functions

A: Skills

Problem 1: Power Function Basics (x)

For each function, state the domain and the range and decide whether the graph is symmetric about the \(y\)-axis (even function), symmetric about the origin (odd function) or neither.

  1. \(f(x) = \frac{1}{2}x^5\)

  2. \(g(x) = -2x^{-4}\)

  3. \(h(x) = x^{4/5}\)

  4. \(p(x) = x^{3/4}\)

  5. \(q(x) = x^{-1/3}\)

  6. \(r(x) = 4x^{-3/2}\)

Write every power with a root or a fraction first. The domain follows from two rules: a negative exponent excludes \(x = 0\), and an even root needs \(x \geq 0\). For the symmetry, compare \(f(-x)\) with \(f(x)\): the graph of \(f(-x)\) is the mirror image of the graph of \(f\) at the \(y\)-axis. If \(f(-x) = -f(x)\), mirroring at the \(y\)-axis and then at the \(x\)-axis gives back the same graph: it is symmetric about the origin.

  1. \(f(x) = \frac{1}{2}x^5\)
    • Domain: all real numbers. Range: all real numbers.
    • Odd: \(f(-x) = \frac{1}{2}(-x)^5 = -\frac{1}{2}x^5 = -f(x)\) (odd exponent)
  2. \(g(x) = -2x^{-4} = -\frac{2}{x^4}\)
    • Domain: all real numbers except \(x = 0\)
    • Range: \(x^4 > 0\), so \(g(x) < 0\). Range: \((-\infty; 0)\)
    • Even: \(g(-x) = -\frac{2}{(-x)^4} = -\frac{2}{x^4} = g(x)\) (even exponent)
  3. \(h(x) = x^{4/5} = (\sqrt[5]{x})^4\)
    • Domain: all real numbers (the fifth root is an odd root)
    • Range: an even power is never negative. Range: \([0; \infty)\)
    • Even: \(h(-x) = (\sqrt[5]{-x})^4 = (-\sqrt[5]{x})^4 = (\sqrt[5]{x})^4 = h(x)\)
  4. \(p(x) = x^{3/4} = (\sqrt[4]{x})^3\)
    • Domain: \(x \geq 0\) (the fourth root is an even root). Range: \([0; \infty)\)
    • Neither: \(p\) is not defined for \(x < 0\), so the graph has no mirror image on the left.
  5. \(q(x) = x^{-1/3} = \frac{1}{\sqrt[3]{x}}\)
    • Domain: all real numbers except \(x = 0\) (negative exponent, odd root)
    • Range: all real numbers except \(0\) (a fraction with numerator 1 is never 0)
    • Odd: \(q(-x) = \frac{1}{\sqrt[3]{-x}} = -\frac{1}{\sqrt[3]{x}} = -q(x)\)
  6. \(r(x) = 4x^{-3/2} = \frac{4}{(\sqrt{x})^3}\)
    • Domain: \(x > 0\) (even root and negative exponent). Range: \((0; \infty)\)
    • Neither: \(r\) is not defined for \(x < 0\).

Problem 2: Evaluating and Comparing Power Functions (x)

Compute the following values without a calculator:

  1. \(f(x) = x^{5/2}\): \(f(4)\) and \(f(9)\)

  2. \(g(x) = 6x^{-1/2}\): \(g(36)\) and \(g\left(\frac{1}{4}\right)\)

  3. \(h(x) = x^{4/3}\): \(h(8)\) and \(h(-27)\)

  4. \(p(x) = -4x^{-3}\): \(p\left(\frac{1}{2}\right)\) and \(p(-2)\)

  5. Order the functions \(u(x) = x^{5/6}\), \(v(x) = x^{-1/3}\), \(w(x) = x^{7/6}\), \(s(x) = x\) and \(z(x) = x^{1/2}\) from the slowest to the fastest growth for large positive \(x\). Check your order by computing all five values at \(x = 64\) without a calculator. Decide whether the values at \(x = \frac{1}{64}\) have the same order.

  1. \(f(x) = x^{5/2} = (\sqrt{x})^5\)

    • \(f(4) = (\sqrt{4})^5 = 2^5 = 32\)
    • \(f(9) = (\sqrt{9})^5 = 3^5 = 243\)

  2. \(g(x) = 6x^{-1/2} = \frac{6}{\sqrt{x}}\)

    • \(g(36) = \frac{6}{\sqrt{36}} = \frac{6}{6} = 1\)
    • \(g\left(\frac{1}{4}\right) = \frac{6}{\sqrt{1/4}} = \frac{6}{1/2} = 12\)

  3. \(h(x) = x^{4/3} = (\sqrt[3]{x})^4\)

    • \(h(8) = (\sqrt[3]{8})^4 = 2^4 = 16\)
    • \(h(-27) = (\sqrt[3]{-27})^4 = (-3)^4 = 81\) (the cube root of a negative number exists)

  4. \(p(x) = -4x^{-3} = -\frac{4}{x^3}\)

    • \(p\left(\frac{1}{2}\right) = -\frac{4}{1/8} = -32\)
    • \(p(-2) = -\frac{4}{-8} = \frac{1}{2}\)

  5. For \(x > 1\), a larger exponent means faster growth. \(v\) has a negative exponent, so it even falls. Order from the slowest to the fastest:

    \[v(x) = x^{-1/3}, \quad z(x) = x^{1/2}, \quad u(x) = x^{5/6}, \quad s(x) = x, \quad w(x) = x^{7/6}\]

    Check at \(x = 64 = 2^6\):

    • \(v(64) = \frac{1}{\sqrt[3]{64}} = \frac{1}{4}\)
    • \(z(64) = \sqrt{64} = 8\)
    • \(u(64) = (\sqrt[6]{64})^5 = 2^5 = 32\)
    • \(s(64) = 64\)
    • \(w(64) = (\sqrt[6]{64})^7 = 2^7 = 128\)

    \(\frac{1}{4} < 8 < 32 < 64 < 128\) ✓

    At \(x = \frac{1}{64}\): \(v = 4\), \(z = \frac{1}{8}\), \(u = \frac{1}{32}\), \(s = \frac{1}{64}\), \(w = \frac{1}{128}\). The order is not the same: for \(0 < x < 1\), a larger positive exponent gives a smaller value (multiplying by a number below 1 makes it smaller), and \(v\) is now the largest. The growth order only describes large \(x\).

Problem 3: Finding Asymptotes (x)

For each rational function, determine all vertical and horizontal asymptotes. If there is no horizontal asymptote, determine the oblique asymptote by polynomial long division.

  1. \(f(x) = \frac{5}{x + 4}\)

  2. \(g(x) = \frac{4x - 1}{2x - 6}\)

  3. \(h(x) = \frac{x^2 + 5}{x^2 + 2x - 15}\)

  4. \(k(x) = \frac{2x^2 + 5x - 1}{x + 2}\)

  5. \(m(x) = \frac{x + 3}{x^2 + 4}\)

Vertical asymptotes lie where the denominator is zero and the numerator is not (otherwise check for a hole). Horizontal and oblique asymptotes follow from the degrees of numerator and denominator.

  1. \(f(x) = \frac{5}{x + 4}\)

    • Denominator zero at \(x = -4\), numerator \(5 \neq 0\): vertical asymptote \(x = -4\)
    • Degree 0 < degree 1: horizontal asymptote \(y = 0\)

  2. \(g(x) = \frac{4x - 1}{2x - 6}\)

    • Denominator zero at \(x = 3\), numerator \(4 \cdot 3 - 1 = 11 \neq 0\): vertical asymptote \(x = 3\)
    • Degree 1 = degree 1: ratio of the leading coefficients, horizontal asymptote \(y = \frac{4}{2} = 2\)

  3. \(h(x) = \frac{x^2 + 5}{x^2 + 2x - 15} = \frac{x^2 + 5}{(x + 5)(x - 3)}\)

    • Denominator zero at \(x = -5\) and \(x = 3\). Numerator: \((-5)^2 + 5 = 30 \neq 0\) and \(3^2 + 5 = 14 \neq 0\): vertical asymptotes \(x = -5\) and \(x = 3\)
    • Degree 2 = degree 2: horizontal asymptote \(y = \frac{1}{1} = 1\)

  4. \(k(x) = \frac{2x^2 + 5x - 1}{x + 2}\)

    • Denominator zero at \(x = -2\), numerator \(8 - 10 - 1 = -3 \neq 0\): vertical asymptote \(x = -2\)
    • Degree 2 > degree 1: no horizontal asymptote. The degree is higher by exactly 1, so there is an oblique asymptote.

    Long division by \((x + 2)\):

    \[\begin{array}{l} (2x^2 + 5x - 1) : (x + 2) = 2x + 1 \;\text{ remainder } -3 \\ \underline{-(2x^2 + 4x)} \\ \phantom{(2x^2}\;\; x - 1 \\ \phantom{(2x^2}\;\; \underline{-(x + 2)} \\ \phantom{(2x^2 + x}\;\; -3 \end{array}\]

    \(k(x) = 2x + 1 - \frac{3}{x + 2}\). For \(x \to \pm\infty\), \(\frac{3}{x + 2} \to 0\): oblique asymptote \(y = 2x + 1\).

    Check: \((x + 2)(2x + 1) - 3 = 2x^2 + 5x + 2 - 3 = 2x^2 + 5x - 1\) ✓

  5. \(m(x) = \frac{x + 3}{x^2 + 4}\)

    • \(x^2 + 4 \geq 4 > 0\) for all \(x\): the denominator is never zero, so there is no vertical asymptote
    • Degree 1 < degree 2: horizontal asymptote \(y = 0\)

Problem 4: Holes vs. Asymptotes (x)

For each function, factor numerator and denominator and decide whether the graph has a hole or a vertical asymptote at the given \(x\)-values. Give the coordinates of every hole.

  1. \(f(x) = \frac{x^2 + 2x - 8}{x - 2}\) at \(x = 2\)

  2. \(g(x) = \frac{x + 1}{x^2 - 6x - 7}\) at \(x = -1\) and at \(x = 7\)

  3. \(h(x) = \frac{x^2 - 6x + 9}{x^2 - 9}\) at \(x = 3\) and at \(x = -3\). Decide also whether \(h\) has a zero at \(x = 3\).

A factor that cancels gives a hole. A factor that stays in the denominator gives a vertical asymptote. The \(y\)-value of a hole comes from the simplified function.

  1. \(f(x) = \frac{(x + 4)(x - 2)}{x - 2}\)
    • \((x - 2)\) cancels: \(f(x) = x + 4\) for \(x \neq 2\)
    • Hole at \(x = 2\) with \(y = 2 + 4 = 6\): \((2, 6)\)
  2. \(g(x) = \frac{x + 1}{(x - 7)(x + 1)}\)
    • \((x + 1)\) cancels: \(g(x) = \frac{1}{x - 7}\) for \(x \neq -1\)
    • \(x = -1\): hole with \(y = \frac{1}{-1 - 7} = -\frac{1}{8}\): \(\left(-1, -\frac{1}{8}\right)\)
    • \(x = 7\): the factor \((x - 7)\) stays in the denominator: vertical asymptote \(x = 7\)
  3. \(h(x) = \frac{(x - 3)^2}{(x - 3)(x + 3)}\)
    • One factor \((x - 3)\) cancels: \(h(x) = \frac{x - 3}{x + 3}\) for \(x \neq 3\)
    • \(x = 3\): hole with \(y = \frac{3 - 3}{3 + 3} = 0\): \((3, 0)\)
    • \(x = -3\): vertical asymptote \(x = -3\)
    • \(x = 3\) is not a zero of \(h\): \(h\) is not defined at \(x = 3\). The hole lies on the \(x\)-axis, but the graph has a gap there. The simplified function has only the zero \(x = 3\), so \(h\) has no zero at all.

Problem 5: Sketching a Rational Function (xx)

Sketch the graph of \(f(x) = \frac{3x + 6}{x - 2}\) by following these steps:

  1. Determine the domain and all asymptotes.

  2. Find the \(x\)- and \(y\)-intercepts.

  3. Determine the sign of \(f(x)\) in each interval with test points.

  4. Describe the asymptotic behavior of \(f\) (what happens to \(f(x)\) near the vertical asymptote and for \(x \to \pm\infty\)). Decide with the test values \(f(10)\) and \(f(-10)\) whether the graph approaches the horizontal asymptote from above or from below.

  5. Sketch the graph.

  1. Domain: \(x - 2 = 0\) at \(x = 2\), so the domain is all real numbers except \(x = 2\).
    • Numerator at \(x = 2\): \(3 \cdot 2 + 6 = 12 \neq 0\): vertical asymptote \(x = 2\)
    • Degree 1 = degree 1: horizontal asymptote \(y = \frac{3}{1} = 3\)
  2. Intercepts:
    • \(x\)-intercept: numerator zero: \(3x + 6 = 0 \iff x = -2\), so \((-2, 0)\)
    • \(y\)-intercept: \(f(0) = \frac{6}{-2} = -3\), so \((0, -3)\)
  3. The sign can only change at the zero \(x = -2\) and at the vertical asymptote \(x = 2\):
Interval \((-\infty; -2)\) \((-2; 2)\) \((2; \infty)\)
Test point \(x = -3\) \(x = 0\) \(x = 3\)
\(f(x)\) \(\frac{-3}{-5} = 0.6\) \(-3\) \(\frac{15}{1} = 15\)
Sign positive negative positive
  1. Near the vertical asymptote: the numerator is close to 12 (positive).
    • For \(x \to 2\) from the left, the denominator is a small negative number: \(f(x) \to -\infty\)
    • For \(x \to 2\) from the right, the denominator is a small positive number: \(f(x) \to +\infty\)
    For \(x \to \pm\infty\): \(f(x) \to 3\).
    • \(f(10) = \frac{36}{8} = 4.5 > 3\): for large \(x\) the graph lies above \(y = 3\)
    • \(f(-10) = \frac{-24}{-12} = 2 < 3\): for very negative \(x\) the graph lies below \(y = 3\)
  2. The graph has two branches:
    • Left branch (\(x < 2\)): comes from the left just below \(y = 3\), falls, crosses the \(x\)-axis at \((-2, 0)\), passes through \((0, -3)\) and drops to \(-\infty\) at the vertical asymptote \(x = 2\)
    • Right branch (\(x > 2\)): comes down from \(+\infty\) next to \(x = 2\), passes through \((3, 15)\) and \((10, 4.5)\), and approaches \(y = 3\) from above

B: Set Up & Check

Problem 6: From Text to Functions (xx)

Translate each statement into mathematics. Then answer the check question: it tells you whether your translation works.

  1. “A cooking school runs evening courses. For a course with \(x\) participants, its costs are \(C(x) = 420 + 22x\) EUR (420 EUR for the instructor and the room, 22 EUR for the food per participant).” Give the average cost \(AC(x)\) per participant (the total costs divided by the number of participants) and compute it for 12 participants.

    Check: Multiply your average cost for 12 participants by 12. Do you get the total costs \(C(12)\)?

  2. “The costs of our customer service grow less than proportionally with the number of customers.” A consultant proposes four cost models (\(x\) in thousand customers, costs in thousand EUR): \(C_1(x) = 40x^{0.8}\), \(C_2(x) = 40x^{1.25}\), \(C_3(x) = 40x^{-0.8}\) and \(C_4(x) = 40x\). Decide which model fits the statement. Give the condition on the exponent \(b\) of a cost model \(C(x) = a \cdot x^b\) with \(a > 0\) that the statement describes.

    Check: Compute \(C(1)\) and \(C(2)\) for all four models. For which model do the costs grow, but by a factor smaller than 2?

  3. “A company’s costs are \(C(x) = 75x^{0.6}\). Its manager says: if we double the output, our costs rise by 60 %.” Write \(C(2x)\) as a multiple of \(C(x)\) and correct the manager’s statement. Round the percentage to one decimal place.

    Check: Compute \(C(10)\) and \(C(20)\) with the calculator. By how many percent do the costs rise?

  4. “A producer of garden hoses says: in the long run, our average cost approaches 45 EUR per hose.” Write this statement as a property of the graph of the average cost function \(AC(x)\).

    Check: The producer’s total costs for \(x\) hoses are \(C(x) = 0.01x^2 + 45x + 900\) EUR. Compute \(AC(100)\), \(AC(1000)\) and \(AC(10000)\). Does the statement hold?

  5. “A factory must clean its exhaust air. Removing \(p\) % of the dust costs \(K(p) = \frac{50p}{100 - p}\) thousand EUR per year. The formula is not defined at \(p = 100\).” Translate the last sentence into a statement about the graph of \(K\) and explain what it means for the factory.

    Check: Compute \(K(90)\), \(K(99)\) and \(K(99.9)\). What happens to the costs?

  1. Divide every term of the total costs by \(x\):

    \(AC(x) = \frac{C(x)}{x} = \frac{420 + 22x}{x} = \frac{420}{x} + 22\) EUR per participant, for \(x > 0\)

    \(AC(12) = \frac{420}{12} + 22 = 35 + 22 = 57\) EUR per participant.

    Check: yes. \(57 \cdot 12 = 684\) and \(C(12) = 420 + 264 = 684\) EUR.

    Typical mistake: \(AC(x) = \frac{420}{x} + 22x\). Only the fixed costs were divided by \(x\). Then \(AC(12) = 299\) EUR, and \(299 \cdot 12 = 3{,}588 \neq 684\).

  2. Only \(C_1\) fits. “Grow less than proportionally” means: the costs grow with \(x\), but more slowly than \(x\) itself. For \(C(x) = a \cdot x^b\) with \(a > 0\), this is the case for \(0 < b < 1\).

    Check: yes.

    • \(C_1\): \(C_1(1) = 40\), \(C_1(2) = 40 \cdot 2^{0.8} \approx 69.64\), factor \(\approx 1.74 < 2\)
    • \(C_2\): \(C_2(2) = 40 \cdot 2^{1.25} \approx 95.14\), factor \(\approx 2.38 > 2\) (more than proportional)
    • \(C_3\): \(C_3(2) = 40 \cdot 2^{-0.8} \approx 22.97 < 40\) (the costs fall)
    • \(C_4\): \(C_4(2) = 80\), factor exactly 2 (proportional)

    Typical mistake: choosing \(C_3\) because its costs “grow slowly”. A negative exponent means that the costs fall when the number of customers grows. They do not grow at all.

  3. By the power rule \((2x)^{0.6} = 2^{0.6} \cdot x^{0.6}\):

    \(C(2x) = 75 \cdot (2x)^{0.6} = 2^{0.6} \cdot 75x^{0.6} = 2^{0.6} \cdot C(x) \approx 1.516 \cdot C(x)\)

    Correct statement: if the output doubles, the costs rise by about 51.6 %, for every output \(x\).

    Check: \(C(10) = 75 \cdot 10^{0.6} \approx 298.58\) and \(C(20) = 75 \cdot 20^{0.6} \approx 452.56\). \(\frac{452.56}{298.58} \approx 1.516\), i.e. \(+51.6\) % ✓

    Typical mistake: reading the exponent 0.6 as 60 %. The exponent is not a percentage. The costs are multiplied by \(2^{0.6}\).

  4. The graph of \(AC\) has the horizontal asymptote \(y = 45\): \(AC(x) \to 45\) for \(x \to \infty\).

    Check: \(AC(x) = \frac{0.01x^2 + 45x + 900}{x} = 0.01x + 45 + \frac{900}{x}\).

    • \(AC(100) = 1 + 45 + 9 = 55\) EUR
    • \(AC(1000) = 10 + 45 + 0.9 = 55.9\) EUR
    • \(AC(10000) = 100 + 45 + 0.09 = 145.09\) EUR

    The statement does not hold. The term \(0.01x\) grows without bound, and the average cost has the oblique asymptote \(y = 0.01x + 45\). The statement would hold for linear costs \(C(x) = F + 45x\), where \(AC(x) = \frac{F}{x} + 45\).

    Typical mistake: taking the coefficient 45 of \(x\) as the long-run average cost and overlooking the term \(0.01x^2\) in the costs.

  5. At \(p = 100\), the denominator \(100 - p\) is zero and the numerator is \(50 \cdot 100 = 5000 \neq 0\). No factor cancels, so the graph of \(K\) has the vertical asymptote \(p = 100\): \(K(p) \to \infty\) for \(p \to 100\) from the left. For the factory: removing all of the dust is impossible at any price, and every further percent close to 100 % becomes extremely expensive.

    Check: \(K(90) = \frac{4500}{10} = 450\), \(K(99) = \frac{4950}{1} = 4{,}950\) and \(K(99.9) = \frac{4995}{0.1} = 49{,}950\) thousand EUR. The costs explode.

    Typical mistake: “not defined, so the graph has a hole at \(p = 100\)”. A hole needs a factor that cancels. Here the numerator is not zero at \(p = 100\), so it is a vertical asymptote.

Problem 7: One Paragraph, All Conditions (xx)

A small company brews kombucha (a fermented tea drink) and sells it in bottles. Its fixed costs are 5,400 EUR per month, and each bottle causes variable costs of 0.90 EUR. A supermarket chain pays 1.50 EUR per bottle. The owner wants to know the average cost per bottle, what happens to it with very few and with very many bottles, and from which output the average cost is below the price. The prices of fermentation tanks follow a power function of their volume: a tank with twice the volume costs 40 % more. (1 QU = 1,000 bottles, 1 MU = 1,000 EUR)

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 average cost as \(AC(x) = \frac{5400}{x} + 0.9x\). Name the two mistakes.

  1. Fixed costs: 5,400 EUR \(= 5.4\) MU
  2. Variable costs: 0.90 EUR per bottle \(= 900\) EUR per QU \(= 0.9\) MU per QU
  3. Total costs: \(C(x) = 5.4 + 0.9x\)
  4. Average cost: \(AC(x) = \frac{C(x)}{x} = \frac{5.4}{x} + 0.9\) for \(x > 0\) (in MU per QU, which is the same as EUR per bottle here)
  5. Very few bottles: \(AC(x) \to \infty\) for \(x \to 0\) from the right, the vertical asymptote \(x = 0\)
  6. Very many bottles: \(AC(x) \to 0.9\) for \(x \to \infty\), the horizontal asymptote \(y = 0.9\)
  7. Price: 1.50 EUR per bottle \(= 1.5\) MU per QU. Average cost below the price: \(AC(x) < 1.5\)
  8. Tank prices: \(T(V) = a \cdot V^b\) with the volume \(V\), and \(T(2V) = 1.4 \cdot T(V)\), i.e. \(2^b = 1.4\)

Check: 5400 is the fixed cost in EUR, but the function must be in MU: \(\frac{5.4}{x}\). And the variable costs must be divided by \(x\) as well: \(\frac{0.9x}{x} = 0.9\). The correct term is \(AC(x) = \frac{5.4}{x} + 0.9\).

Typical mistake: writing the price condition as \(C(x) < 1.5\). The price per bottle must be compared with the average cost per bottle, not with the total costs.

C: Applied

Problem 8: Costs of a Glass Recycling Plant (xx)

A recycling plant cleans and crushes waste glass from bottle banks (containers where people drop off their empty glass bottles) and sells the glass pieces to glassworks (factories that make new glass). Its monthly total costs are

\[C(x) = 90 + 60x^{2/3}\]

in thousand EUR, where \(x\) is the amount of glass in thousand tonnes per month and \(0 \leq x \leq 30\). Round percentages to one decimal place and amounts in EUR per tonne to two decimal places.

  1. Give the fixed costs in EUR. Compute the total costs and the average cost per tonne for 8,000 tonnes and for 27,000 tonnes per month.

(for verification: 41.25 EUR per tonne for 8,000 tonnes)

  1. The variable costs are \(V(x) = 60x^{2/3}\). Explain with the exponent why they grow less than proportionally with the amount of glass. Compute by how many percent the variable costs rise when the amount doubles. Compute by how many percent the total costs rise from 8,000 to 16,000 tonnes and explain why this percentage is smaller.

  2. The plant’s budget allows total costs of at most 450,000 EUR per month. Determine the largest amount of glass the plant can process within this budget, in whole tonnes.

Hint: bring the equation into the form \(x^{2/3} = c\) and raise both sides to the power \(\frac{3}{2}\).

  1. Show that the average cost in EUR per tonne is \(AC(x) = \frac{90}{x} + 60x^{-1/3}\). The towns pay the plant 30 EUR for each tonne of glass. Ignore the income from the glassworks. Compute the profit or loss for 8,000 and for 27,000 tonnes per month. Explain with the two terms of \(AC(x)\) why larger amounts of glass lower the average cost.

a) Fixed costs: \(C(0) = 90\) thousand EUR \(= 90{,}000\) EUR per month.

  • 8,000 tonnes \(= 8\) thousand tonnes: \(8^{2/3} = (\sqrt[3]{8})^2 = 4\), so \(C(8) = 90 + 60 \cdot 4 = 330\) thousand EUR. Average cost: \(\frac{330{,}000}{8{,}000} = 41.25\) EUR per tonne.
  • 27,000 tonnes: \(27^{2/3} = (\sqrt[3]{27})^2 = 9\), so \(C(27) = 90 + 540 = 630\) thousand EUR. Average cost: \(\frac{630{,}000}{27{,}000} \approx 23.33\) EUR per tonne.

b) The exponent \(\frac{2}{3}\) lies between 0 and 1. So \(V\) grows with \(x\), but more slowly than \(x\) itself: twice the glass does not need twice the variable costs (the machines and the staff are used better).

Doubling: \(V(2x) = 60 \cdot (2x)^{2/3} = 2^{2/3} \cdot 60x^{2/3} = 2^{2/3} \cdot V(x) \approx 1.587 \cdot V(x)\). The variable costs rise by about 58.7 %, for every amount \(x\).

Total costs: \(C(16) = 90 + 60 \cdot 16^{2/3} \approx 90 + 60 \cdot 6.3496 \approx 470.98\) thousand EUR. \(\frac{470.98}{330} \approx 1.427\), i.e. the total costs rise by about 42.7 %.

The fixed costs of 90 thousand EUR stay the same when the amount doubles. Only the variable part (240 of the 330 thousand EUR) rises by 58.7 %, so the total costs rise by less.

c) Budget: \(C(x) \leq 450\). \(C\) grows with \(x\) (the power \(x^{2/3}\) grows), so the largest amount is where \(C(x) = 450\):

\[90 + 60x^{2/3} = 450 \iff 60x^{2/3} = 360 \iff x^{2/3} = 6\]

Raise both sides to the power \(\frac{3}{2}\). The amount \(x\) cannot be negative, so only the positive solution counts:

\[x = 6^{3/2} = \sqrt{6^3} = \sqrt{216} \approx 14.697\]

Check: \((6^{3/2})^{2/3} = 6^1 = 6\) ✓. The plant can process at most 14,696 tonnes per month (14,697 tonnes would already exceed the budget slightly, so round down).

d) Divide every term by \(x\) (quotient rule for powers):

\[AC(x) = \frac{C(x)}{x} = \frac{90}{x} + \frac{60x^{2/3}}{x} = \frac{90}{x} + 60x^{2/3 - 1} = \frac{90}{x} + 60x^{-1/3}\]

\(C\) is in thousand EUR and \(x\) in thousand tonnes, so \(AC\) is in EUR per tonne ✓.

  • 8,000 tonnes: revenue \(30 \cdot 8 = 240\) thousand EUR, costs 330 thousand EUR. Loss of 90,000 EUR. (Or: \(8 \cdot (30 - 41.25) = -90\).)
  • 27,000 tonnes: revenue \(30 \cdot 27 = 810\) thousand EUR, costs 630 thousand EUR. Profit of 180,000 EUR.

Both terms of \(AC\) have a negative exponent, so both fall when \(x\) grows. \(\frac{90}{x}\): the fixed costs are spread over more tonnes. \(60x^{-1/3} = \frac{60}{\sqrt[3]{x}}\): the variable costs per tonne fall as well, because the variable costs grow less than proportionally (b)). These are economies of scale.

Problem 9: Injection Moulding or 3D Printing? (xx)

A company makes replacement knobs for kitchen stoves. For each new knob model, it can choose between two methods:

  • Injection moulding: a metal mould (a form into which liquid plastic is pressed) costs 7,200 EUR, and each knob costs another 0.60 EUR.
  • 3D printing: preparing the printer costs 240 EUR, and each knob costs another 3.50 EUR.

Let \(x\) be the number of knobs of one model. Round amounts to two decimal places.

  1. Give the total cost functions \(C_M(x)\) and \(C_P(x)\) and the average cost functions \(AC_M(x)\) and \(AC_P(x)\). Compute both average costs for 500 knobs and for 6,000 knobs.

  2. Determine the vertical and the horizontal asymptote of \(AC_M\) and of \(AC_P\). Explain their meaning for the company.

  3. Determine the number of knobs for which both methods have the same average cost. Show that the equation \(C_M(x) = C_P(x)\) gives the same number and explain why.

(for verification: 2,400 knobs)

  1. A chain of hardware stores orders 1,500 knobs of a new model and pays 4.20 EUR per knob. Decide which method the company should use and compute its profit.

  2. A sales representative promises another customer 3.40 EUR per knob for a very large 3D-printed order: “The average cost keeps falling with the size of the order.” Assess this promise. Determine from which number of knobs injection moulding allows a price of 3.40 EUR per knob without a loss.

Hint: The denominator is positive, so you can multiply both sides by it without changing the sign.

a) Total costs (fixed costs + variable costs):

\[C_M(x) = 7200 + 0.6x, \qquad C_P(x) = 240 + 3.5x\]

Average costs (divide every term by \(x\), \(x > 0\)):

\[AC_M(x) = \frac{7200}{x} + 0.6, \qquad AC_P(x) = \frac{240}{x} + 3.5\]

Knobs \(AC_M\) \(AC_P\)
500 \(14.40 + 0.60 = 15.00\) EUR \(0.48 + 3.50 = 3.98\) EUR
6,000 \(1.20 + 0.60 = 1.80\) EUR \(0.04 + 3.50 = 3.54\) EUR

For small orders 3D printing is much cheaper per knob, for large orders injection moulding.

b) Both functions have the form \(\frac{F}{x} + v\):

  • Vertical asymptote \(x = 0\) for both: \(\frac{7200}{x}\) and \(\frac{240}{x}\) grow without bound for \(x \to 0\) from the right. With very few knobs, the fixed costs (mould or preparation) are spread over very few knobs, so each knob becomes extremely expensive. This hits injection moulding much harder.
  • Horizontal asymptotes \(y = 0.6\) for \(AC_M\) and \(y = 3.5\) for \(AC_P\): for very large orders, the share of the fixed costs per knob goes to 0, and the average cost approaches the variable costs per knob.

c) Condition \(AC_M(x) = AC_P(x)\) with \(x > 0\):

\[\frac{7200}{x} + 0.6 = \frac{240}{x} + 3.5 \iff \frac{6960}{x} = 2.9 \;\big|\cdot x \iff 6960 = 2.9x \iff x = 2400\]

Both methods have the same average cost for 2,400 knobs: \(AC_M(2400) = 3 + 0.6 = 3.60\) EUR and \(AC_P(2400) = 0.10 + 3.50 = 3.60\) EUR ✓.

With the total costs: \(7200 + 0.6x = 240 + 3.5x \iff 6960 = 2.9x \iff x = 2400\) ✓.

Reason: \(AC(x) = \frac{C(x)}{x}\). Multiplying the equation \(\frac{C_M(x)}{x} = \frac{C_P(x)}{x}\) by \(x \neq 0\) gives \(C_M(x) = C_P(x)\). For the same number of knobs, equal costs per knob mean equal total costs.

d) 1,500 knobs lie below 2,400, so 3D printing is cheaper:

  • \(AC_P(1500) = 0.16 + 3.50 = 3.66\) EUR per knob
  • \(AC_M(1500) = 4.80 + 0.60 = 5.40\) EUR per knob

Profit with 3D printing: \(1500 \cdot (4.20 - 3.66) = 810\) EUR. Check: revenue \(6{,}300\) EUR, costs \(C_P(1500) = 240 + 5250 = 5{,}490\) EUR ✓. (With injection moulding, the company would make a loss of \(1500 \cdot (5.40 - 4.20) = 1{,}800\) EUR.)

e) The average cost of 3D printing falls, but never below its horizontal asymptote: \(\frac{240}{x} > 0\), so \(AC_P(x) > 3.50\) EUR for every order. At 3.40 EUR per knob, the company loses more than 0.10 EUR on every knob. The promise must be withdrawn.

For large orders, injection moulding is the right method. Condition \(AC_M(x) \leq 3.4\) with \(x > 0\):

\[\frac{7200}{x} + 0.6 \leq 3.4 \iff \frac{7200}{x} \leq 2.8 \;\big|\cdot x \;(x > 0) \iff 7200 \leq 2.8x \iff x \geq 2571.43\]

From 2,572 knobs on, injection moulding makes 3.40 EUR per knob possible without a loss.

Problem 10: Steel Tanks for Wineries (xx)

A company builds closed cylindrical steel tanks for wineries. In all its tanks, the height equals the diameter: \(h = 2r\), where \(r\) is the radius in m. The volume of a cylinder is \(V = \pi r^2 h\), and its surface (bottom, lid and side wall) is \(S = 2\pi r^2 + 2\pi r h\). Round lengths and areas to two decimal places, percentages to one decimal place and amounts of money to whole euros.

  1. Show that \(V(r) = 2\pi r^3\) and \(S(r) = 6\pi r^2\). Compute the volume in litres and the surface of a tank with \(r = 0.5\) m. (1 m³ = 1,000 litres)

  2. The company doubles the radius. Determine the factors by which the volume and the surface grow. Decide how the amount of steel per litre of volume changes.

  3. Show that the surface as a function of the volume is \(S(V) = 3\sqrt[3]{2\pi} \cdot V^{2/3} \approx 5.54 \cdot V^{2/3}\). Explain with the exponent why the steel grows less than proportionally with the volume.

Hint: solve \(V = 2\pi r^3\) for \(r\) first.

  1. Stainless steel for the tanks (including welding) costs 100 EUR per m² of surface. A winery needs 12 m³ of tank volume. Compare the steel costs of one tank with 12 m³ and of two tanks with 6 m³ each.

  2. A competitor sells cube-shaped tanks. Compute the surface of a cube-shaped tank with 12 m³ and decide by how many percent the cylindrical tank from d) needs less steel.

a) Insert \(h = 2r\):

\[V = \pi r^2 \cdot 2r = 2\pi r^3, \qquad S = 2\pi r^2 + 2\pi r \cdot 2r = 2\pi r^2 + 4\pi r^2 = 6\pi r^2 \quad \checkmark\]

For \(r = 0.5\): \(V = 2\pi \cdot 0.125 = 0.25\pi \approx 0.79\) m³ \(\approx 785\) litres, and \(S = 6\pi \cdot 0.25 = 1.5\pi \approx 4.71\) m².

b) By the power rule \((2r)^n = 2^n r^n\):

  • \(V(2r) = 2\pi (2r)^3 = 2^3 \cdot 2\pi r^3 = 8 \cdot V(r)\): the volume grows by the factor 8
  • \(S(2r) = 6\pi (2r)^2 = 2^2 \cdot 6\pi r^2 = 4 \cdot S(r)\): the surface grows by the factor 4

Steel per litre: \(\frac{S(2r)}{V(2r)} = \frac{4}{8} \cdot \frac{S(r)}{V(r)} = \frac{1}{2} \cdot \frac{S(r)}{V(r)}\). The larger tank needs only half as much steel per litre.

c) Solve \(V = 2\pi r^3\) for \(r\): \(r^3 = \frac{V}{2\pi}\), so \(r = \left(\frac{V}{2\pi}\right)^{1/3}\). Insert into \(S = 6\pi r^2\):

\[S = 6\pi \left(\frac{V}{2\pi}\right)^{2/3} = \frac{6\pi}{(2\pi)^{2/3}} \cdot V^{2/3} = 3 \cdot \frac{2\pi}{(2\pi)^{2/3}} \cdot V^{2/3} = 3 (2\pi)^{1/3} \cdot V^{2/3} = 3\sqrt[3]{2\pi} \cdot V^{2/3}\]

\(3\sqrt[3]{2\pi} \approx 3 \cdot 1.845 \approx 5.54\) ✓

The exponent \(\frac{2}{3}\) lies between 0 and 1, so \(S\) grows with \(V\), but more slowly: twice the volume needs only \(2^{2/3} \approx 1.59\) times the steel. Large tanks are cheaper per litre.

d) One tank with 12 m³: \(S(12) = 3\sqrt[3]{2\pi} \cdot 12^{2/3} \approx 5.5358 \cdot 5.2415 \approx 29.02\) m², steel costs \(\approx 2{,}902\) EUR.

Two tanks with 6 m³: \(S(6) \approx 5.5358 \cdot 3.3019 \approx 18.28\) m² each, together \(\approx 36.56\) m², steel costs \(\approx 3{,}656\) EUR.

The single tank saves about 754 EUR, i.e. about 20.6 % of the steel. This does not depend on the constant: \(\frac{S(12)}{2 \cdot S(6)} = \frac{2^{2/3}}{2} = 2^{-1/3} \approx 0.794\).

e) Cube with side \(s\): \(s^3 = 12\), so \(s = \sqrt[3]{12} \approx 2.29\) m. Surface \(6s^2 = 6 \cdot 12^{2/3} \approx 31.45\) m².

\(\frac{29.02}{31.45} \approx 0.923\): the cylindrical tank needs about 7.7 % less steel than the cube-shaped tank with the same volume.

D: FSP Problem

Work on this problem as in the exam: write down every step. The points show what each part is worth.

FSP Problem: Mattresses from a Family Business [21 pts. total]

A family business makes mattresses and sells them to furniture stores. Its monthly total costs are

\[C(x) = 0.5x^2 + 15x + 164\]

with \(x\) in QU and \(C\) in MU. The furniture stores pay 350 EUR per mattress. The company plans to make and sell 2,000 mattresses per month.

(QU = quantity units, MU = monetary units. 1 QU = 100 mattresses, 1 MU = 1,000 EUR. Average cost = total costs divided by the output.)

Part a)

Show that the average cost function is given by \(AC(x) = 0.5x + 15 + \frac{164}{x}\). Calculate the average cost per mattress in EUR for the planned output. [3 pts.]

Average cost: divide every term of \(C(x)\) by \(x\) (\(x \neq 0\)):

\[AC(x) = \frac{C(x)}{x} = \frac{0.5x^2 + 15x + 164}{x} = 0.5x + 15 + \frac{164}{x} \quad \checkmark\]

Planned output: 2,000 mattresses \(= 20\) QU:

\[AC(20) = 0.5 \cdot 20 + 15 + \frac{164}{20} = 10 + 15 + 8.2 = 33.2 \text{ MU per QU}\]

Conversion: 1 MU per QU \(= \frac{1{,}000 \text{ EUR}}{100 \text{ mattresses}} = 10\) EUR per mattress. So $AC(20) = 33.2 = $ 332 EUR per mattress.

Check: \(C(20) = 200 + 300 + 164 = 664\) MU \(= 664{,}000\) EUR, and \(\frac{664{,}000}{2{,}000} = 332\) EUR ✓

Grading: 1 Pt. termwise division, 1 Pt. \(AC(20) = 33.2\) MU per QU, 1 Pt. conversion to 332 EUR per mattress.

Part b)

Investigate the asymptotic behavior of \(AC\), and determine the \(x\) and \(y\) intercepts. [6 pts.]

Domain: the denominator \(x\) must not be zero, so \(D = \mathbb{R} \setminus \{0\}\).

Vertical asymptote: at \(x = 0\) the numerator is \(C(0) = 164 \neq 0\), so no factor cancels and \(x = 0\) is a vertical asymptote. Near \(x = 0\), the terms \(0.5x + 15\) stay close to 15, but \(\frac{164}{x}\) grows without bound:

\[AC(x) \to +\infty \text{ for } x \to 0 \text{ from the right}, \qquad AC(x) \to -\infty \text{ for } x \to 0 \text{ from the left}\]

Behavior for \(x \to \pm\infty\): the numerator has degree 2, the denominator degree 1. So there is no horizontal asymptote, but an oblique asymptote. From a):

\[AC(x) = \underbrace{0.5x + 15}_{\text{linear part}} + \frac{164}{x}, \qquad \frac{164}{x} \to 0 \text{ for } x \to \pm\infty\]

Oblique asymptote: \(y = 0.5x + 15\). For \(x > 0\), \(\frac{164}{x} > 0\), so the graph lies above the asymptote. For \(x < 0\) it lies below.

\(y\)-intercept: \(x = 0\) is not in the domain (it is the vertical asymptote). The graph has no \(y\)-intercept.

\(x\)-intercepts: \(AC(x) = 0 \iff C(x) = 0\) (for \(x \neq 0\)):

\[0.5x^2 + 15x + 164 = 0 \;\big|\cdot 2 \iff x^2 + 30x + 328 = 0, \qquad \Delta = 30^2 - 4 \cdot 328 = 900 - 1312 = -412 < 0\]

There is no real solution, so the graph has no \(x\)-intercept.

Grading: 1 Pt. domain and vertical asymptote \(x = 0\) (numerator \(\neq 0\)), 1 Pt. behavior on both sides of \(x = 0\), 1 Pt. degree argument (no horizontal asymptote), 1 Pt. oblique asymptote \(y = 0.5x + 15\) with the remainder term \(\to 0\), 1 Pt. no \(y\)-intercept with reason, 1 Pt. no \(x\)-intercept with \(\Delta < 0\).

Part c)

Explain the business meaning of both asymptotes. The owner says: “The more mattresses we make, the cheaper each mattress becomes.” Assess this statement. [3 pts.]

Vertical asymptote \(x = 0\): with very few mattresses, the fixed costs of 164,000 EUR per month are spread over very few mattresses, so the average cost becomes huge. Negative outputs (\(x < 0\)) have no business meaning.

Oblique asymptote \(y = 0.5x + 15\): with large outputs, the share of the fixed costs per mattress \(\frac{164}{x}\) becomes small. The average cost then behaves like \(0.5x + 15\) and keeps growing: every further 100 mattresses make each mattress about \(0.5\) MU per QU \(= 5\) EUR more expensive (overtime, extra shifts). Unlike with linear costs, the average cost does not approach a constant.

Assessment: \(AC(8) = 4 + 15 + 20.5 = 39.5\) (395 EUR), \(AC(20) = 33.2\) (332 EUR), \(AC(40) = 20 + 15 + 4.1 = 39.1\) (391 EUR). So \(AC(20)\) is lower than \(AC(8)\), but \(AC(40)\) is higher than \(AC(20)\). Since \(AC(x) > 0.5x + 15\) for \(x > 0\), the average cost even grows without bound. The statement is wrong for large outputs.

Grading: 1 Pt. meaning of the vertical asymptote, 1 Pt. meaning of the oblique asymptote, 1 Pt. assessment with values or the asymptote as a lower bound.

Part d)

Determine the outputs at which the average cost is exactly equal to the price of 350 EUR per mattress. Explain what these outputs mean for the company and decide in which output range the company makes a profit. [5 pts.]

(for verification: \(x = 20 \pm 6\sqrt{2}\))

Price in MU per QU: 350 EUR per mattress \(= 35{,}000\) EUR per QU \(= 35\) MU per QU.

Fractional equation (domain \(x \neq 0\)), multiply by \(x\):

\[0.5x + 15 + \frac{164}{x} = 35 \;\big|\cdot x \iff 0.5x^2 + 15x + 164 = 35x \iff 0.5x^2 - 20x + 164 = 0 \;\big|\cdot 2 \iff x^2 - 40x + 328 = 0\]

Quadratic formula: \(\Delta = 40^2 - 4 \cdot 328 = 1600 - 1312 = 288\) and \(\sqrt{288} = 12\sqrt{2}\):

\[x_{1,2} = \frac{40 \pm 12\sqrt{2}}{2} = 20 \pm 6\sqrt{2}, \qquad x_1 \approx 11.51, \qquad x_2 \approx 28.49\]

Both solutions are positive, so both lie in the domain.

Meaning: \(AC(x) = 35 \iff C(x) = 35x = R(x)\). At these outputs, revenue and total costs are equal, and the profit is zero. \(x_1 \approx 11.51\) QU is the break-even point, \(x_2 \approx 28.49\) QU the profit limit (upper break-even point).

Profit zone: each mattress brings a profit if the average cost is below the price. Test values: \(AC(8) = 39.5 > 35\), \(AC(20) = 33.2 < 35\), \(AC(40) = 39.1 > 35\). So the company makes a profit for \(11.51 < x < 28.49\), i.e. with 1,152 to 2,848 mattresses per month (rounded inward to whole mattresses). The planned 2,000 mattresses lie inside: profit \(20 \cdot (35 - 33.2) = 36\) MU \(= 36{,}000\) EUR.

Grading: 1 Pt. price in MU per QU and equation with \(x \neq 0\), 1 Pt. multiplying by \(x\) and zero form, 1 Pt. quadratic formula \(20 \pm 6\sqrt{2}\), 1 Pt. meaning as break-even point and profit limit, 1 Pt. profit zone with test values in whole mattresses.

Part e)

To grow, the company considers buying a second foam-cutting machine. A supplier offers a machine with a capacity (the maximum possible output) of 2,000 mattresses per month for 175,000 EUR and a machine with a capacity of 4,000 mattresses per month for 280,000 EUR. The supplier’s prices follow a power function \(I(K) = a \cdot K^b\), with the capacity \(K\) in QU per month and the price \(I\) in MU.

Determine the exponent \(b\), rounded to three decimal places. Explain what \(b\) says about the price when the capacity grows. Compute the price of a machine for 8,000 mattresses per month and compare the price per mattress of monthly capacity for the three machines. [4 pts.]

(for verification: \(b \approx 0.678\))

Conditions: \(I(20) = 175\) and \(I(40) = 280\). Divide the two equations, \(a\) cancels:

\[\frac{I(40)}{I(20)} = \frac{a \cdot 40^b}{a \cdot 20^b} = \left(\frac{40}{20}\right)^b = 2^b = \frac{280}{175} = 1.6\]

Logarithm: \(b \ln 2 = \ln 1.6\), so

\[b = \frac{\ln 1.6}{\ln 2} \approx 0.678\]

Meaning: \(0 < b < 1\), so the price grows less than proportionally with the capacity. Doubling the capacity multiplies the price only by \(2^b = 1.6\), i.e. +60 %.

Machine for 8,000 mattresses (\(K = 80\), twice 40): \(I(80) = 2^b \cdot I(40) = 1.6 \cdot 280 = 448\) MU \(= 448{,}000\) EUR.

Price per mattress of monthly capacity:

  • 2,000 mattresses: \(\frac{175{,}000}{2{,}000} = 87.50\) EUR
  • 4,000 mattresses: \(\frac{280{,}000}{4{,}000} = 70\) EUR
  • 8,000 mattresses: \(\frac{448{,}000}{8{,}000} = 56\) EUR

The larger the machine, the cheaper each mattress of capacity (economies of scale).

Grading: 1 Pt. ratio \(2^b = 1.6\), 1 Pt. \(b \approx 0.678\) with the logarithm, 1 Pt. meaning of \(b < 1\), 1 Pt. 448,000 EUR and the prices per mattress of capacity.