Tasks 04-01 - Polynomial Functions

Section 04: Advanced Functions

A: Skills

Problem 1: Polynomial Identification and End Behavior (x)

Decide which of the following are polynomial functions. For each polynomial, give the degree and the leading coefficient and describe the end behavior without graphing.

  1. \(f(x) = 5x^4 - 2x^3 + 7x - 1\)

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

  3. \(h(x) = \sqrt{2}x^5 - \pi x^2 + e\)

  4. \(p(x) = (3 - x)(x + 2)(x - 1)\)

  5. \(q(x) = x^{2.5} + 3x - 1\)

  6. \(r(x) = \frac{6 - x^6 + 3x^2}{2}\)

For the end behavior, only the leading term (the term with the highest power) counts: even or odd degree, positive or negative leading coefficient.

  1. Polynomial. Degree 4, leading coefficient 5.

    • Even degree, positive leading coefficient: both ends up (↗↗)
    • \(f(x) \to +\infty\) for \(x \to -\infty\) and for \(x \to +\infty\)

  2. Not a polynomial. \(\frac{2}{x} = 2x^{-1}\) has a negative exponent. Polynomials only have whole-number exponents \(0, 1, 2, \ldots\)

  3. Polynomial. Degree 5, leading coefficient \(\sqrt{2}\) (a coefficient can be any real number, also \(\pi\) or \(e\)).

    • Odd degree, positive leading coefficient: down-up (↘↗)
    • \(h(x) \to -\infty\) for \(x \to -\infty\) and \(h(x) \to +\infty\) for \(x \to +\infty\)

  4. Polynomial. Expand step by step:

    • \((x + 2)(x - 1) = x^2 + x - 2\)
    • \((3 - x)(x^2 + x - 2) = 3x^2 + 3x - 6 - x^3 - x^2 + 2x = -x^3 + 2x^2 + 5x - 6\)
    • Degree 3, leading coefficient \(-1\). The factor \((3 - x)\) contributes \(-x\), not \(+x\).
    • Odd degree, negative leading coefficient: up-down (↗↘)

  5. Not a polynomial. The exponent 2.5 is not a whole number.

  6. Polynomial. Sorted by descending powers: \(r(x) = -\frac{1}{2}x^6 + \frac{3}{2}x^2 + 3\). Degree 6, leading coefficient \(-\frac{1}{2}\). The leading term does not have to be written first.

    • Even degree, negative leading coefficient: both ends down (↘↘)

Problem 2: Zeros and Multiplicities (xx)

For each factored polynomial:

  • List all zeros and their multiplicities
  • Describe whether the graph crosses or touches the \(x\)-axis at each zero
  • Find the \(y\)-intercept
  • Give the degree and the end behavior

  1. \(P(x) = 3(x - 2)^2(x + 1)(x - 4)\)

  2. \(Q(x) = -2x^3(x + 3)^2\)

  3. \(R(x) = \frac{1}{2}(x + 4)(x - 1)^2(x - 2)^3\)

  1. \(P(x) = 3(x - 2)^2(x + 1)(x - 4)\)
    • \(x = 2\) (multiplicity 2, even: touches the \(x\)-axis)
    • \(x = -1\) (multiplicity 1, odd: crosses)
    • \(x = 4\) (multiplicity 1, odd: crosses)
    • \(y\)-intercept: \(P(0) = 3 \cdot (-2)^2 \cdot 1 \cdot (-4) = 3 \cdot 4 \cdot (-4) = -48\)
    • Degree \(2 + 1 + 1 = 4\) (even), leading coefficient \(3 \cdot 1 \cdot 1 \cdot 1 = 3\) (positive): both ends up (↗↗)
  2. \(Q(x) = -2x^3(x + 3)^2\)
    • \(x = 0\) (multiplicity 3, odd: crosses, flattened)
    • \(x = -3\) (multiplicity 2, even: touches)
    • \(y\)-intercept: \(Q(0) = 0\)
    • Degree \(3 + 2 = 5\) (odd), leading coefficient \(-2\) (negative): up-down (↗↘)
  3. \(R(x) = \frac{1}{2}(x + 4)(x - 1)^2(x - 2)^3\)
    • \(x = -4\) (multiplicity 1, odd: crosses)
    • \(x = 1\) (multiplicity 2, even: touches)
    • \(x = 2\) (multiplicity 3, odd: crosses, flattened)
    • \(y\)-intercept: \(R(0) = \frac{1}{2} \cdot 4 \cdot (-1)^2 \cdot (-2)^3 = \frac{1}{2} \cdot 4 \cdot 1 \cdot (-8) = -16\)
    • Degree \(1 + 2 + 3 = 6\) (even), leading coefficient \(\frac{1}{2}\) (positive): both ends up (↗↗)

Problem 3: Rational Zeros and Long Division (xx)

For each polynomial, list the candidates of the Rational Root Theorem and test them in a smart order until you find a zero. Divide by polynomial long division, check the division by multiplying back and determine all real zeros.

  1. \(f(x) = x^3 - 4x^2 - 11x + 30\)

  2. \(g(x) = 3x^3 - 11x^2 - 6x + 8\)

  3. \(h(x) = x^3 - 2x^2 - 2x - 3\). Explain with the Fundamental Theorem of Algebra how many zeros \(h\) has and how many of them are real.

  1. Candidates: leading coefficient 1, so the candidates are the divisors of 30: \(\pm 1, \pm 2, \pm 3, \pm 5, \pm 6, \pm 10, \pm 15, \pm 30\).

    Test: \(f(1) = 1 - 4 - 11 + 30 = 16\), \(\; f(-1) = -1 - 4 + 11 + 30 = 36\), \(\; f(2) = 8 - 16 - 22 + 30 = 0\) ✓

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

    \[\begin{array}{l} (x^3 - 4x^2 - 11x + 30) : (x - 2) = x^2 - 2x - 15 \\ \underline{-(x^3 - 2x^2)} \\ \phantom{(x^3}\;\; -2x^2 - 11x \\ \phantom{(x^3}\;\; \underline{-(-2x^2 + 4x)} \\ \phantom{(x^3 - 2x^2}\;\; -15x + 30 \\ \phantom{(x^3 - 2x^2}\;\; \underline{-(-15x + 30)} \\ \phantom{(x^3 - 2x^2 - 15x + 3}\;\; 0 \end{array}\]

    Check: \((x - 2)(x^2 - 2x - 15) = x^3 - 2x^2 - 15x - 2x^2 + 4x + 30 = x^3 - 4x^2 - 11x + 30\) ✓

    \(x^2 - 2x - 15 = (x - 5)(x + 3)\), so \(f(x) = (x - 2)(x - 5)(x + 3)\). Zeros: \(x = -3\), \(x = 2\), \(x = 5\).

  2. Candidates: \(p\) divides 8: \(\pm 1, \pm 2, \pm 4, \pm 8\). \(q\) divides 3: \(\pm 1, \pm 3\). So the candidates are \(\pm 1, \pm 2, \pm 4, \pm 8, \pm \frac{1}{3}, \pm \frac{2}{3}, \pm \frac{4}{3}, \pm \frac{8}{3}\).

    Test (whole numbers first): \(g(1) = 3 - 11 - 6 + 8 = -6\), \(\; g(-1) = -3 - 11 + 6 + 8 = 0\) ✓

    Long division by \((x + 1)\):

    \[\begin{array}{l} (3x^3 - 11x^2 - 6x + 8) : (x + 1) = 3x^2 - 14x + 8 \\ \underline{-(3x^3 + 3x^2)} \\ \phantom{(3x^3}\;\; -14x^2 - 6x \\ \phantom{(3x^3}\;\; \underline{-(-14x^2 - 14x)} \\ \phantom{(3x^3 - 14x^2}\;\; 8x + 8 \\ \phantom{(3x^3 - 14x^2}\;\; \underline{-(8x + 8)} \\ \phantom{(3x^3 - 14x^2 - 8x + }\;\; 0 \end{array}\]

    Check: \((x + 1)(3x^2 - 14x + 8) = 3x^3 - 14x^2 + 8x + 3x^2 - 14x + 8 = 3x^3 - 11x^2 - 6x + 8\) ✓

    Quadratic formula for \(3x^2 - 14x + 8 = 0\): \(\Delta = 196 - 96 = 100\), \(x = \frac{14 \pm 10}{6}\), so \(x = 4\) or \(x = \frac{2}{3}\). Both are on the candidate list. Zeros: \(x = -1\), \(x = \frac{2}{3}\), \(x = 4\).

  3. Candidates: the divisors of 3: \(\pm 1, \pm 3\).

    Test: \(h(1) = 1 - 2 - 2 - 3 = -6\), \(\; h(-1) = -1 - 2 + 2 - 3 = -4\), \(\; h(3) = 27 - 18 - 6 - 3 = 0\) ✓

    Long division by \((x - 3)\):

    \[\begin{array}{l} (x^3 - 2x^2 - 2x - 3) : (x - 3) = x^2 + x + 1 \\ \underline{-(x^3 - 3x^2)} \\ \phantom{(x^3}\;\; x^2 - 2x \\ \phantom{(x^3}\;\; \underline{-(x^2 - 3x)} \\ \phantom{(x^3 - 2x^2}\;\; x - 3 \\ \phantom{(x^3 - 2x^2}\;\; \underline{-(x - 3)} \\ \phantom{(x^3 - 2x^2 - 2x}\;\; 0 \end{array}\]

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

    \(x^2 + x + 1 = 0\): \(\Delta = 1 - 4 = -3 < 0\), so there is no further real zero. Only real zero: \(x = 3\).

    By the Fundamental Theorem of Algebra, \(h\) has degree 3 and therefore exactly 3 zeros, counting multiplicities and complex zeros. Here one zero is real (\(x = 3\)), and the other two are the complex zeros of \(x^2 + x + 1\).

Problem 4: Intermediate Value Theorem (xx)

Consider the polynomial \(P(x) = x^4 - 5x^2 + 2x + 3\).

  1. Compute \(P(-3)\), \(P(-2)\), \(P(-1)\), \(P(0)\), \(P(1)\) and \(P(2)\).

  2. Use the Intermediate Value Theorem to give two intervals of length 1 that must each contain a zero.

  3. \(P(1)\) and \(P(2)\) are both positive. Compute \(P(1.5)\) and decide what the Intermediate Value Theorem now tells you about the interval \((1; 2)\).

  4. Give the largest possible number of zeros of \(P\). Decide whether you have now located all zeros of \(P\).

  1. Function values:

    • \(P(-3) = 81 - 45 - 6 + 3 = 33\)
    • \(P(-2) = 16 - 20 - 4 + 3 = -5\)
    • \(P(-1) = 1 - 5 - 2 + 3 = -3\)
    • \(P(0) = 3\)
    • \(P(1) = 1 - 5 + 2 + 3 = 1\)
    • \(P(2) = 16 - 20 + 4 + 3 = 3\)

  2. \(P\) is a polynomial, so it is continuous. Look for a change of sign:

    • \(P(-3) = 33 > 0\) and \(P(-2) = -5 < 0\): at least one zero in \((-3; -2)\)
    • \(P(-1) = -3 < 0\) and \(P(0) = 3 > 0\): at least one zero in \((-1; 0)\)

  3. \(P(1.5) = 5.0625 - 11.25 + 3 + 3 = -0.1875 < 0\).

    • \(P(1) = 1 > 0\) and \(P(1.5) < 0\): at least one zero in \((1; 1.5)\)
    • \(P(1.5) < 0\) and \(P(2) = 3 > 0\): at least one zero in \((1.5; 2)\)

    So \((1; 2)\) contains at least two zeros. The table from a) could not show them: equal signs at both ends do not mean “no zero”. The theorem only guarantees a zero where the sign changes.

  4. \(P\) has degree 4, so it has at most 4 zeros. We have found four intervals \((-3; -2)\), \((-1; 0)\), \((1; 1.5)\) and \((1.5; 2)\), each with at least one zero. So each interval contains exactly one zero, and these are all zeros of \(P\).

Problem 5: Polynomial Sketching (xx)

Sketch the graph of \(P(x) = -\frac{1}{4}(x + 2)(x - 2)(x - 4)^2\) by following these steps:

  1. Identify all zeros and their multiplicities.

  2. Determine the end behavior.

  3. Find the \(y\)-intercept.

  4. Determine the sign of \(P(x)\) in each interval between the zeros with test points.

  5. Sketch the complete graph.

  1. Zeros and multiplicities:

    • \(x = -2\) (multiplicity 1: crosses)
    • \(x = 2\) (multiplicity 1: crosses)
    • \(x = 4\) (multiplicity 2: touches)

  2. Degree \(1 + 1 + 2 = 4\) (even), leading coefficient \(-\frac{1}{4}\) (negative): both ends down (↘↘).

  3. \(P(0) = -\frac{1}{4} \cdot 2 \cdot (-2) \cdot (-4)^2 = -\frac{1}{4} \cdot 2 \cdot (-2) \cdot 16 = 16\), so the \(y\)-intercept is \((0, 16)\).

  4. Test points:

    • \((-\infty; -2)\), \(x = -3\): \(P(-3) = -\frac{1}{4} \cdot (-1) \cdot (-5) \cdot 49 = -61.25\), negative
    • \((-2; 2)\), \(x = 0\): \(P(0) = 16\), positive
    • \((2; 4)\), \(x = 3\): \(P(3) = -\frac{1}{4} \cdot 5 \cdot 1 \cdot 1 = -1.25\), negative
    • \((4; \infty)\), \(x = 5\): \(P(5) = -\frac{1}{4} \cdot 7 \cdot 3 \cdot 1 = -5.25\), negative

  5. The graph:

    • comes from below on the left (\(P(x) \to -\infty\))
    • crosses the \(x\)-axis upward at \(x = -2\)
    • is positive between \(-2\) and \(2\) and passes through \((0, 16)\)
    • crosses the \(x\)-axis downward at \(x = 2\)
    • is negative between \(2\) and \(4\), touches the \(x\)-axis from below at \(x = 4\) and turns back down
    • goes down on the right (\(P(x) \to -\infty\))

B: Set Up & Check

Problem 6: From Text to Polynomials (xx)

Translate each statement into mathematics. Then answer the check question: it tells you whether your translation works. In all items, \(x\) is the output in QU (quantity units) and costs and profits are in MU (monetary units).

  1. “A skateboard manufacturer models its monthly total costs by \(C(x) = 0.4x^3 - 6x^2 + 35x + d\) with 1 QU = 100 skateboards and 1 MU = 100 EUR. Its fixed costs (the costs that arise even without production) are 2,600 EUR per month.” Give \(d\).

    Check: What does your cost function give for \(x = 0\)? Is this the amount from the statement?

  2. “The profit function of a headlamp producer is \(P(x) = -x^3 + 5x^2 + 12x - 36\) with 1 QU = 1,000 headlamps. The company breaks even at 2,000 headlamps.” Write this statement as an equation and give the linear factor of \(P\) that follows from it.

    Check: Insert your \(x\)-value into \(P\). Do you get 0?

  3. An FSP task says: “The fixed costs are 20 MU. Determine the cost function assuming it is a polynomial of third degree.” Write down the approach (ansatz) for \(C(x)\) with all the information from this sentence.

    Check: How many unknowns are left in your ansatz? How many further conditions does the task have to give you?

  4. “The profit function of a snowboard producer is a polynomial of third degree with a negative leading coefficient.” Describe the end behavior of the profit function and explain what it means for the company.

    Check: The producer’s profit function is \(P(x) = -0.5x^3 + 5x^2 + 4x - 30\). Compute \(P(8)\) and \(P(12)\). Do the two values fit your answer?

  5. “The profit function of a water filter producer is \(P(x) = -2(x + 1)(x - 4)(x - 8)\). The company makes a profit between 4 QU and 8 QU.” Write the second sentence as an inequality with the output range as an interval.

    Check: Compute \(P(2)\), \(P(6)\) and \(P(9)\). Do the signs confirm the statement?

  1. Fixed costs are the costs at \(x = 0\): \(C(0) = d\). In MU: 2,600 EUR \(= 26\) MU, so \(d = 26\) and \(C(x) = 0.4x^3 - 6x^2 + 35x + 26\).

    Check: yes. \(C(0) = 26\) MU \(= 2{,}600\) EUR.

    Typical mistake: \(d = 2600\). The cost function is in MU, so the EUR amount must be divided by 100 first. A second mistake is the condition \(C(1) = 26\): fixed costs belong to the output 0, not to the first QU.

  2. 2,000 headlamps \(= 2\) QU. Breaking even means that the profit is zero: \(P(2) = 0\). So \(x = 2\) is a zero, and \((x - 2)\) is a factor: \(P(x) = (x - 2) \cdot Q(x)\) with a quadratic \(Q(x)\).

    Check: yes. \(P(2) = -8 + 20 + 24 - 36 = 0\).

    Typical mistake: \(P(2000) = 0\) (the output must be in QU) or the factor \((x + 2)\). A zero \(x = c\) gives the factor \((x - c)\), so the zero 2 gives \((x - 2)\).

  3. “Polynomial of third degree” means that all powers up to \(x^3\) can occur, and the fixed costs give the constant term:

    \(C(x) = ax^3 + bx^2 + cx + 20\)

    Check: three unknowns \(a\), \(b\), \(c\) are left, so the task has to give three further conditions (for example three cost values).

    Typical mistake: \(C(x) = ax^3 + 20\) or \(C(x) = x^3 + bx^2 + cx + 20\). The first ansatz drops the terms \(bx^2\) and \(cx\), and the second assumes a leading coefficient of 1. Neither is part of “third degree”.

  4. Odd degree and negative leading coefficient: up-down (↗↘). For \(x \to \infty\), \(P(x) \to -\infty\). Only \(x \geq 0\) matters here: at very large outputs, the costs grow faster than the revenue, and the company makes a loss that keeps growing. So there must be an output above which the company no longer makes a profit.

    Check: yes. \(P(8) = -256 + 320 + 32 - 30 = 66 > 0\) and \(P(12) = -864 + 720 + 48 - 30 = -126 < 0\). At 8 QU the company makes a profit, at 12 QU a loss.

    Typical mistake: “The leading coefficient is negative, so the company always makes a loss.” The sign of the leading coefficient only says what happens at very large outputs. \(P(8) > 0\) shows that the profit can be positive in between.

  5. \(P(x) > 0\) for \(4 < x < 8\), i.e. for \(x \in (4; 8)\).

    Check: yes.

    • \(P(2) = -2 \cdot 3 \cdot (-2) \cdot (-6) = -72 < 0\)
    • \(P(6) = -2 \cdot 7 \cdot 2 \cdot (-2) = 56 > 0\)
    • \(P(9) = -2 \cdot 10 \cdot 5 \cdot 1 = -100 < 0\)

    All three zeros have multiplicity 1, so the sign changes at each zero, and the test points give the sign in each interval.

    Typical mistake: “The profit is positive between any two neighbouring zeros.” The signs alternate, so between \(-1\) and \(4\) the profit is negative (test point \(P(2) < 0\)). Also, \(P\) is positive for \(x < -1\) (for example \(P(-2) = 120\)), but a negative output is not possible, so this interval does not count.

Problem 7: One Paragraph, All Conditions (xx)

A small company makes wooden toy trains. It models its monthly total costs by a polynomial of third degree. Its fixed costs are 5,000 EUR per month. In a month with 200 toy trains, the total costs were 7,600 EUR, with 400 toy trains 8,600 EUR and with 600 toy trains 12,800 EUR. Toy shops pay 38 EUR per train. The owner wants to know the cost function, the profit function, the break-even point and the profit limit (upper break-even point), and whether the planned output of 500 toy trains per month brings a profit. (1 QU = 100 toy trains, 1 MU = 100 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 condition for 400 toy trains as \(C(400) = 8600\). Name the two mistakes.

  1. Ansatz (polynomial of third degree): \(C(x) = ax^3 + bx^2 + cx + d\)
  2. Fixed costs: \(C(0) = 50\), so \(d = 50\)
  3. 200 toy trains: \(C(2) = 76\)
  4. 400 toy trains: \(C(4) = 86\)
  5. 600 toy trains: \(C(6) = 128\)
  6. Price: 38 EUR per train \(= 3{,}800\) EUR per QU \(= 38\) MU per QU, so \(R(x) = 38x\)
  7. Profit function: \(P(x) = R(x) - C(x)\)
  8. Break-even point and profit limit: the zeros of \(P\) with \(x \geq 0\), the smaller one is the break-even point and the larger one the profit limit
  9. Planned output: 500 toy trains \(= 5\) QU. Profit if \(P(5) > 0\)

Check: \(x\) must be in QU (400 toy trains \(= 4\) QU), and the costs must be in MU (8,600 EUR \(= 86\) MU). The correct condition is \(C(4) = 86\).

Typical mistake: only three conditions. “Polynomial of third degree” has four unknowns, and the fixed costs are the fourth condition, \(C(0) = 50\).

C: Applied

Problem 8: Costs of a Greenhouse Manufacturer (xx)

A company builds small greenhouses for gardens (glass houses for growing tomatoes and herbs). Its monthly total costs are

\[C(x) = x^3 - 9x^2 + 30x + 60\]

with \(C\) in MU and \(x\) in QU per month (1 QU = 100 greenhouses, 1 MU = 1,000 EUR). The company can build at most 900 greenhouses per month (its capacity, the maximum possible output), so \(0 \leq x \leq 9\).

  1. Give the fixed costs in EUR. Compute the total costs for the planned output of 500 greenhouses and the average cost per greenhouse (total costs divided by the number of greenhouses).

(for verification: 220 EUR per greenhouse)

  1. Compute the additional costs of 100 more greenhouses in four cases: from 100 to 200, from 300 to 400, from 600 to 700 and from 800 to 900 greenhouses. Describe the pattern and explain it.

  2. The owner’s budget allows total costs of at most 236,000 EUR per month. Determine the largest output within the budget with the Rational Root Theorem and polynomial long division. Show that the cost equation has no other real solution.

Hint: your results from a) and b) tell you between which outputs the solution must lie.

(for verification: 800 greenhouses)

  1. Garden centres pay 250 EUR per greenhouse. Show that the company breaks even at exactly 400 greenhouses. Explain with the average cost from a) why the company makes a profit with 500 greenhouses and compute this profit in EUR.

a) Fixed costs: \(C(0) = 60\) MU \(= 60{,}000\) EUR per month.

500 greenhouses \(= 5\) QU: \(C(5) = 125 - 225 + 150 + 60 = 110\) MU \(= 110{,}000\) EUR.

Average cost: \(\frac{110{,}000}{500} = 220\) EUR per greenhouse.

b) The needed cost values: \(C(1) = 82\), \(C(2) = 92\), \(C(3) = 96\), \(C(4) = 100\), \(C(6) = 132\), \(C(7) = 172\), \(C(8) = 236\), \(C(9) = 330\).

  • 100 to 200: \(C(2) - C(1) = 10\) MU \(= 10{,}000\) EUR
  • 300 to 400: \(C(4) - C(3) = 4\) MU \(= 4{,}000\) EUR
  • 600 to 700: \(C(7) - C(6) = 40\) MU \(= 40{,}000\) EUR
  • 800 to 900: \(C(9) - C(8) = 94\) MU \(= 94{,}000\) EUR

The total costs rise in every step, but the additional costs first fall and then rise sharply. At small outputs, each further batch is cheaper: machines and staff are used better, and material is bought in larger amounts (economies of scale). At large outputs, overtime and the capacity limit make each further batch more expensive. This is the shape of a cost function with scale effects.

c) Condition \(C(x) = 236\) (236,000 EUR \(= 236\) MU):

\[x^3 - 9x^2 + 30x + 60 = 236 \iff f(x) = x^3 - 9x^2 + 30x - 176 = 0\]

Where is the solution? \(f(5) = C(5) - 236 = -126 < 0\) and \(f(9) = C(9) - 236 = 94 > 0\). \(f\) is a polynomial and continuous, so by the Intermediate Value Theorem it has a zero between 5 and 9.

Rational Root Theorem: leading coefficient 1, so the candidates are the divisors of 176: \(\pm 1, \pm 2, \pm 4, \pm 8, \pm 11, \pm 16, \pm 22, \pm 44, \pm 88, \pm 176\). The only candidate between 5 and 9 is 8:

\[f(8) = 512 - 576 + 240 - 176 = 0 \quad \checkmark\]

(Testing in the usual order also works: \(f(1) = -154\), \(f(2) = -144\), \(f(4) = -136\), \(f(8) = 0\).)

Long division by \((x - 8)\):

\[\begin{array}{l} (x^3 - 9x^2 + 30x - 176) : (x - 8) = x^2 - x + 22 \\ \underline{-(x^3 - 8x^2)} \\ \phantom{(x^3}\;\; -x^2 + 30x \\ \phantom{(x^3}\;\; \underline{-(-x^2 + 8x)} \\ \phantom{(x^3 - 9x^2}\;\; 22x - 176 \\ \phantom{(x^3 - 9x^2}\;\; \underline{-(22x - 176)} \\ \phantom{(x^3 - 9x^2 + 22x - 1}\;\; 0 \end{array}\]

Check: \((x - 8)(x^2 - x + 22) = x^3 - x^2 + 22x - 8x^2 + 8x - 176 = x^3 - 9x^2 + 30x - 176\) ✓

\(x^2 - x + 22 = 0\): \(\Delta = 1 - 88 = -87 < 0\), no real solution. So \(x = 8\) is the only real solution.

Decision: \(f\) has no zero between 0 and 8, so its sign does not change there: \(f(0) = -176 < 0\) means \(C(x) < 236\) for \(0 \leq x < 8\). After 8, \(f(9) = 94 > 0\), so the budget is exceeded. The largest output within the budget is \(x = 8\) QU \(=\) 800 greenhouses, with \(C(8) = 236\) MU \(= 236{,}000\) EUR.

d) 250 EUR per greenhouse \(= 250 \cdot 100 = 25{,}000\) EUR per QU \(= 25\) MU per QU, so \(R(x) = 25x\) and

\[P(x) = 25x - (x^3 - 9x^2 + 30x + 60) = -x^3 + 9x^2 - 5x - 60\]

400 greenhouses \(= 4\) QU: \(P(4) = -64 + 144 - 20 - 60 = 0\) ✓. Revenue and costs are both 100 MU \(= 100{,}000\) EUR, so the company breaks even.

With 500 greenhouses, the average cost is 220 EUR per greenhouse (from a)), but each greenhouse sells for 250 EUR. So each greenhouse brings in 30 EUR more than it costs on average, and the profit is \(500 \cdot 30 = 15{,}000\) EUR. Check: \(P(5) = 125 - 110 = 15\) MU \(= 15{,}000\) EUR ✓

Problem 9: Market Share of a New Energy Drink (xx)

A beverage company launches a new energy drink in the supermarkets of a region. Its market share (its percentage of all energy drinks sold in the region) \(t\) years after the launch is modelled by

\[S(t) = -0.2t^3 + 1.8t^2 \text{ (in \%)}, \qquad 0 \leq t \leq 9\]

  1. Compute \(S(1)\), \(S(3)\) and \(S(8)\) and interpret the values.

  2. Factor \(S(t)\) completely. Give the zeros with their multiplicities, describe how the graph behaves at each zero and explain what the zeros mean for the drink.

  3. Describe the end behavior of \(S\). Explain why the model can only be used for \(0 \leq t \leq 9\).

  4. The supermarkets keep the drink on the shelves at eye level as long as its market share is at least 10.8 %. Show that the equation \(S(t) = 10.8\) can be written as \(t^3 - 9t^2 + 54 = 0\). Determine all its solutions with your result from a) and polynomial long division, rounded to two decimal places. Decide for how long the drink stays on the shelves at eye level.

Hint: the equation has no \(t\)-term. Write it as \(0t\) in the long division, so that every power has its own column.

(for verification: about 8.20 years after the launch)

  1. Show with the Intermediate Value Theorem that the market share passes 18 % between \(t = 4\) and \(t = 5\). Find a second interval of length 1 in which the market share is 18 % again.

Hint: apply the theorem to \(f(t) = S(t) - 18\).

a) Values:

  • \(S(1) = -0.2 + 1.8 = 1.6\): one year after the launch, 1.6 % of all energy drinks sold in the region are the new drink
  • \(S(3) = -5.4 + 16.2 = 10.8\): after three years, the market share is 10.8 %
  • \(S(8) = -102.4 + 115.2 = 12.8\): after eight years, it is 12.8 %

b) Factor out the common factor \(-0.2t^2\):

\[S(t) = -0.2t^2(t - 9) = 0.2t^2(9 - t)\]

  • \(t = 0\), multiplicity 2 (even): the graph touches the \(t\)-axis. At the launch, the market share is 0, and it grows only slowly at first
  • \(t = 9\), multiplicity 1 (odd): the graph crosses the \(t\)-axis. Nine years after the launch, the market share has fallen back to 0, and the drink has disappeared from the market

c) Degree 3 (odd), leading coefficient \(-0.2\) (negative): up-down (↗↘). For \(t \to \infty\), \(S(t) \to -\infty\).

For \(0 < t < 9\), \(t^2 > 0\) and \(9 - t > 0\), so \(S(t) > 0\). For \(t > 9\), the factor \(9 - t\) is negative, so \(S(t) < 0\). A market share cannot be negative, so the model only makes sense up to \(t = 9\). Before the launch (\(t < 0\)) there is no market share at all.

d) Bring the equation into zero form and multiply by \(-5\):

\[-0.2t^3 + 1.8t^2 = 10.8 \iff -0.2t^3 + 1.8t^2 - 10.8 = 0 \;\big|\cdot(-5) \iff t^3 - 9t^2 + 54 = 0 \quad \checkmark\]

From a), \(S(3) = 10.8\), so \(t = 3\) is a solution. Check: \(27 - 81 + 54 = 0\) ✓

Long division by \((t - 3)\):

\[\begin{array}{l} (t^3 - 9t^2 + 0t + 54) : (t - 3) = t^2 - 6t - 18 \\ \underline{-(t^3 - 3t^2)} \\ \phantom{(t^3}\;\; -6t^2 + 0t \\ \phantom{(t^3}\;\; \underline{-(-6t^2 + 18t)} \\ \phantom{(t^3 - 9t^2}\;\; -18t + 54 \\ \phantom{(t^3 - 9t^2}\;\; \underline{-(-18t + 54)} \\ \phantom{(t^3 - 9t^2 - 18t + 5}\;\; 0 \end{array}\]

The missing \(t\)-term is written as \(0t\). Check: \((t - 3)(t^2 - 6t - 18) = t^3 - 6t^2 - 18t - 3t^2 + 18t + 54 = t^3 - 9t^2 + 54\) ✓

Quadratic formula for \(t^2 - 6t - 18 = 0\): \(\Delta = 36 + 72 = 108\) and \(\sqrt{108} = 6\sqrt{3}\):

\[t_{1,2} = \frac{6 \pm 6\sqrt{3}}{2} = 3 \pm 3\sqrt{3}, \qquad t_1 \approx 8.20, \qquad t_2 \approx -2.20\]

\(t_2\) lies before the launch and is rejected. So \(S(t) = 10.8\) at \(t = 3\) and at \(t \approx 8.20\).

Test value between them: \(S(6) = -43.2 + 64.8 = 21.6 \geq 10.8\) ✓. Outside: \(S(2) = 5.6 < 10.8\).

The drink stays on the shelves at eye level from 3 years to about 8.20 years after the launch, i.e. for about 5.20 years.

e) \(S\) is a polynomial, so \(f(t) = S(t) - 18\) is continuous.

  • \(S(4) = -12.8 + 28.8 = 16\), so \(f(4) = -2 < 0\)
  • \(S(5) = -25 + 45 = 20\), so \(f(5) = 2 > 0\)

\(f\) changes its sign, so by the Intermediate Value Theorem there is a \(t\) between 4 and 5 with \(f(t) = 0\), i.e. \(S(t) = 18\).

Second interval: \(S(7) = -68.6 + 88.2 = 19.6\), so \(f(7) = 1.6 > 0\), and \(S(8) = 12.8\), so \(f(8) = -5.2 < 0\). Between \(t = 7\) and \(t = 8\), the market share falls back to 18 %.

Problem 10: Profit of a Bike Shop During Construction Work (xx)

A bike shop is located on a street where the city builds a new tram line. The owner models the shop’s profit by

\[P(t) = 0.5t^3 - 4t^2 + 0.5t + 21\]

in thousand EUR per quarter, where \(t\) is the time in quarters since the beginning of 2026 and \(0 \leq t \leq 8\).

  1. Compute \(P(0)\), \(P(2)\), \(P(5)\) and \(P(8)\) and interpret the values.

  2. Determine all zeros of \(P\) with the Rational Root Theorem and polynomial long division. Give \(P\) in factored form.

Hint: multiply the equation \(P(t) = 0\) by 2 first, so that all coefficients are integers.

(for verification: \(P(t) = 0.5(t + 2)(t - 3)(t - 7)\))

  1. Decide with test points in which period the shop makes a loss and how long this period lasts. Explain why the zero \(t = -2\) plays no role.

  2. Describe the end behavior of \(P\). Compute \(P(10)\) and explain why the owner should not use the model after 2027.

a) Values:

  • \(P(0) = 21\): at the beginning of 2026, the shop earns 21,000 EUR per quarter
  • \(P(2) = 4 - 16 + 1 + 21 = 10\): after two quarters, only 10,000 EUR per quarter
  • \(P(5) = 62.5 - 100 + 2.5 + 21 = -14\): after five quarters, the shop makes a loss of 14,000 EUR per quarter
  • \(P(8) = 256 - 256 + 4 + 21 = 25\): at the end of 2027, the shop earns 25,000 EUR per quarter, more than before the construction work

b) Multiply \(P(t) = 0\) by 2:

\[t^3 - 8t^2 + t + 42 = 0\]

Candidates: leading coefficient 1, so the divisors of 42: \(\pm 1, \pm 2, \pm 3, \pm 6, \pm 7, \pm 14, \pm 21, \pm 42\).

Test: with \(g(t) = t^3 - 8t^2 + t + 42\): \(g(1) = 1 - 8 + 1 + 42 = 36\), \(\; g(-1) = -1 - 8 - 1 + 42 = 32\), \(\; g(2) = 8 - 32 + 2 + 42 = 20\), \(\; g(-2) = -8 - 32 - 2 + 42 = 0\) ✓

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

\[\begin{array}{l} (t^3 - 8t^2 + t + 42) : (t + 2) = t^2 - 10t + 21 \\ \underline{-(t^3 + 2t^2)} \\ \phantom{(t^3}\;\; -10t^2 + t \\ \phantom{(t^3}\;\; \underline{-(-10t^2 - 20t)} \\ \phantom{(t^3 - 10t^2}\;\; 21t + 42 \\ \phantom{(t^3 - 10t^2}\;\; \underline{-(21t + 42)} \\ \phantom{(t^3 - 10t^2 + 21t + }\;\; 0 \end{array}\]

Check: \((t + 2)(t^2 - 10t + 21) = t^3 - 10t^2 + 21t + 2t^2 - 20t + 42 = t^3 - 8t^2 + t + 42\) ✓

\(t^2 - 10t + 21 = (t - 3)(t - 7)\). Zeros: \(t = -2\), \(t = 3\), \(t = 7\), and

\[P(t) = 0.5(t + 2)(t - 3)(t - 7)\]

Note: a) gives a shortcut. \(P(2) > 0\) and \(P(5) < 0\), so there is a zero between 2 and 5, and the only candidate there is 3.

c) All three zeros have multiplicity 1, so the sign changes at each of them. Test points in the domain:

  • \(0 \leq t < 3\): \(P(1) = 0.5 \cdot 3 \cdot (-2) \cdot (-6) = 18 > 0\), profit
  • \(3 < t < 7\): \(P(5) = -14 < 0\), loss
  • \(7 < t \leq 8\): \(P(8) = 25 > 0\), profit

The shop makes a loss for \(3 < t < 7\), i.e. from the beginning of October 2026 to the beginning of October 2027. The loss period lasts four quarters, i.e. one year.

\(t = -2\) means two quarters before the beginning of 2026. This lies outside the domain \(0 \leq t \leq 8\) of the model, so it has no meaning here.

d) Degree 3 (odd), leading coefficient \(0.5\) (positive): down-up (↘↗). For \(t \to \infty\), \(P(t) \to \infty\): the model predicts profits that keep growing.

\(P(10) = 500 - 400 + 5 + 21 = 126\), i.e. 126,000 EUR per quarter, six times the profit at the beginning of 2026. The new tram stop can bring more customers, but not without limit. The model describes the construction period and the time shortly after it, so the owner should only use it for \(0 \leq t \leq 8\).

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: Treadmills for Home Use [20 pts. total]

A young company builds treadmills for home use and sells them to sports retailers. Its fixed costs (rent for the production hall, machines, base salaries) are 144,000 EUR per month. In a month with 100 treadmills, the total costs were 170,000 EUR, in a month with 300 treadmills 186,000 EUR, and in a month with 500 treadmills 234,000 EUR. The retailers pay 900 EUR per treadmill. The company can build at most 1,100 treadmills per month (its capacity) and plans to build and sell 600 treadmills per month.

(QU = quantity units, MU = monetary units. 1 QU = 100 treadmills, 1 MU = 1,000 EUR. Revenue = quantity × price. Profit = revenue − total costs.)

Part a)

Determine the cost function \(C(x)\), assuming it is a polynomial of third degree. [6 pts.]

(for verification: \(C(x) = 2x^3 - 14x^2 + 38x + 144\))

Ansatz: \(C(x) = ax^3 + bx^2 + cx + d\).

Convert to QU and MU. Fixed costs: 144,000 EUR \(= 144\) MU, so \(d = C(0) = 144\). 100 treadmills \(= 1\) QU cost 170 MU, 300 treadmills \(= 3\) QU cost 186 MU, and 500 treadmills \(= 5\) QU cost 234 MU.

Insert the data points into \(C(x) = ax^3 + bx^2 + cx + 144\), subtract 144 and divide by the \(x\)-value:

\[\begin{aligned} \text{(I)} \quad C(1) &= a + b + c + 144 = 170 &&\implies a + b + c = 26 \\ \text{(II)} \quad C(3) &= 27a + 9b + 3c + 144 = 186 &&\implies 9a + 3b + c = 14 \\ \text{(III)} \quad C(5) &= 125a + 25b + 5c + 144 = 234 &&\implies 25a + 5b + c = 18 \end{aligned}\]

Eliminate \(c\):

\[\text{(II)} - \text{(I)}: \; 8a + 2b = -12 \;\big|: 2 \implies \text{(IV)} \; 4a + b = -6\]

\[\text{(III)} - \text{(II)}: \; 16a + 2b = 4 \;\big|: 2 \implies \text{(V)} \; 8a + b = 2\]

Eliminate \(b\): \(\text{(V)} - \text{(IV)}: \; 4a = 8 \implies a = 2\).

Back-substitution: (IV): \(8 + b = -6 \implies b = -14\). (I): \(2 - 14 + c = 26 \implies c = 38\).

Check with the data: \(C(3) = 54 - 126 + 114 + 144 = 186\) ✓, \(\; C(5) = 250 - 350 + 190 + 144 = 234\) ✓

\[C(x) = 2x^3 - 14x^2 + 38x + 144 \quad \checkmark\]

Grading: 1 Pt. ansatz and \(d = 144\) from the fixed costs, 1 Pt. conversion to QU and MU, 1 Pt. the three equations, 2 Pts. elimination of \(c\) and \(b\), 1 Pt. back-substitution and check.

Part b)

Show that the profit function is given by \(P(x) = -2x^3 + 14x^2 + 52x - 144\). Calculate the profit in EUR for the planned output. [3 pts.]

Price in MU per QU: 900 EUR per treadmill \(= 900 \cdot 100 = 90{,}000\) EUR per QU \(= 90\) MU per QU. So \(R(x) = 90x\).

Profit = revenue − total costs. The minus sign changes every sign in the bracket:

\[P(x) = 90x - (2x^3 - 14x^2 + 38x + 144) = -2x^3 + 14x^2 + 52x - 144 \quad \checkmark\]

Planned output: 600 treadmills \(= 6\) QU:

\[P(6) = -2 \cdot 216 + 14 \cdot 36 + 52 \cdot 6 - 144 = -432 + 504 + 312 - 144 = 240 \text{ MU} = 240{,}000 \text{ EUR per month}\]

Grading: 1 Pt. price of 90 MU per QU and \(R(x)\), 1 Pt. \(P = R - C\) with sign change, 1 Pt. profit of 240,000 EUR.

Part c)

Investigate the asymptotic behavior of \(P\) (what happens to \(P(x)\) for \(x \to \infty\) and for \(x \to -\infty\)) and determine the intercept of the graph of \(P\) with the vertical axis. Explain the business meaning of both results. Explain with your result from b) why the company must have a break-even point and a profit limit. [3 pts.]

Asymptotic behavior: degree 3 (odd), leading coefficient \(-2\) (negative): up-down (↗↘).

\[P(x) \to -\infty \text{ for } x \to \infty, \qquad P(x) \to +\infty \text{ for } x \to -\infty\]

Meaning: at very large outputs, the costs (with the term \(2x^3\)) grow much faster than the revenue \(90x\), so the losses keep growing. Negative outputs are not possible, so \(x \to -\infty\) has no business meaning.

Intercept: \(P(0) = -144\), i.e. the point \((0, -144)\). Without production, the company loses its fixed costs of 144,000 EUR per month.

Break-even point and profit limit: \(P\) is a polynomial and continuous. \(P(0) = -144 < 0\) and \(P(6) = 240 > 0\), so by the Intermediate Value Theorem there is a zero between 0 and 6: the break-even point. After \(x = 6\), the profit must fall to \(-\infty\), e.g. \(P(11) = -540 < 0\). So there is a second zero between 6 and 11: the profit limit.

Grading: 1 Pt. asymptotic behavior with meaning, 1 Pt. intercept \((0, -144)\) with meaning, 1 Pt. change of sign as reason.

Part d)

Determine the (lower) break-even point and the profit limit (upper break-even point). [5 pts.]

(for verification: break-even point at 2 QU)

Zero form with integer coefficients: \(P(x) = 0\), divide by \(-2\):

\[-2x^3 + 14x^2 + 52x - 144 = 0 \;\big|: (-2) \iff x^3 - 7x^2 - 26x + 72 = 0\]

Rational Root Theorem: leading coefficient 1, so the candidates are the divisors of 72: \(\pm 1, \pm 2, \pm 3, \pm 4, \pm 6, \pm 8, \pm 9, \pm 12, \pm 18, \pm 24, \pm 36, \pm 72\). By c), a zero lies between 0 and 6.

Test: \(x = 1\): \(1 - 7 - 26 + 72 = 40 \neq 0\). \(\; x = 2\): \(8 - 28 - 52 + 72 = 0\) ✓

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

\[\begin{array}{l} (x^3 - 7x^2 - 26x + 72) : (x - 2) = x^2 - 5x - 36 \\ \underline{-(x^3 - 2x^2)} \\ \phantom{(x^3}\;\; -5x^2 - 26x \\ \phantom{(x^3}\;\; \underline{-(-5x^2 + 10x)} \\ \phantom{(x^3 - 7x^2}\;\; -36x + 72 \\ \phantom{(x^3 - 7x^2}\;\; \underline{-(-36x + 72)} \\ \phantom{(x^3 - 7x^2 - 36x + 7}\;\; 0 \end{array}\]

Check: \((x - 2)(x^2 - 5x - 36) = x^3 - 5x^2 - 36x - 2x^2 + 10x + 72 = x^3 - 7x^2 - 26x + 72\) ✓

Remaining zeros: \(x^2 - 5x - 36 = (x - 9)(x + 4) = 0\), so \(x = 9\) or \(x = -4\). (Quadratic formula: \(x = \frac{5 \pm \sqrt{25 + 144}}{2} = \frac{5 \pm 13}{2}\).)

\[P(x) = -2(x - 2)(x - 9)(x + 4)\]

\(x = -4\) is not a possible output. Break-even point: \(x = 2\) QU \(= 200\) treadmills. Profit limit: \(x = 9\) QU \(= 900\) treadmills.

Grading: 1 Pt. zero form with integer coefficients, 1 Pt. candidates and test of \(x = 2\), 1 Pt. long division, 1 Pt. remaining zeros 9 and \(-4\), 1 Pt. break-even point and profit limit in treadmills with \(x = -4\) rejected.

Part e)

Decide in which output range the company makes a profit. The sales department suggests using the full capacity. Assess this suggestion. [3 pts.]

Test points: all zeros have multiplicity 1, so the sign of \(P\) can only change at \(x = 2\) and \(x = 9\).

  • \(0 \leq x < 2\): \(P(1) = -2 + 14 + 52 - 144 = -80 < 0\), loss
  • \(2 < x < 9\): \(P(6) = 240 > 0\), profit
  • \(9 < x \leq 11\): \(P(10) = -2000 + 1400 + 520 - 144 = -224 < 0\), loss

Profit zone: \(2 < x < 9\), i.e. the company makes a profit with more than 200 and fewer than 900 treadmills per month (201 to 899 treadmills). The planned 600 treadmills lie inside.

Full capacity: 1,100 treadmills \(= 11\) QU:

\[P(11) = -2 \cdot 1331 + 14 \cdot 121 + 52 \cdot 11 - 144 = -2662 + 1694 + 572 - 144 = -540 \text{ MU}\]

This is a loss of 540,000 EUR per month. The suggestion is bad: the capacity lies above the profit limit of 900 treadmills, and every output above it brings a loss.

Grading: 1 Pt. test points in the three intervals, 1 Pt. profit zone 201 to 899 treadmills, 1 Pt. \(P(11) = -540\) with decision.