Tasks 04-02 - Power Functions & Roots

Section 04: Advanced Functions

Problem 1: Power Function Basics (x)

For each function, state the domain and identify whether it’s even, odd, or neither:

  1. \(f(x) = x^4\)

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

  3. \(h(x) = x^{2/3}\)

  4. \(p(x) = \sqrt{x}\)

  5. \(q(x) = x^{-2}\)

  6. \(r(x) = \sqrt[3]{x}\)

  1. \(f(x) = x^4\)
    • Domain: All real numbers
    • Even function (symmetric about y-axis since \(f(-x) = (-x)^4 = x^4 = f(x)\))
  2. \(g(x) = x^{-3} = \frac{1}{x^3}\)
    • Domain: All real numbers except \(x = 0\)
    • Odd function (symmetric about origin since \(g(-x) = \frac{1}{(-x)^3} = -\frac{1}{x^3} = -g(x)\))
  3. \(h(x) = x^{2/3} = \sqrt[3]{x^2}\)
    • Domain: All real numbers (cube root accepts all values)
    • Even function (since \(h(-x) = \sqrt[3]{(-x)^2} = \sqrt[3]{x^2} = h(x)\))
  4. \(p(x) = \sqrt{x} = x^{1/2}\)
    • Domain: \(x \geq 0\) (square root requires non-negative values)
    • Neither even nor odd (not defined for negative x)
  5. \(q(x) = x^{-2} = \frac{1}{x^2}\)
    • Domain: All real numbers except \(x = 0\)
    • Even function (since \(q(-x) = \frac{1}{(-x)^2} = \frac{1}{x^2} = q(x)\))
  6. \(r(x) = \sqrt[3]{x} = x^{1/3}\)
    • Domain: All real numbers (odd root accepts all values)
    • Odd function (since \(r(-x) = \sqrt[3]{-x} = -\sqrt[3]{x} = -r(x)\))

Problem 2: Evaluating Power Functions (x)

Calculate the following values without a calculator:

  1. \(f(x) = x^{3/2}\), find \(f(4)\) and \(f(9)\)

  2. \(g(x) = x^{-1/2}\), find \(g(16)\) and \(g(25)\)

  3. \(h(x) = x^{2/3}\), find \(h(8)\) and \(h(27)\)

  4. \(p(x) = 2x^{-2}\), find \(p(1/2)\) and \(p(-3)\)

  1. \(f(x) = x^{3/2} = (\sqrt{x})^3\)
    • \(f(4) = (\sqrt{4})^3 = 2^3 = 8\)
    • \(f(9) = (\sqrt{9})^3 = 3^3 = 27\)
  2. \(g(x) = x^{-1/2} = \frac{1}{\sqrt{x}}\)
    • \(g(16) = \frac{1}{\sqrt{16}} = \frac{1}{4}\)
    • \(g(25) = \frac{1}{\sqrt{25}} = \frac{1}{5}\)
  3. \(h(x) = x^{2/3} = (\sqrt[3]{x})^2\)
    • \(h(8) = (\sqrt[3]{8})^2 = 2^2 = 4\)
    • \(h(27) = (\sqrt[3]{27})^2 = 3^2 = 9\)
  4. \(p(x) = 2x^{-2} = \frac{2}{x^2}\)
    • \(p(1/2) = \frac{2}{(1/2)^2} = \frac{2}{1/4} = 8\)
    • \(p(-3) = \frac{2}{(-3)^2} = \frac{2}{9}\)

Problem 3: Economies of Scale Application (xx)

A manufacturing company has a cost function:

\[C(x) = 2000 + 150x^{0.75}\]

where \(x\) is the number of units produced (in thousands) and \(C(x)\) is the cost in thousands of euros.

  1. Calculate the total cost of producing 8,000 units.

  2. Calculate the average cost per unit when producing 8,000 units.

  3. Compare the average cost per unit for 1,000 units versus 16,000 units. What does this tell you about economies of scale?

  4. If the company sells each unit for €0.50, find the production level where revenue equals €3,000 (in thousands).

  1. Total cost for 8,000 units (x = 8):

    \(C(8) = 2000 + 150(8)^{0.75}\)

    To calculate \(8^{0.75} = 8^{3/4} = (8^{1/4})^3 = (\sqrt[4]{8})^3\): Since \(8 = 2^3\), we have \(8^{1/4} = (2^3)^{1/4} = 2^{3/4} \approx 1.68\) Therefore \(8^{0.75} = (1.68)^3 \approx 4.74\)

    \(C(8) = 2000 + 150(4.74) = 2000 + 711 = 2711\)

    Total cost: €2,711,000

  2. Average cost per unit for 8,000 units:

    Average cost per thousand = \(\frac{2711}{8} = 338.88\) thousand euros

    Per individual unit: €338.88

  3. Average cost comparison:

    For 1,000 units (x = 1): \(C(1) = 2000 + 150(1)^{0.75} = 2000 + 150 = 2150\) Average = \(\frac{2150}{1} = 2150\) thousand euros per thousand = €2,150 per unit

    For 16,000 units (x = 16): \(16^{0.75} = (16^{1/4})^3 = 2^3 = 8\) \(C(16) = 2000 + 150(8) = 2000 + 1200 = 3200\) Average = \(\frac{3200}{16} = 200\) thousand euros per thousand = €200 per unit

    Interpretation: Average cost decreases dramatically as production increases (€2,150 → €338.88 → €200), demonstrating strong economies of scale.

  4. Finding production level for €3,000 revenue:

    Revenue in thousands: R = 3000 Price per unit: €0.50 = €500 per thousand units Production needed: \(x = \frac{3000}{500} = 6\) thousand units

    Therefore, the company needs to produce 6,000 units.

Problem 4: Growth Rate Comparison (xx)

Order the following functions from slowest to fastest growth for large positive values of x:

\(f(x) = x^{0.5}\), \(g(x) = x^{1.5}\), \(h(x) = x\), \(p(x) = x^{-0.5}\), \(q(x) = x^2\)

Then evaluate each function at \(x = 4\) and \(x = 100\) to verify your ordering.

Ordering from slowest to fastest growth:

  1. \(p(x) = x^{-0.5}\) (decreases as x increases)
  2. \(f(x) = x^{0.5}\) (square root)
  3. \(h(x) = x\) (linear)
  4. \(g(x) = x^{1.5}\)
  5. \(q(x) = x^2\) (quadratic)

Verification at x = 4:

  • \(p(4) = 4^{-0.5} = \frac{1}{\sqrt{4}} = \frac{1}{2} = 0.5\)
  • \(f(4) = 4^{0.5} = \sqrt{4} = 2\)
  • \(h(4) = 4\)
  • \(g(4) = 4^{1.5} = 4 \times \sqrt{4} = 4 \times 2 = 8\)
  • \(q(4) = 4^2 = 16\)

Order confirmed: 0.5 < 2 < 4 < 8 < 16 ✓

Verification at x = 100:

  • \(p(100) = 100^{-0.5} = \frac{1}{10} = 0.1\)
  • \(f(100) = 100^{0.5} = 10\)
  • \(h(100) = 100\)
  • \(g(100) = 100^{1.5} = 100 \times 10 = 1000\)
  • \(q(100) = 100^2 = 10000\)

Order confirmed: 0.1 < 10 < 100 < 1000 < 10000 ✓

The ordering remains consistent, with differences becoming more pronounced for larger x.

Problem 5: Container Design (xxx)

A company manufactures cubic and cylindrical containers.

Cubic container: Side length \(s\)

  • Surface area: \(S_c = 6s^2\)
  • Volume: \(V_c = s^3\)

Cylindrical container: Radius \(r\) and height \(h = 2r\)

  • Surface area: \(S_{cyl} = 2\pi r^2 + 4\pi r^2 = 6\pi r^2\)
  • Volume: \(V_{cyl} = \pi r^2 \times 2r = 2\pi r^3\)

  1. For a cubic container with volume 64 cubic meters, find the surface area.

  2. For a cylindrical container with the same volume (64 cubic meters), find the radius and surface area.

  3. Which shape uses less material (smaller surface area) for the same volume?

  4. The material costs €10 per square meter. Calculate the material cost difference between the two containers.

  1. Cubic container with V = 64:

    From \(V_c = s^3 = 64\): \(s = \sqrt[3]{64} = 4\) meters

    Surface area: \(S_c = 6s^2 = 6(16) = 96\) square meters

  2. Cylindrical container with V = 64:

    From \(V_{cyl} = 2\pi r^3 = 64\): \(r^3 = \frac{64}{2\pi} = \frac{32}{\pi}\) \(r = \sqrt[3]{\frac{32}{\pi}} \approx \sqrt[3]{10.19} \approx 2.17\) meters

    Surface area: \(S_{cyl} = 6\pi r^2 = 6\pi(2.17)^2 \approx 6\pi(4.71) \approx 88.6\) square meters

  3. Comparison:

    • Cubic: 96 square meters
    • Cylindrical: 88.6 square meters

    The cylindrical container uses less material (about 7.4 square meters less).

  4. Cost difference:

    Material saved: \(96 - 88.6 = 7.4\) square meters Cost savings: \(7.4 \times 10 = €74\)

    The cylindrical container saves €74 in material costs.

Problem 6: Power Function Application (xxxx)

A company’s profit function combines multiple power terms:

\[P(x) = -x^2 + 8x^{1.5} - 12x^{0.5} + 100\]

where \(x\) is the production level in thousands of units \((x > 0)\) and \(P(x)\) is profit in thousands of euros.

  1. Rewrite the function by factoring out \(x^{0.5}\) from the first three terms.

  2. Calculate the profit for production levels of 1, 4, and 9 thousand units.

  3. Determine which term dominates for small x (near 0) and which dominates for large x.

  4. Based on your calculations, estimate the production level that might maximize profit.

  1. Factoring out \(x^{0.5}\):

    \(P(x) = -x^2 + 8x^{1.5} - 12x^{0.5} + 100\) \(P(x) = x^{0.5}(-x^{1.5} + 8x - 12) + 100\) \(P(x) = \sqrt{x}(-x\sqrt{x} + 8x - 12) + 100\)

  2. Profit calculations:

    For x = 1: \(P(1) = -(1)^2 + 8(1)^{1.5} - 12(1)^{0.5} + 100\) \(P(1) = -1 + 8 - 12 + 100 = 95\) thousand euros

    For x = 4: \(P(4) = -(4)^2 + 8(4)^{1.5} - 12(4)^{0.5} + 100\) \(P(4) = -16 + 8(8) - 12(2) + 100\) \(P(4) = -16 + 64 - 24 + 100 = 124\) thousand euros

    For x = 9: \(P(9) = -(9)^2 + 8(9)^{1.5} - 12(9)^{0.5} + 100\) \(P(9) = -81 + 8(27) - 12(3) + 100\) \(P(9) = -81 + 216 - 36 + 100 = 199\) thousand euros

  3. Dominant terms:

    • For small x (near 0): The \(-12x^{0.5}\) term dominates among variable terms (lowest power)
    • For large x: The \(-x^2\) term dominates (highest power, negative coefficient)
    • This means profit initially drops from 100, then rises, but eventually falls again for very large x

  4. Estimating maximum profit:

    From our calculations:

    • P(1) = 95 (decreasing from 100)
    • P(4) = 124 (increasing)
    • P(9) = 199 (still increasing)

    Since the \(-x^2\) term eventually dominates, profit will decrease for large enough x. Testing x = 16: \(P(16) = -256 + 8(64) - 12(4) + 100 = -256 + 512 - 48 + 100 = 308\)

    Testing x = 25: \(P(25) = -625 + 8(125) - 12(5) + 100 = -625 + 1000 - 60 + 100 = 415\)

    The maximum appears to be beyond x = 25, likely around x = 30-35 thousand units.