Practice Maths

Solutions — Curve Sketching — Maxima, Minima and End Behaviour

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  1. Q1 — Sketch y = x³ − 3x²

    Step 1 — y-intercept: f(0) = 0. Point: (0, 0).

    Step 2 — x-intercepts: x³ − 3x² = x²(x − 3) = 0 → x = 0 (double root, touches axis) and x = 3. Intercepts: (0, 0) and (3, 0).

    Step 3 — Stationary points: f′(x) = 3x² − 6x = 3x(x − 2) = 0 → x = 0 or x = 2.
    f(2) = 8 − 12 = −4. Points: (0, 0) and (2, −4).
    Sign: f′(−1) = 9 > 0; f′(1) = −3 < 0; f′(3) = 9 > 0.
    (0, 0): +→− → local max. (2, −4): −→+ → local min.

    Step 4 — End behaviour: Leading term x³ (odd, positive): y → −∞ as x → −∞; y → +∞ as x → +∞.

    Step 5 — Sketch features: Curve descends from bottom-left, rises to local max at (0, 0), falls to local min at (2, −4), then rises to top-right. Touches x-axis at origin (double root), crosses at (3, 0).

  2. Q2 — Sketch y = −x³ + 4x

    Step 1 — y-intercept: f(0) = 0. Point: (0, 0).

    Step 2 — x-intercepts: −x³ + 4x = −x(x² − 4) = −x(x − 2)(x + 2) = 0 → x = 0, x = 2, x = −2. Intercepts: (−2, 0), (0, 0), (2, 0).

    Step 3 — Stationary points: f′(x) = −3x² + 4 = 0 → x² = 4⁄3 → x = ±2⁄√3 = ±2√3⁄3 ≈ ±1.155.
    f(2⁄√3) = −(8⁄3√3) + 8⁄√3 = −8⁄(3√3) + 8⁄√3 = (8⁄√3)(1 − 1⁄3) = (8⁄√3)(2⁄3) = 16⁄(3√3) = 16√3⁄9 ≈ 3.08.
    By symmetry (odd function): f(−2⁄√3) ≈ −3.08.
    Sign: f′(0) = 4 > 0 (between the two roots); f′(2) = −8 < 0 (outside).
    x = −2⁄√3: −→+ → local min. x = 2⁄√3: +→− → local max.

    Step 4 — End behaviour: Leading term −x³ (odd, negative): y → +∞ as x → −∞; y → −∞ as x → +∞.

    Step 5 — Sketch features: Curve descends from top-left, reaches local min at x ≈ −1.15, rises to local max at x ≈ 1.15, then falls to bottom-right. Crosses x-axis at −2, 0, and 2.

  3. Q3 — End behaviour

    (a) y = 2x4: even degree, positive leading coefficient.
    As x → +∞, y → +∞. As x → −∞, y → +∞. (U-shape ends)

    (b) y = −3x³: odd degree, negative leading coefficient.
    As x → +∞, y → −∞. As x → −∞, y → +∞.

    (c) y = x5 − 2x: odd degree, positive leading coefficient (x5 dominates).
    As x → +∞, y → +∞. As x → −∞, y → −∞.

  4. Q4 — Full sketch of f(x) = x4 − 8x² + 7

    Step 1 — y-intercept: f(0) = 7. Point: (0, 7).

    Step 2 — x-intercepts: Let u = x²: u² − 8u + 7 = (u − 7)(u − 1) = 0 → x² = 7 or x² = 1.
    x = ±√7 ≈ ±2.65 and x = ±1. Four x-intercepts: (−√7, 0), (−1, 0), (1, 0), (√7, 0).

    Step 3 — Stationary points: f′(x) = 4x³ − 16x = 4x(x² − 4) = 4x(x − 2)(x + 2) = 0 → x = 0, ±2.
    f(0) = 7; f(2) = 16 − 32 + 7 = −9; f(−2) = −9.
    Sign: f′(−3) = 4(−27)+48 = −60 < 0; f′(−1) = −4+16 = 12 > 0; f′(1) = −12 < 0; f′(3) = 60 > 0.
    x = −2: −→+ → local min (−2, −9).
    x = 0: +→− → local max (0, 7).
    x = 2: −→+ → local min (2, −9).

    Step 4 — End behaviour: Even degree, positive: y → +∞ at both ends.

    Sketch description: W-shaped quartic. Rises from top-left, local min at (−2, −9), rises to local max at (0, 7), falls to local min at (2, −9), rises to top-right. Crosses x-axis at x = −√7, −1, 1, √7.

  5. Q5 — Fencing paddock with 120 m, 3 sides

    Let x = length of each of the two sides perpendicular to the river (m). Then the side parallel to the river = 120 − 2x.

    Area A(x) = x(120 − 2x) = 120x − 2x². Domain: 0 < x < 60.

    A′(x) = 120 − 4x = 0 → x = 30 m.

    A′(10) = 80 > 0; A′(40) = −40 < 0. Sign +→−: confirmed maximum.

    Side along river = 120 − 2(30) = 60 m.

    Maximum area = 30 × 60 = 1800 m².

  6. Q6 — Open box from 20 × 20 cm sheet

    (a) After cutting squares of side x: base dimensions = (20 − 2x) × (20 − 2x), height = x.
    V = x(20 − 2x)² = x(400 − 80x + 4x²) = 400x − 80x² + 4x³ = 4x³ − 80x² + 400x. □

    (b) V′(x) = 12x² − 160x + 400 = 4(3x² − 40x + 100) = 4(3x − 10)(x − 10).
    V′ = 0 at x = 10⁄3 or x = 10. Since 0 < x < 10, the only interior critical point is x = 10⁄3 cm.
    Verify: V′(2) = 4(6−10)(2−10) = 4(−4)(−8) = 128 > 0; V′(5) = 4(15−10)(5−10) = −100 < 0. Maximum at x = 10⁄3.

    (c) V(10⁄3) = 4(10⁄3)³ − 80(10⁄3)² + 400(10⁄3)
    = 4(1000⁄27) − 80(100⁄9) + 4000⁄3
    = 4000⁄27 − 8000⁄9 + 4000⁄3
    = 4000⁄27 − 24000⁄27 + 36000⁄27
    = 16000⁄27 ≈ 592.6 cm³.

  7. Q7 — f(x) = x³ + ax² − 9x + b, stationary point at (1, −5)

    f′(x) = 3x² + 2ax − 9.

    Condition 1 — f′(1) = 0:
    3 + 2a − 9 = 0 → 2a = 6 → a = 3.

    Condition 2 — f(1) = −5:
    1 + 3 − 9 + b = −5 → −5 + b = −5 → b = 0.

    So f(x) = x³ + 3x² − 9x and f′(x) = 3x² + 6x − 9 = 3(x² + 2x − 3) = 3(x + 3)(x − 1).

    (b) Other stationary point: f′ = 0 also at x = −3. f(−3) = −27 + 27 + 27 = 27. Other stationary point: (−3, 27).

    (c) Sign: f′(−4) = 3(5)(−3) < 0 (−); f′(0) = 3(−3)(1) < 0... wait: f′(0) = 3(3)(−1) = −9 < 0 (−); f′(2) = 3(5)(1) = 15 > 0 (+). And for x = −3: test x = −4: f′(−4) = 3(1)(−5) = −15 < 0 (−); test x = −2: f′(−2) = 3(1)(−3) = −9 < 0... Hmm, re-check sign chart.
    f′(x) = 3(x+3)(x−1). Zeros at x = −3 and x = 1.
    x < −3: test x = −4: (−1)(−5) = 5 > 0 (+).
    −3 < x < 1: test x = 0: (3)(−1) = −3 < 0 (−).
    x > 1: test x = 2: (5)(1) = 5 > 0 (+).
    x = −3: +→− → local maximum at (−3, 27).
    x = 1: −→+ → local minimum at (1, −5).

  8. Q8 — Profit P(x) = −x³ + 15x² − 48x − 10 for 0 ≤ x ≤ 10

    P′(x) = −3x² + 30x − 48 = −3(x² − 10x + 16) = −3(x − 2)(x − 8).

    (a) P′ = 0 at x = 2 and x = 8 (both in [0, 10]).

    P(2) = −8 + 60 − 96 − 10 = −54 (thousand dollars — a loss).
    P(8) = −512 + 960 − 384 − 10 = 54 (thousand dollars).

    (b) Sign: P′(1) = −3(−1)(−7) = −21 < 0; P′(5) = −3(3)(−3) = 27 > 0; P′(9) = −3(7)(1) = −21 < 0.
    x = 2: −→+ (but P(2) is a minimum). x = 8: +→− → local maximum.

    (c) Endpoints: P(0) = −10. P(10) = −1000 + 1500 − 480 − 10 = 10.
    Compare all candidates: P(0) = −10; P(2) = −54; P(8) = 54; P(10) = 10.
    Maximum profit = $54 000 at x = 8 hundred items (800 items).

  9. Q9 — Cylinder: volume = 100π cm³, minimise surface area

    (a) Volume of cylinder: V = πr²h = 100π → r²h = 100 → h = 100⁄r².

    (b) Surface area (closed cylinder): S = 2πr² + 2πrh = 2πr² + 2πr · (100⁄r²) = 2πr² + 200π⁄r. □

    (c) dS⁄dr = 4πr − 200π⁄r². Set to zero: 4πr = 200π⁄r² → 4r³ = 200 → r³ = 50 → r = ³√50 = 5³√(2⁄5) ≈ 3.68 cm.

    Verify minimum: dS⁄dr changes −→+ at r = ³√50 (check r = 2: dS/dr = 8π − 50π < 0; r = 5: dS/dr = 20π − 8π > 0). ✓

    Minimum S = 2π(50)2/3 + 200π⁄501/3 = 2π·502/3 + 4π·502/3 = 6π·502/3 ≈ 254.5 cm².

  10. Q10 — Two numbers summing to 12, minimise sum of squares

    Let the two numbers be x and 12 − x (with x > 0 and 12 − x > 0, so 0 < x < 12).

    Sum of squares: S = x² + (12 − x)² = x² + 144 − 24x + x² = 2x² − 24x + 144.

    dS⁄dx = 4x − 24 = 0 → x = 6.

    d²S⁄dx² = 4 > 0 (or sign test: dS/dx(3) = −12 < 0; dS/dx(9) = 12 > 0 → minimum).

    The two numbers are both 6. Minimum sum of squares = 2(36) = 72.