Practice Maths

Solutions — The Derivative as a Function

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  1. Q1 — Sign of f′(x)

    (a) f′(3) = 5 > 0 → f is increasing at x = 3.

    (b) f′(−1) = −2 < 0 → f is decreasing at x = −1.

    (c) f′(0) = 0 → f has a stationary point at x = 0 (tangent is horizontal).

    (d) f′(4) = −7 < 0 → f is decreasing at x = 4.

  2. Q2 — Stationary points from f′

    (a) 2x − 6 = 0 → x = 3

    (b) 3x² − 12 = 0 → x² = 4 → x = ±2

    (c) x² − x − 6 = (x − 3)(x + 2) = 0 → x = 3 or x = −2

    (d) 4x(x − 3) = 0 → x = 0 or x = 3

  3. Q3 — f(x) = x² − 4x + 1

    (a) f′(x) = 2x − 4

    (b) 2x − 4 = 0 → x = 2 (stationary point)

    (c) f′(x) > 0 when 2x − 4 > 0 → x > 2 → f is increasing on (2, ∞)

    (d) f′(x) < 0 when x < 2 → f is decreasing on (−∞, 2)

  4. Q4 — Sign chart: f′(x) = (x − 1)(x + 3)

    Zeros: x = −3 and x = 1.

    Test x = −4: f′(−4) = (−5)(−1) = 5 > 0 → increasing on (−∞, −3)

    Test x = 0: f′(0) = (−1)(3) = −3 < 0 → decreasing on (−3, 1)

    Test x = 2: f′(2) = (1)(5) = 5 > 0 → increasing on (1, ∞)

    Summary: Increasing on (−∞, −3) ∪ (1, ∞); decreasing on (−3, 1). Stationary points at x = −3 and x = 1.

  5. Q5 — f(x) = x³ − 3x² − 9x + 2

    (a) f′(x) = 3x² − 6x − 9 = 3(x² − 2x − 3) = 3(x − 3)(x + 1)

    (b) 3(x − 3)(x + 1) = 0 → x = 3 or x = −1

    (c) Sign chart: test x = −2: 3(−5)(−1) = 15 > 0; test x = 0: 3(−3)(1) = −9 < 0; test x = 4: 3(1)(5) = 15 > 0.

    Increasing on (−∞, −1) ∪ (3, ∞); decreasing on (−1, 3).

  6. Q6 — Sketching f′ from turning points

    (a) Turning points of f at x = −2 and x = 3 → f′(−2) = 0 and f′(3) = 0. x-intercepts of f′ are x = −2 and x = 3.

    (b) f is a degree 3 polynomial, so f′ is degree 2 — a parabola. Since f increases on (−∞, −2) and (3, ∞), f′ > 0 outside the roots → parabola opening upward.

    (c) Sketch: upward parabola with x-intercepts at x = −2 and x = 3, vertex approximately at x = 0.5, below x-axis between −2 and 3.

  7. Q7 — Matching f to derivative description

    (a) f(x) = 3x + 1 → f′(x) = 3 (constant) → matches (i) horizontal line at y = 3

    (b) f(x) = x² − 2x → f′(x) = 2x − 2 (line, zero at x = 1) → matches (iii) line with x-intercept at x = 1

    (c) f(x) = x³ − 3x → f′(x) = 3x² − 3 = 3(x − 1)(x + 1) (parabola, zeros at ±1, opens up) → matches (ii) upward parabola crossing x-axis at −1 and 1

  8. Q8 — f′(x) = −2x + 4

    (a) −2x + 4 = 0 → x = 2

    (b) f′ > 0 when −2x + 4 > 0 → x < 2: increasing on (−∞, 2). f′ < 0 when x > 2: decreasing on (2, ∞).

    (c) f′(0) = 4 (gradient = 4 at x = 0); f′(5) = −10 + 4 = −6 (gradient = −6 at x = 5).

    (d) f′ is linear (degree 1) so f is a quadratic (degree 2). Since f increases then decreases, f has a local maximum at x = 2. It opens downward.

  9. Q9 — Sketching f′ from described f

    (a) f′(x) crosses x-axis at x = −1 and x = 2. f′ > 0 for x < −1 (f increasing) and x > 2 (f increasing). f′ < 0 for −1 < x < 2 (f decreasing). Sketch: upward parabola with roots at x = −1 and x = 2.

    (b) The steepest part of f is around the midpoint of the decreasing interval, approximately x = 0.5 (vertex of f′ parabola, which is the minimum of f′).

    (c) f′(x) > 0 for x ∈ (−∞, −1) ∪ (2, ∞).

  10. Q10 — Water level application

    (a) W′(t) = −3t² + 12t

    (b) W′ is a downward parabola: W′′(t) = −6t + 12 = 0 → t = 2. W′(2) = −12 + 24 = 12 L/min. Rising fastest at t = 2 minutes.

    (c) W′(t) = −3t(t − 4). Zeros at t = 0 and t = 4. W′ > 0 for 0 < t < 4 (rising); W′ < 0 for 4 < t ≤ 6 (falling).

    (d) When W′(t) = 0, the water level is momentarily neither rising nor falling — it is at a maximum level (at t = 4 min).