Mathematics Advanced • Year 11 • Module 3 • Lesson 11
Increasing and Decreasing Functions
HSC-style writing on sign analysis of f′(x) — including an extended response that combines sign tables, parameters, and curve sketching.
1. Short-answer questions
1.1 Find all intervals on which f(x) = x³ − 6x² + 9x + 1 is increasing. 3 marks Band 3-4
1.2 The derivative of a function is f′(x) = (x − 1)(x + 2)². Determine where f is increasing and where it is decreasing. State, with reason, what happens to f at x = −2. 3 marks Band 4
1.3 Find all values of k for which f(x) = x³ + kx² + 3x is increasing for every real x. 4 marks Band 4-5
Stuck on 1.3? f′(x) is a quadratic in x with positive leading coefficient — for f′ ≥ 0 everywhere, you need the discriminant ≤ 0.2. Extended response
2.1 Let f(x) = x⁴ − 4x³.
(a) Find f′(x) and factorise.
(b) Hence find all stationary points and determine the intervals on which f is increasing and decreasing. Present your sign analysis in a clearly labelled sign table.
(c) Show that x = 0 is a stationary point but NOT a turning point, and identify what kind of point it is. Justify by referring to your sign table.
(d) Sketch the curve y = f(x), showing the stationary points, intervals of increase/decrease, and the behaviour as x → ±∞. Mark the x-intercepts. 8 marks Band 5-6
Explicit marking criteria
Part (a) — 1 mark
• 1 mark — correctly differentiates and factorises: f′(x) = 4x³ − 12x² = 4x²(x − 3).
Part (b) — 3 marks
• 1 mark — correctly identifies stationary points at x = 0 and x = 3.
• 1 mark — correct sign table with test points in each region and signs of each factor (4x² ≥ 0, (x − 3)).
• 1 mark — correct intervals: f is decreasing on (−∞, 0) ∪ (0, 3) and increasing on (3, ∞), using open intervals.
Part (c) — 2 marks
• 1 mark — observes that f′ does not change sign at x = 0 (negative on both sides because 4x² ≥ 0 and (x − 3) is negative).
• 1 mark — concludes x = 0 is a stationary point of inflection (horizontal inflection), not a turning point.
Part (d) — 2 marks
• 1 mark — sketch shows the horizontal inflection at the origin (curve flattens then continues to fall), the minimum turning point at x = 3, and x-intercepts at x = 0 and x = 4.
• 1 mark — correct end behaviour (f(x) → ∞ as x → ±∞) and overall shape consistent with sign analysis.
Your response:
Stuck on (c)? Look at the sign of f′(x) just to the left and just to the right of x = 0. If the signs are the same, it is NOT a turning point.How did this worksheet feel?
What I'll revisit before next class:
1.1 — Intervals where f(x) = x³ − 6x² + 9x + 1 is increasing (3 marks)
Sample response. f′(x) = 3x² − 12x + 9 = 3(x² − 4x + 3) = 3(x − 1)(x − 3). f′(x) = 0 at x = 1 and x = 3. Test: f′(0) = 9 (+); f′(2) = 3(1)(−1) = −3 (−); f′(4) = 3(3)(1) = 9 (+). Hence f is increasing on (−∞, 1) ∪ (3, ∞).
Marking notes. 1 mark — correct derivative AND factorisation. 1 mark — correct stationary points and a sign test in each region. 1 mark — correct intervals stated as open intervals using union notation. Common error: writing [−∞, 1] ∪ [3, ∞) (square brackets); also forgetting the union and writing two separate inequalities loses 0.5.
1.2 — Sign analysis of f′(x) = (x − 1)(x + 2)² (3 marks)
Sample response. Stationary points where f′ = 0: x = 1 and x = −2. Since (x + 2)² ≥ 0 always, the sign of f′(x) matches the sign of (x − 1). For x < 1: f′ ≤ 0; for x > 1: f′ > 0. So f is decreasing on (−∞, 1) and increasing on (1, ∞). At x = −2: f′(−2) = 0 but the squared factor does not change sign and (x − 1) is negative either side of x = −2, so f′ does not change sign there — x = −2 is a stationary point of inflection, not a turning point.
Marking notes. 1 mark — identifies both stationary points AND uses the squared factor argument. 1 mark — correct intervals of increase/decrease. 1 mark — correctly classifies x = −2 as a stationary point of inflection. Common errors: claiming x = −2 is a min or max (no sign change!); ignoring the (x + 2)² and writing inverted intervals.
1.3 — Values of k making x³ + kx² + 3x everywhere increasing (4 marks)
Sample response. f′(x) = 3x² + 2kx + 3. For f to be (weakly) increasing for every real x we need f′(x) ≥ 0 for all x. Since f′ is a quadratic opening upward (leading coefficient 3 > 0), this requires no two distinct real roots, i.e. discriminant ≤ 0. Δ = (2k)² − 4(3)(3) = 4k² − 36 ≤ 0 ⇒ k² ≤ 9 ⇒ −3 ≤ k ≤ 3.
(For strictly increasing you would need Δ < 0, i.e. −3 < k < 3; at k = ±3 the cubic has a horizontal point of inflection rather than a turning point and is still increasing in the weak sense.)
Marking notes. 1 mark — correct derivative. 1 mark — recognises "everywhere increasing ⇒ f′ ≥ 0 ⇒ discriminant ≤ 0" using the positive leading coefficient. 1 mark — correctly computes Δ. 1 mark — correctly states final inequality. Common errors: using Δ < 0 only (losing the endpoints); reversing the inequality; ignoring the leading-coefficient direction.
2.1 — Extended response (8 marks): sample Band-6 response with annotations
Sample Band-6 response.
Part (a). f(x) = x⁴ − 4x³ ⇒ f′(x) = 4x³ − 12x² = 4x²(x − 3). [1 mark — differentiation and factorisation.]
Part (b). Stationary points where f′(x) = 0: from 4x²(x − 3) = 0 we get x = 0 (double root) and x = 3. [1 mark — stationary points.]
Sign table for f′(x) = 4x² · (x − 3):
| Region | Test x | 4x² | (x − 3) | f′(x) | f is… |
|---|---|---|---|---|---|
| x < 0 | x = −1 | + (= 4) | − (= −4) | − (= −16) | decreasing |
| 0 < x < 3 | x = 1 | + (= 4) | − (= −2) | − (= −8) | decreasing |
| x > 3 | x = 4 | + (= 64) | + (= 1) | + (= 64) | increasing |
[1 mark — correct sign table with test points.] Hence f is decreasing on (−∞, 0) ∪ (0, 3) and increasing on (3, ∞). [1 mark — correct intervals using open intervals.]
Part (c). From the sign table, f′(x) is negative both immediately to the left of x = 0 (test x = −1: f′ = −16) and immediately to the right of x = 0 (test x = 1: f′ = −8). Because f′ does not change sign at x = 0, the point is not a turning point. The function is momentarily level at x = 0 (since f′(0) = 0) but continues to decrease through it, so x = 0 is a stationary (horizontal) point of inflection. [1 mark — no sign change observation; 1 mark — correctly classified as horizontal point of inflection.]
Part (d). f(x) = x⁴ − 4x³ = x³(x − 4), so x-intercepts at x = 0 (triple root) and x = 4. f(0) = 0; f(3) = 81 − 108 = −27 (minimum value); as x → ±∞, f(x) → ∞ because the leading term is x⁴ with positive coefficient. Sketch: descending from +∞ on the far left, flattens through the origin without crossing direction (horizontal inflection), continues descending to a minimum at (3, −27), then rises back through the x-intercept at x = 4 and continues to +∞.
+∞ \ / +∞
\____ (0,0) ────_ /
\___(3,-27)__/ passes (4, 0)
[1 mark — sketch shows horizontal inflection at origin, minimum at (3, −27), correct x-intercepts; 1 mark — correct end behaviour and overall shape consistent with sign analysis.]
Total: 8/8.
Band descriptors for marker.
Band 3: Differentiates and identifies stationary points but errs on the sign table, e.g. claims x = 0 is a minimum because f′(0) = 0. Sketch is rough and inconsistent. ≈ 3-4 marks.
Band 4: Correctly identifies sign of f′ in each interval and concludes the intervals of increase/decrease. Misses the horizontal inflection at x = 0 or includes endpoints in intervals (uses [ ] instead of ( )). ≈ 5-6 marks.
Band 5: Completes (a)-(c). Sketch lacks the horizontal-tangent character at the origin, or misses the x-intercept at x = 4. ≈ 6-7 marks.
Band 6: All four parts complete, sign table presented with test points and signs of each factor, x = 0 explicitly identified as a horizontal point of inflection, sketch correctly shows horizontal inflection at origin, minimum at (3, −27), and correct end behaviour. 8/8.