Mathematics Advanced • Year 11 • Module 3 • Lesson 11
Increasing and Decreasing Functions
Build procedural fluency in sign analysis — using f′(x) to decide where a function rises, falls, or is momentarily level.
1. Quick recall
Answer each question in the space provided. 1 mark each
Q1.1 Complete each statement with one of: increasing / decreasing / stationary.
(a) If f′(x) > 0 on an interval, f is ____________ on that interval.
(b) If f′(x) < 0 on an interval, f is ____________ on that interval.
(c) A point where f′(x) = 0 is called a ____________ point.
Q1.2 True or false (circle one):
(a) A function can be both increasing and decreasing on the same open interval. T / F
(b) Intervals of increase are written as open intervals (e.g. (2, ∞) not [2, ∞)). T / F
(c) Every point where f′(x) = 0 is a turning point. T / F
Q1.3 For f(x) = x² − 6x, find f′(x) and the value of x at the stationary point.
f′(x) = ____________ ; stationary point at x = ____________ .
2. Worked example — where is f(x) = x² − 4x + 3 increasing/decreasing?
Follow each line of algebra. Every step has a reason on the right.
Step 1 — Differentiate.
f′(x) = 2x − 4
Reason: apply the power rule to each term.
Step 2 — Solve f′(x) = 0.
2x − 4 = 0 ⇒ x = 2
Reason: stationary points separate intervals of increase from intervals of decrease.
Step 3 — Test signs in each interval.
For x < 2: f′(1) = 2(1) − 4 = −2 < 0, so f is decreasing on (−∞, 2).
For x > 2: f′(3) = 2(3) − 4 = 2 > 0, so f is increasing on (2, ∞).
Reason: pick any convenient test value in each interval; the sign of f′ is the same throughout the interval.
Conclusion. f is decreasing on (−∞, 2) and increasing on (2, ∞). x = 2 is a minimum turning point.
3. Faded example — fill in the missing steps
Find the intervals on which f(x) = x³ − 3x is increasing. 3 marks
Step 1 — Differentiate and factorise:
f′(x) = ____________ = 3(x − ____)(x + ____)
Step 2 — Solve f′(x) = 0:
x = ____________ or x = ____________ .
Step 3 — Sign-test in each region:
x < −1: f′(−2) = ____________ (sign ____)
−1 < x < 1: f′(0) = ____________ (sign ____)
x > 1: f′(2) = ____________ (sign ____)
Conclusion. f is increasing on ____________ ∪ ____________ .
4. Graduated practice — where is f increasing/decreasing?
Show f′(x), the stationary points, and a one-line sign test for each interval.
Foundation — clean quadratics (4 questions)
| Q | Function | f′(x) | Increasing on |
|---|---|---|---|
| 4.1 1 | f(x) = x² − 6x + 5 | ||
| 4.2 1 | f(x) = x² + 4x − 1 | ||
| 4.3 1 | f(x) = −x² + 8x | ||
| 4.4 1 | f(x) = 3x² − 12x + 7 |
Standard — cubics and reciprocals (6 questions)
State the stationary points, then the intervals of increase AND decrease.
4.5 f(x) = x³ − 12x. Find all intervals of increase and decrease. 2 marks
4.6 f(x) = 2x³ − 9x² + 12x. Find all intervals of increase. 2 marks
4.7 f(x) = x³ − 3x² + 3x − 1. Show that f is increasing for all real x. 2 marks
4.8 f(x) = 1/x (x ≠ 0). State where f is decreasing. 2 marks
4.9 f(x) = x⁴ − 4x³. Find all intervals of increase and decrease. 2 marks
4.10 f(x) = x + 4/x (x > 0). Find where f is decreasing on (0, ∞). 2 marks
Extension — parameters and squared factors (2 questions)
4.11 The derivative of a function is f′(x) = (x − 4)(x + 1)². Identify all stationary points and state on what intervals f is increasing. Then explain why x = −1 is not a turning point even though f′(−1) = 0. 3 marks
4.12 Find the values of k for which f(x) = x³ + kx² + 3x is increasing for every real x. 3 marks
5. Self-check the easy 3
Tick the first three once you have checked your method works.
How did this worksheet feel?
What I'll revisit before next class:
Q1.1 — Sign rules
(a) increasing. (b) decreasing. (c) stationary.
Q1.2 — True/false
(a) F — direction is unique on any interval where f′ has constant sign. (b) T — at the stationary point itself, f is neither increasing nor decreasing. (c) F — e.g. f′(x) = x² has f′(0) = 0 but no sign change, so x = 0 is a horizontal point of inflection, not a turning point.
Q1.3 — f(x) = x² − 6x
f′(x) = 2x − 6; stationary at x = 3.
Q3 — Faded example f(x) = x³ − 3x
f′(x) = 3x² − 3 = 3(x − 1)(x + 1). x = −1 or 1.
Test points: f′(−2) = 3(4) − 3 = 9 (+); f′(0) = −3 (−); f′(2) = 9 (+).
f is increasing on (−∞, −1) ∪ (1, ∞).
Q4.1 — f(x) = x² − 6x + 5
f′ = 2x − 6 = 0 at x = 3. f′(0) = −6 (decr); f′(4) = 2 (incr). Increasing on (3, ∞).
Q4.2 — f(x) = x² + 4x − 1
f′ = 2x + 4 = 0 at x = −2. f′(−3) = −2 (decr); f′(0) = 4 (incr). Increasing on (−2, ∞).
Q4.3 — f(x) = −x² + 8x
f′ = −2x + 8 = 0 at x = 4. f′(0) = 8 (incr); f′(5) = −2 (decr). Increasing on (−∞, 4).
Q4.4 — f(x) = 3x² − 12x + 7
f′ = 6x − 12 = 0 at x = 2. f′(0) = −12 (decr); f′(3) = 6 (incr). Increasing on (2, ∞).
Q4.5 — f(x) = x³ − 12x
f′ = 3x² − 12 = 3(x − 2)(x + 2) = 0 at x = ±2. f′(−3) = 15 (+); f′(0) = −12 (−); f′(3) = 15 (+). Increasing on (−∞, −2) ∪ (2, ∞); decreasing on (−2, 2).
Q4.6 — f(x) = 2x³ − 9x² + 12x
f′ = 6x² − 18x + 12 = 6(x − 1)(x − 2) = 0 at x = 1, 2. f′(0) = 12 (+); f′(1.5) = −1.5 (−); f′(3) = 12 (+). Increasing on (−∞, 1) ∪ (2, ∞).
Q4.7 — f(x) = x³ − 3x² + 3x − 1
f′ = 3x² − 6x + 3 = 3(x − 1)². Since (x − 1)² ≥ 0 for all real x, f′(x) ≥ 0 everywhere, with equality only at x = 1. Hence f is (weakly) increasing for all real x; the single stationary point at x = 1 is a horizontal point of inflection, not a turning point.
Q4.8 — f(x) = 1/x
f′(x) = −1/x², which is negative for every x ≠ 0. Hence f is decreasing on (−∞, 0) and on (0, ∞) (but not on the whole real line because of the asymptote — you must state the two open intervals separately).
Q4.9 — f(x) = x⁴ − 4x³
f′ = 4x³ − 12x² = 4x²(x − 3) = 0 at x = 0 and x = 3. Since 4x² ≥ 0, the sign of f′ matches (x − 3): for x < 3, f′ ≤ 0; for x > 3, f′ > 0. Decreasing on (−∞, 3); increasing on (3, ∞). (x = 0 is a stationary point of inflection.)
Q4.10 — f(x) = x + 4/x for x > 0
f′(x) = 1 − 4/x² = 0 when x² = 4, so x = 2 (taking x > 0). f′(1) = 1 − 4 = −3 (−); f′(3) = 1 − 4/9 > 0 (+). Decreasing on (0, 2).
Q4.11 — f′(x) = (x − 4)(x + 1)²
Stationary at x = 4 and x = −1. Since (x + 1)² ≥ 0 always, sign of f′ matches (x − 4): negative for x < 4, positive for x > 4. So f is increasing on (4, ∞) only. At x = −1, f′ = 0 but the squared factor does not switch sign, so f′ does not change sign — x = −1 is a stationary point of inflection, not a turning point.
Q4.12 — k such that x³ + kx² + 3x is everywhere increasing
f′(x) = 3x² + 2kx + 3. For f to be increasing for all x we need f′(x) > 0 for all x. f′ is an upward-opening quadratic (leading coefficient 3 > 0), so we need no real roots: discriminant < 0. Δ = (2k)² − 4(3)(3) = 4k² − 36 < 0 ⇒ k² < 9 ⇒ −3 < k < 3.