SL 5.2 — Increasing & decreasing functions

Core Concept	IB AI HL Definition & Application
Increasing function	For f increasing on interval I, if x₁ < x₂ in I, then f(x₁) ≤ f(x₂) (strictly increasing if <). AI HL requires analytical proof using derivatives for real-world modelling.
Decreasing function	For f decreasing on I, if x₁ < x₂ in I, then f(x₁) ≥ f(x₂) (strictly if >). Essential for optimization problems in applications.
Derivative test	f'(x) > 0 ⇒ increasing; f'(x) < 0 ⇒ decreasing; f'(x) = 0 ⇒ stationary point. Sign analysis drives decision-making in HL applications.
Critical points	Solutions to f'(x) = 0 or points where f'(x) undefined. These identify optimization opportunities.

📌 1. Formal Definitions & Real-World Context

Precise definition: f is increasing on (a,b) if ∀ x₁, x₂ ∈ (a,b) with x₁ < x₂, then f(x₁) ≤ f(x₂)
Strictly increasing: f(x₁) < f(x₂) — no horizontal segments
AI HL focus: Behaviour analysis for profit maximization, population growth modelling
Interval notation: Always specify: “increasing on (-∞, 2) ∪ (4, ∞)”
Modelling principle: Real functions change monotonicity only at critical points

🧠 Examiner Tip

Always write “since f'(x) > 0 for all x in interval…“
HL marks: (1) correct derivative, (2) critical points, (3) complete sign chart, (4) real-world interpretation
Never rely on “graph shows increasing” — zero application marks
Test one value per interval and state sign explicitly

📌 2. First Derivative Test — HL Procedure

Differentiate: Compute f'(x) with chain/quotient/product rules
Solve: f'(x) = 0 → identify all critical points
Partition: Domain using critical points + domain restrictions
Test: Evaluate f'(x) sign in each subinterval
Interpret: Link to optimization (max profit when f’ changes – to +)

🌍 Real-World Connection

Business: Revenue R(q) increasing when marginal revenue MR(q) > 0
Environment: Pollution P(t) increasing during industrial growth phase
Finance: Investment growth when rate of return > 0
Medicine: Drug concentration C(t) increases then decreases in bloodstream

📌 3. HL Applications & Complex Cases

Optimization: Local max (profit peak) when f’ changes + to –
Constant functions: f'(x) = 0 everywhere → neither strictly increasing nor decreasing
Discontinuities: Vertical asymptotes partition analysis intervals
Parametric: For r(t) = (x(t), y(t)), analyse dx/dt and dy/dt signs
Modelling validation: Check if predicted intervals match real data trends

📐 IA Spotlight

HL RQ: “Using calculus, optimize [real company] production for maximum profit”
Data modelling: Fit cubic/quartic to actual revenue data, find critical points
Evaluation: Compare predicted optimum with industry reports
HL strength: Multiple critical points + second derivative confirmation
Visuals: Sign diagrams + actual vs predicted revenue graphs

🔍 TOK Perspective

Does mathematical certainty about monotonic intervals transfer to real-world predictions?
How do computational approximations affect our confidence in optimization results?
Can graphical intuition ever replace rigorous derivative analysis in applications?

📝 Paper 2 HL Strategy

10-14 marks: Complex derivative + multiple critical points + economic interpretation
Time: 18 minutes maximum
Always include sign diagram — method marks even if arithmetic errors
State context: “Profit increases when MR > 0, i.e. on (0, 150) units”

📱 GDC Tips

TI-Nspire CX II: Calculus > Derivative > f'(x) > solve(f'(x)=0,x)
Casio fx-CG50: RUN-MATRIX > CALC > d/dx > solve f'(x)=0
Sign testing: Evaluate f'(test value) directly on calculator
Table mode: Generate f'(x) table across critical points

📌 HL PRACTICE QUESTIONS

Multiple Choice Questions

Q1. If f'(x) = 2(x-1)(x+3), on which interval is f(x) decreasing?

A. (-∞, -3) B. (-3, 1) C. (1, ∞) D. (-3, ∞)

Answer: B
Critical points: x = -3, 1. Test x = 0: f'(0) = 2(-1)(3) = -6 < 0 so decreasing on (-3, 1).

Q2. f has f'(3) = 0 and f'(x) > 0 for x ≠ 3. Then x=3 is:

A. Local maximum B. Local minimum C. Point of inflection D. Neither

Answer: D
No sign change across critical point → neither local max nor min (like f(x) = x³).

Q3. For f(x) = ln(x) + x, f'(x) = 1/x + 1 > 0 for all x > 0. Thus:

A. Decreasing on (0,∞) B. Increasing on (0,∞) C. Constant D. Undefined

Answer: B
f'(x) > 0 everywhere in domain → increasing on entire domain.

Q4. Number of intervals to test for f'(x) = (x+1)²(x-2):

A. 2 B. 3 C. 4 D. 1

Answer: B
Critical points x = -1 (double root), x = 2 → intervals: (-∞, -1), (-1, 2), (2, ∞).

Short Answer Questions

SAQ 1: Explain why f'(x) = 0 doesn’t guarantee a turning point. 📱

Consider f(x) = x³. Then f'(x) = 3x² = 0 at x = 0, but f'(x) > 0 for x ≠ 0, so no sign change → point of inflection.

SAQ 2: Why test intervals separately? 📱

Critical points divide domain where derivative sign may change. f'(x) can only change sign at roots, so monotonicity constant within each interval.

Long Answer Questions

LAQ 1 (14 marks): Comprehensive analysis of f(x) = x³ – 6x² + 9x

(a) Find f'(x) and all critical points [3 marks]

f'(x) = 3x² – 12x + 9 = 3(x² – 4x + 3) = 3(x-1)(x-3)
Critical points: x = 1, x = 3

(b) Complete sign chart and state monotonicity intervals [5 marks]

(-∞,1)	x=1	(1,3)	x=3	(3,∞)
+	0	–	0	+

Increasing: (-∞,1) ∪ (3,∞); Decreasing: (1,3)

(c) Classify turning points [3 marks]

x=1: + to – → Local maximum, f(1) = 4
x=3: – to + → Local minimum, f(3) = 0

(d) Business interpretation [3 marks]

If f(x) = profit, produce between x=1 and x=3 units to minimize losses (decreasing phase), optimal at x=1 (max profit).

LAQ 2 (16 marks): Analyse f(x) = (x² + 1)/(x-2) completely

(a) State domain and find f'(x) [4 marks]

Domain: x ≠ 2
f'(x) = [2x(x-2) – (x²+1)(1)]/(x-2)² = (x² – 6x – 1)/(x-2)²

(b) Solve f'(x) = 0 and identify analysis points [3 marks]

x² – 6x – 1 = 0 → x = 3 ± 2√2 ≈ x₁ = -0.83, x₂ = 6.83
Points: x ≈ -0.83, x = 2 (asymptote), x ≈ 6.83

(c) Sign analysis across all intervals [5 marks]

(-∞,-0.83)	-0.83	(-0.83,2)	2	(2,6.83)	6.83	(6.83,∞)
–	0	+	∅	–	0	+

(d) Monotonicity + optimization conclusion [4 marks]

Decreasing: (-∞,-0.83) ∪ (2,6.83)
Increasing: (-0.83,2) ∪ (6.83,∞)
Local max x≈-0.83; local min x≈6.83