The core rule: • f′(x) > 0 at a point → the function is increasing there (rising from left to right) • f′(x) < 0 at a point → the function is decreasing there (falling from left to right) • f′(x) = 0 at a point → the function is stationary there (flat, neither rising nor falling)
Increasing where the curve rises (f′ > 0), decreasing where it falls (f′ < 0); the turning points are where f′ = 0.
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This is not a new calculation — it is simply reading the sign of the derivative you already know how to find.
| Gradient f′(x) | What the curve does | Picture |
|---|---|---|
| Positive (e.g. 3) | Rising — going uphill | ↗ |
| Zero | Flat — peak or valley | → |
| Negative (e.g. −5) | Falling — going downhill | ↘ |
Example
Step by step
- Given — f(x) = x² − 4x
- Differentiate — f′(x) = 2x − 4
- At x = 5 — f′(5) = 10 − 4 = 6 > 0 → increasing at x = 5
- At x = 1 — f′(1) = 2 − 4 = −2 < 0 → decreasing at x = 1
- At x = 2 — f′(2) = 4 − 4 = 0 → stationary at x = 2
Common mistake: Don't confuse where a function is large with where it is increasing.
A function can be at a high value but still be decreasing (e.g. a ball at 40 m but falling).
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To find where f is increasing or decreasing over a range, you need to know where f′(x) = 0 (the crossover points), then test the sign of f′ in each region.
Method
- Find f′(x).
- Set f′(x) = 0 and solve to get the critical x-values.
- Draw a number line divided by those critical values.
- Pick one test value in each region and check the sign of f′.
- State the interval(s) where f′ > 0 (increasing) and f′ < 0 (decreasing).
Full example
Step by step
- f(x) = x³ − 3x
- Step 1 — Differentiate — f′(x) = 3x² − 3
- Step 2 — Set = 0 — 3x² − 3 = 0 → x² = 1 → x = ±1
- Step 3 — Test x = −2 (region x < −1) — f′(−2) = 12 − 3 = 9 > 0 → increasing
- Step 4 — Test x = 0 (region −1 < x < 1) — f′(0) = −3 < 0 → decreasing
- Step 5 — Test x = 2 (region x > 1) — f′(2) = 12 − 3 = 9 > 0 → increasing
- Conclusion — Increasing for x < −1 and x > 1. Decreasing for −1 < x < 1.
Interval notation: IB accepts either inequality notation (x < −1) or interval notation (−∞, −1).
Use whichever you find clearer.
Always include the direction — 'increasing' or 'decreasing'.
IB-style question — where is a 1/x function increasing?
A function is f(x) = ½x² + 8⁄x, for x ≠ 0.
Find the values of x for which f is increasing.
Step by step
- Rewrite 8⁄x as 8x⁻¹ so you can differentiate it.
- f increases where f′(x) > 0. Combine over x² (which is always positive, x ≠ 0), so the sign is the top's sign.
Final answer
f is increasing for x > 2.
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A sign diagram is a quick visual tool that shows the sign of f′(x) across the x-axis.
It replaces writing several sentences and is expected in many IB solutions.
How to draw a sign diagram: 1.
Draw a horizontal line (this represents the x-axis). 2.
Mark the x-values where f′(x) = 0. 3.
Write + or − in each region based on a test value. 4.
Under the x-axis, write ↗ for + and ↘ for −.
Sign diagram for f(x) = x³ − 3x
Step by step
- Critical values — x = −1 and x = 1
- Diagram — −1 1 −−−−−−+−−−|−−−−−−−|−−−+−−−−− f′: + − + ↗ ↘ ↗
This single diagram immediately shows: increasing, decreasing, increasing.
Sign diagrams at endpoints: If the domain is restricted (e.g. 0 ≤ x ≤ 4), only consider the sign within that domain.
Ignore what happens outside the given interval.
In IB exam questions, increasing/decreasing analysis often appears inside a real-world context.
The language changes but the maths is identical.
| Context word | Mathematical meaning | Test |
|---|---|---|
| Revenue is growing | R′(t) > 0 | Check sign of R′ |
| Temperature is falling | T′(t) < 0 | Check sign of T′ |
| Population is stable | P′(t) = 0 | f′ = 0 |
| Speed is increasing | v′(t) > 0 | Check sign of v′ |
Contextual example
Step by step
- Model — P(t) = −t³ + 6t² + 15 is a company's profit (thousands) for t years, 0 ≤ t ≤ 5.
- Differentiate — P′(t) = −3t² + 12t = −3t(t − 4)
- Critical values — t = 0 and t = 4
- Sign test at t = 2 — P′(2) = −12 + 24 = 12 > 0 → profit increasing
- Sign test at t = 5 — P′(5) = −75 + 60 = −15 < 0 → profit decreasing
- Answer — Profit is increasing for 0 < t < 4. Profit is decreasing for 4 < t ≤ 5.
IB exam language: If asked 'when is profit increasing?', answer with an interval AND a direction word: 'Profit is increasing for 0 < t < 4.' Just writing '0 < t < 4' with no context word may lose a mark.