Key Idea: The sign of the derivative tells you which way a curve is heading. Questions ask where is f increasing / decreasing? or read this graph of f′ — pure non-calculator work on Paper 1.
📈 The sign of f′ decides everything
- the gradient (derivative) at that x
- stationary — the curve is momentarily flat
| Sign of f′(x) | What the curve does | Picture |
|---|---|---|
| f′(x) > 0 (positive) | increasing — going uphill | ↗ |
| f′(x) < 0 (negative) | decreasing — going downhill | ↘ |
| f′(x) = 0 | stationary — flat (max, min or inflexion) | → |
To test a point, find f′ there and check its sign — not its size. f′(2) = −7 and f′(2) = −0.1 both just mean decreasing at x = 2.
🧭 Finding the intervals
| Want to find… | What to do |
|---|---|
| Where f is increasing | Differentiate, then solve f′(x) > 0. |
| Where f is decreasing | Differentiate, then solve f′(x) < 0. |
| The boundaries | Solve f′(x) = 0, then test the sign of f′ in each region between them. |
👀 Reading a graph of f′
| On the graph of f′ … | … means for f |
|---|---|
| f′ is above the x-axis | f is increasing |
| f′ is below the x-axis | f is decreasing |
| f′ crosses zero + → − | local maximum of f |
| f′ crosses zero − → + | local minimum of f |
✏️ IB-style worked examples
IB-style question — increasing or decreasing at a point?
For f(x) = x² − 6x, the gradient function is f′(x) = 2x − 6. State whether f is increasing or decreasing at x = 1 and at x = 4.
Step by step:
Evaluate f′ at each point.
Read the signs.
Decreasing at x = 1; increasing at x = 4.
IB-style question — find where a function is increasing
Find the values of x for which f(x) = x² − 8x + 3 is increasing.
Step by step:
Differentiate, then set f′(x) > 0.
Solve the inequality.
Increasing for x > 4 (and decreasing for x < 4); the boundary x = 4 is the vertex.
IB-style question — intervals from a cubic
For f(x) = x³ − 12x, the gradient function is f′(x) = 3x² − 12. Find where f is increasing and where it is decreasing.
Step by step:
Stationary points: solve f′(x) = 0.
Test the sign of f′ in each region.
Increasing for x < −2 and x > 2; decreasing for −2 < x < 2.
IB-style question — read the graph of f′
The graph of f′ crosses the x-axis at x = 3, going from positive to negative. Explain why f has a local maximum at x = 3.
Step by step:
Left of 3, f′ > 0 → f increasing; right of 3, f′ < 0 → f decreasing.
Increasing then decreasing makes a peak.
f changes from increasing to decreasing at x = 3 (f′ goes + → −), so there is a local maximum there.
Important: Solving f′(x) = 0 only gives the boundaries, not the answer. You must test the sign of f′ between them — a cubic's f′ can be + − +, so f is increasing, then decreasing, then increasing again. Don't assume one side is increasing just because the other is.
Tap each card to reveal the answer.
f′(5) = −2. Is f increasing or decreasing at x = 5? Decreasing — f′ is negative, and only the sign matters.
f(x) = x² − 10x. Where is f decreasing? x < 5 — f′ = 2x − 10 < 0 gives x < 5.
What do you solve to find the boundaries of the intervals? f′(x) = 0 — the stationary points separate increasing from decreasing.
On a graph of f′, what does 'below the x-axis' mean for f? f is decreasing there (f′ < 0).
f′ crosses zero going − → +. Max or min of f? Local minimum — f goes decreasing → increasing.
Exam tips
- Increasing where f′(x) > 0; decreasing where f′(x) < 0; stationary where f′(x) = 0.
- To find intervals: differentiate, solve f′(x) = 0 for the boundaries, then test the sign of f′ in each region.
- Only the sign of f′ matters — never its size.
- Reading a graph of f′: above the axis = increasing, below = decreasing.
- A + → − crossing of f′ is a maximum of f; a − → + crossing is a minimum.