The sign of f'(x) decides it: A function is increasing where its gradient is positive (f'(x) > 0) and decreasing where its gradient is negative (f'(x) < 0).
So to test a point, find f' there and check its sign.
Where the curve goes uphill (left→right) it's increasing (f′ > 0); downhill it's decreasing (f′ < 0); the turning points are where f′ = 0.
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IB-style question — test two points
For f(x) = x² − 4x, the gradient function is f'(x) = 2x − 4.
State whether f is increasing or decreasing at x = 1 and at x = 3.
Step by step
- Evaluate f' at each point.
- Read the signs.
Final answer
Decreasing at x = 1; increasing at x = 3.
It's all about the sign: You don't need the size of f'(x) — just whether it is positive or negative at the point.
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Solve the inequality f'(x) > 0: To find where a function increases, differentiate, then solve f'(x) > 0 (and f'(x) < 0 for decreasing).
For a linear f', this is a simple inequality.
IB-style question — increasing interval
Find the values of x for which f(x) = x² − 6x + 5 is increasing.
Step by step
- Differentiate, then set f'(x) > 0.
- Solve.
Final answer
f is increasing for x > 3 (and decreasing for x < 3).
The boundary is where f' = 0: The increasing and decreasing parts meet where f'(x) = 0 — here at x = 3, the vertex.
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f'(x) = 0 splits the number line: Where f'(x) = 0 the curve is stationary; these points separate the increasing and decreasing parts.
Solve f'(x) = 0, then test the sign of f' in each region between them.
IB-style question — a cubic
For f(x) = x³ − 3x, the gradient function is f'(x) = 3x² − 3.
Find where f is increasing and where it is decreasing.
Step by step
- Stationary points: f'(x) = 0.
- Test the sign of f' in each region.
Final answer
Increasing for x < −1 and x > 1; decreasing for −1 < x < 1.
Test a point in each region: After finding the stationary x-values, test a value of f' in each interval to see if it's + or −.
Where f' is above the axis, f is increasing: Given the graph of the derivative f': where f' is above the x-axis, f is increasing; where f' is below, f is decreasing; where f' crosses zero, f has a stationary point (max if f' goes + → −, min if − → +).
IB-style question — explain a maximum
The graph of f' crosses the x-axis at x = 2, going from positive to negative.
Explain why f has a local maximum at x = 2.
Step by step
- Left of 2, f' > 0 → f increasing; right of 2, f' < 0 → f decreasing.
- Increasing then decreasing ⇒ a peak.
Final answer
Because f changes from increasing to decreasing at x = 2 (f' goes + → −), there is a local maximum there.
+ → − is a max; − → + is a min: The way f' crosses zero tells you the type: down-crossing (+→−) = maximum, up-crossing (−→+) = minimum.
IB-style question — reading the graph of f′
The graph shown is y = f′(x), where f′(x) = (x + 1)(x − 3).
State the interval(s) on which f is increasing.
Step by step
- f increases exactly where f′ is ABOVE the x-axis (f′ > 0). The parabola f′ is positive outside its roots x = −1 and x = 3.
- So f is increasing on those two intervals.
Final answer
f is increasing for x < −1 and for x > 3 (where f′ > 0).
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