Use derivative signs to locate increasing and decreasing intervals
Factor the derivative, test signs on the curve's actual domain and interpret turning or rising behaviour in a restricted model.
Section 1.7 · Differentiation
Pure Mathematics 1 (Paper 1) · AS & A Level Mathematics 9709, 2026–2027. Use an informal understanding of limits; formal general differentiation from first principles and points of inflexion are not required in section 1.7. Written practice is self-checked.
Before you begin
- Differentiate polynomials and rational powers.
- Use a sign chart and retain domain restrictions.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
What does f′(x) > 0 throughout an interval tell you?
What you will learn
- Use positive and negative derivative values to identify increasing and decreasing intervals.
- Separate a stationary input from a domain break and respect the supplied model interval.
Test the gradient over intervals, not just at one point
A positive derivative throughout an interval means the function increases there; a negative derivative means it decreases. Find inputs where the derivative is zero or undefined, split the permitted domain at them and test a value in each interval. A zero derivative at a single input does not by itself prove that the direction changes. A constant function has zero derivative throughout and is neither strictly increasing nor strictly decreasing.
For f(x) = x³ − 6x² + 9x + 1, f′(x) = 3(x − 1)(x − 3). It is positive for x < 1, negative for 1 < x < 3 and positive for x > 3. The curve therefore rises, falls and rises across those intervals. Inputs 1 and 3 are stationary, with heights f(1) = 5 and f(3) = 1; the original function remains defined at both.
A missing input is not a stationary point
For f(x) = x + 4/x on the supplied domain x > 0, f′(x) = 1 − 4/x². Since x² is positive, its sign is the sign of x² − 4: negative on 0 < x < 2 and positive on x > 2. The domain excludes x = 0, so it cannot become a stationary point or be filled into a sign chart as if the curve were continuous there. The supplied positive domain also means negative inputs are outside this problem.
A practical model may permit only a closed time interval. For h(t) = −t² + 8t + 2 on 0 ≤ t ≤ 8, the rate h′(t) = 8 − 2t is positive before t = 4 and negative after it. Thus the model rises to height 18 at t = 4 and then falls. Include physical endpoint values when interpreting a maximum over a closed interval; a mathematical rule's behaviour outside the supplied time range is not part of the stated model.
Pause and explain
For f(x) = x + 4/x on x > 0, is x = 0 a stationary point?
Worked example
Find where f(x) = x³ − 6x² + 9x + 1 has positive, zero or negative gradient.
Show the worked solution
- Differentiate and factor: f′(x) = 3x² − 12x + 9 = 3(x − 1)(x − 3).
- The two factors have the same sign for x < 1 and x > 3, giving positive gradient; between them their signs differ, giving negative gradient.
- At x = 1 and 3 the gradient is zero; the curve is defined there and the corresponding heights are 5 and 1.
Answer Positive for x < 1 and x > 3; negative for 1 < x < 3; zero at x = 1 and 3.
Try a topic test
Choose Easy, Normal or Difficult. Save your answers and open worked review after finishing.
Test this topic →From guided practice to a new situation
Write your answer and reasoning first. Use a hint only if you are stuck. The model solution is for self-checking; your written work is not automatically marked.
Find the increasing and decreasing intervals of y = x² − 4x + 7.
Give me a hint
Differentiate, solve for the zero gradient and test its sign on each side.
Compare my reasoning
- The derivative is y′ = 2x − 4, which is zero at x = 2.
- For x < 2 it is negative, so the curve decreases there.
- For x > 2 it is positive, so the curve increases there; the zero-gradient point is (2, 3).
Decreasing for x < 2; increasing for x > 2; stationary point (2, 3).
Look for these in your work
- Use derivative signs on both intervals.
- Evaluate the curve separately for the stationary point's height.
On the supplied domain x > 0, find increasing and decreasing intervals for f(x) = x + 4/x.
Give me a hint
Use the positive denominator x² to read the sign of x² − 4.
Compare my reasoning
- f′(x) = 1 − 4/x² = (x² − 4)/x².
- On x > 0, the numerator is negative for 0 < x < 2 and positive for x > 2.
- Thus the curve decreases on (0, 2) and increases on (2, ∞); x = 0 remains excluded and f(2) = 4.
Decreasing for 0 < x < 2; increasing for x > 2; stationary point (2, 4).
Look for these in your work
- Respect the supplied positive domain and non-zero denominator.
- Separate the genuine stationary point from an excluded input.
For the supplied height model h(t) = −t² + 8t + 2 metres on 0 ≤ t ≤ 8 seconds, find when the rate is positive or negative and the greatest height in that interval.
Give me a hint
Use the derivative sign, then compare the stationary height with the two included endpoints.
Compare my reasoning
- h′(t) = 8 − 2t, positive for 0 ≤ t < 4 and negative for 4 < t ≤ 8.
- At t = 4 the rate is zero and h(4) = 18 metres.
- The endpoint heights h(0) and h(8) are both 2 metres, so the greatest permitted height is 18 metres at 4 seconds.
Positive rate for 0 ≤ t < 4; negative for 4 < t ≤ 8; maximum 18 m at t = 4 s.
Look for these in your work
- Keep the rate intervals within the supplied time domain.
- Compare the stationary value with both closed endpoints.
Common mistakes
- Find signs on intervals rather than inferring them from the derivative's roots alone.
- An undefined original input is not a stationary curve point.
- Restrict the answer to the stated real domain or physical time interval.
What can you explain now?
For y = x² − 4x + 7, find where the derivative is negative, zero and positive.
Compare with the explanation
Negative for x < 2; zero at x = 2; positive for x > 2.
The derivative 2x − 4 changes sign from negative to positive at 2.
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Make it stick
Construct a derivative sign chart, identify domain breaks separately from zero gradients and explain how a restricted model changes the intervals and endpoint checks.
If you needed the solution, close it and solve the problem again from a blank page. A correct answer is stronger when you can explain the reason for each step.