Finding where a function increases or decreases
Solve f′(x) = 0, split the number line at those points, and test the sign of f′ in each part. Positive means increasing, negative means decreasing.
The method at a glance
The method, step by step
- 1Find f′(x) and factorise it fully.
- 2Solve f′(x) = 0. Also note any point where f′(x) does not exist.
- 3Mark these points on a number line. They split it into intervals.
- 4In each interval, put one test value of x into the factors of f′(x) to get its sign.
- 5★Where f′(x) > 0, f is strictly increasing. Where f′(x) < 0, f is strictly decreasing.
- 6Write the answer as intervals, inside the domain of f (for log x, only x > 0).
★ key step
Example 1: a cubic
Find the intervals in which f(x) = 2x³ − 21x² + 60x + 3 is strictly increasing or strictly decreasing.
- f′(x) = 6x² − 42x + 60
- f′(x) = 6(x² − 7x + 10) = 6(x − 2)(x − 5)
- f′(x) = 0 gives x = 2 or x = 5.
- These split R into (−∞, 2), (2, 5) and (5, ∞).
- In (−∞, 2), take x = 0: f′(0) = 6(−2)(−5) = 60 > 0.
- In (2, 5), take x = 3: f′(3) = 6(1)(−2) = −12 < 0.
- In (5, ∞), take x = 6: f′(6) = 6(4)(1) = 24 > 0.
f is strictly increasing on (−∞, 2) and (5, ∞), and strictly decreasing on (2, 5).
Sign chart for Example 1
| Interval | Sign of (x − 2) | Sign of (x − 5) | Sign of f′(x) | f is |
|---|---|---|---|---|
| (−∞, 2) | − | − | + | strictly increasing |
| (2, 5) | + | − | − | strictly decreasing |
| (5, ∞) | + | + | + | strictly increasing |
Example 2: a trigonometric function
Find the intervals in which f(x) = sin x − cos x, 0 ≤ x ≤ 2π, is strictly increasing or strictly decreasing.
- f′(x) = cos x + sin x
- f′(x) = 0 gives sin x = −cos x, that is tan x = −1.
- In [0, 2π], x = 3π/4 or x = 7π/4.
- In (0, 3π/4), take x = π/2: f′ = 0 + 1 = 1 > 0.
- In (3π/4, 7π/4), take x = π: f′ = −1 + 0 = −1 < 0.
- In (7π/4, 2π), take x = 11π/6: f′ = √3/2 − 1/2 > 0.
f is strictly increasing on [0, 3π/4) and (7π/4, 2π], and strictly decreasing on (3π/4, 7π/4).
Traps that cost marks
- Do not stop at the solutions of f′(x) = 0. The answer is the intervals, not the points.
- Check the domain first. For log x or √x, work only where the function is defined.
- A squared factor such as (x − 1)² never changes sign. If f′(x) = (x − 1)²(x + 2), f′ is negative only for x < −2.
- Write strictly increasing or decreasing when f′ is positive or negative on the interval.
- In a trigonometric question, keep only the solutions that lie in the given interval.
A dietitian tracking blood sugar after a breakfast of idli asks when the level rises and when it falls. That is the same question as finding the intervals where a function increases or decreases.
2 marks: “Find the intervals in which f(x) = 3x² − 12x + 7 is strictly increasing or strictly decreasing.” f′(x) = 6x − 12 = 6(x − 2). f′(x) > 0 for x > 2 and f′(x) < 0 for x < 2. Strictly increasing on (2, ∞); strictly decreasing on (−∞, 2).
3 marks: “Find the intervals in which f(x) = (x − 1)³(x − 2)² is strictly increasing or strictly decreasing.” f′(x) = (x − 1)²(x − 2)(5x − 8). Zeros: 1, 8/5, 2. (x − 1)² ≥ 0 does not change sign. Strictly increasing on (−∞, 8/5) and (2, ∞); strictly decreasing on (8/5, 2).
3 marks: “Find the intervals in which f(x) = cos 2x, 0 ≤ x ≤ π, is strictly increasing or strictly decreasing.” f′(x) = −2 sin 2x. On (0, π/2), sin 2x > 0, so f′ < 0: strictly decreasing on [0, π/2]. On (π/2, π), sin 2x < 0, so f′ > 0: strictly increasing on [π/2, π].
Factor, mark, test, write. Four moves for every interval question.