Increasing and Decreasing Functions
A function's direction tells you whether it is climbing or falling. The sign of the first derivative reveals this at a glance: positive means rising, negative means falling, zero means momentarily level. In this lesson you will master sign analysis — the core tool for sketching curves and solving optimisation problems.
Practise this lesson
Three printable worksheets that build from foundations to mastery — or build your own from any module’s questions.
Before we start, what do you already know about when functions go up or down? How would you describe a function that is increasing at a particular point?
There are only two moves in this entire lesson. Lock them into muscle memory and the rest is just calculation.
Move 1 — Differentiate and find stationary points. Compute $f'(x)$, then solve $f'(x) = 0$ to identify the points that separate regions of different behaviour.
Move 2 — Test signs in each interval. Pick a test value in each region between stationary points. If $f'(x) > 0$, the function is increasing there; if $f'(x) < 0$, it is decreasing.
Key facts
- $f'(x) > 0$ means increasing; $f'(x) < 0$ means decreasing
- Stationary points are where $f'(x) = 0$
- Intervals use open bracket notation at stationary points
Concepts
- Why the sign of $f'(x)$ determines the direction of $f$
- How stationary points separate intervals of increase from intervals of decrease
- The connection between sign tables and curve sketching
Skills
- Determine where a function is increasing or decreasing using the first derivative
- Find intervals of increase and decrease algebraically
- Use sign tables to analyse the behaviour of a function
The sign of the first derivative tells us the direction of a function. When $f'(x) > 0$, the function is increasing (going uphill). When $f'(x) < 0$, it is decreasing (going downhill).
Stationary points occur where $f'(x) = 0$, and these points separate intervals of increase from intervals of decrease. A sign table for $f'(x)$ is the clearest way to show this analysis.
$f'(x) > 0$: increasing (teal). $f'(x) < 0$: decreasing (red). Stationary points at $f'(x) = 0$.
The sign table method
A sign table lists the critical $x$-values (where $f'(x) = 0$ or undefined) and the sign of $f'(x)$ in each interval between them:
- Find $f'(x)$ and set it equal to zero to find stationary points.
- Mark these points on a number line, dividing it into intervals.
- Test one value of $x$ in each interval by substituting into $f'(x)$.
- Record $+$ (increasing) or $-$ (decreasing) for each interval.
$f'(x) > 0$ on an interval $\Rightarrow$ $f$ is increasing on that interval; $f'(x) < 0$ on an interval $\Rightarrow$ $f$ is decreasing on that interval
Pause — copy the sign rules: $f'(x) > 0$ on an interval means $f$ is increasing there; $f'(x) < 0$ means $f$ is decreasing, into your book.
Quick check: True or false — a function can be classified as increasing at a stationary point where $f'(x) = 0$.
Worked examples · 3 in a row, reveal as you go
Find where $f(x) = x^2 - 4x + 3$ is increasing and decreasing.
Test $x = 3$: $f'(3) = 2 > 0$ → increasing on $(2, \infty)$.
Find the intervals of increase and decrease for $f(x) = x^3 - 3x^2$.
Test $x = 1$: $f'(1) = 3(1)(-1) = -3 < 0$ → decreasing.
Test $x = 3$: $f'(3) = 3(3)(1) = 9 > 0$ → increasing.
For $f(x) = x^4 - 4x^3$, find intervals of increase and decrease.
For $x < 3$: $(x - 3) < 0$ so $f'(x) \le 0$ → decreasing on $(-\infty, 3)$.
For $x > 3$: $(x - 3) > 0$ so $f'(x) > 0$ → increasing on $(3, \infty)$.
Quick check: For $f(x) = x^3 - 3x$, on which interval is $f$ decreasing?
Common errors · the 3 traps that cost marks
Odd one out: Three of the following are true statements about increasing/decreasing functions. Which one is NOT correct?
Quick-fire practice · 5 problems
Find where $f(x) = x^2 - 6x + 5$ is increasing.
Find the intervals where $f(x) = x^3 - 3x$ is decreasing.
For $f(x) = 2x^3 - 9x^2 + 12x$, find where the function is increasing.
Find where $f(x) = \frac{1}{x}$ is decreasing.
Sketch $y = x^3 - 3x^2$ showing intervals of increase and decrease.
Fill the blanks: drag each token into the matching blank.
When $f'(x)$ is ___, the function is increasing. When $f'(x)$ is ___, the function is decreasing. Stationary points occur where $f'(x)$ is ___, and we use ___ intervals to describe the regions of increase and decrease.
Match each function to its correct description of increasing/decreasing behaviour.
- $f(x) = x^2$, interval $(-\infty, 0)$
- $f(x) = x^2$, interval $(0, \infty)$
- $f(x) = -x^2$, interval $(-\infty, 0)$
- $f(x) = x^3$, interval $(-\infty, \infty)$
- Always increasing ($f' = 3x^2 \ge 0$)
- Increasing ($f' = -2x > 0$)
- Increasing ($f' = 2x > 0$)
- Decreasing ($f' = 2x < 0$)
Earlier you were asked about when functions go up or down. The key insight: the first derivative is the key. Positive derivative means the function is increasing, negative means decreasing. Stationary points mark the exact boundaries between these regions — and we exclude them from the intervals.
Pick your answer, then rate your confidence — that tells the system what to drill next. Each retry pulls a fresh mix from the bank.
Q1. Find the intervals where $f(x) = x^3 - 6x^2 + 9x + 1$ is increasing. Show all working. 3 MARKS
Q2. The derivative of a function is $f'(x) = (x - 1)(x + 2)^2$. Find where $f$ is increasing and where it is decreasing. Show all working. 3 MARKS
Q3. Find the values of $k$ for which $f(x) = x^3 + kx^2 + 3x$ is increasing for all $x$. 4 MARKS
📖 Comprehensive answers (click to reveal)
Drill 1: $f'(x) = 2x - 6 = 0 \Rightarrow x = 3$. Increasing on $(3, \infty)$.
Drill 2: $f'(x) = 3x^2 - 3 = 3(x-1)(x+1)$. Decreasing on $(-1, 1)$.
Drill 3: $f'(x) = 6x^2 - 18x + 12 = 6(x-1)(x-2)$. Increasing on $(-\infty, 1)$ and $(2, \infty)$.
Drill 4: $f'(x) = -1/x^2 < 0$ for all $x \ne 0$. Decreasing on $(-\infty, 0)$ and $(0, \infty)$.
Q1 (3 marks): $f'(x) = 3x^2 - 12x + 9 = 3(x-1)(x-3)$ [1]. $f'(x) = 0$ at $x = 1, 3$ [0.5]. Test signs: increasing on $(-\infty, 1) \cup (3, \infty)$ [1.5].
Q2 (3 marks): $(x+2)^2 \ge 0$ always, so sign depends on $(x-1)$ [0.5]. Decreasing for $x < 1$ (i.e. $f' < 0$ except at $x = -2$) [1]. Increasing for $x > 1$ [1]. Note: $x = -2$ is a stationary point of inflection [0.5].
Q3 (4 marks): $f'(x) = 3x^2 + 2kx + 3$ [0.5]. For always increasing, need $f'(x) > 0$ for all $x$, so discriminant $< 0$ [1]. $\Delta = 4k^2 - 36 < 0$ [1]. $k^2 < 9 \Rightarrow -3 < k < 3$ [1.5].
Five timed questions. Beat the boss to bank a tier — gold (90% + speed), silver (75%), or bronze (50%). Replays welcome.
⚔ Enter the arenaClimb platforms by answering increasing/decreasing questions. Lighter alternative to the boss.
Mark lesson as complete
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