Stationary Points
A stationary point is where a function pauses: the gradient is momentarily zero. Like a ball thrown upward that stops at its peak before falling back down. Identifying these points — and classifying them as maxima, minima, or inflections — reveals the entire shape of any curve.
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 this topic? Think about curves you have sketched before: where do they peak or dip?
There are only two moves in this entire lesson. Lock them into muscle memory and the rest is just calculation.
Move 1: Solve then classify. First solve $f'(x) = 0$ to find the $x$-coordinates of stationary points. Then substitute back into $f(x)$ to get the $y$-coordinates.
Move 2: Test the nature. Substitute into $f''(x)$: negative means concave down (local max), positive means concave up (local min), zero means further testing needed via a sign table.
Key facts
- A stationary point satisfies $f'(x) = 0$
- $f''(x) < 0$ at a local maximum; $f''(x) > 0$ at a local minimum
- Three types: local max, local min, horizontal inflection
Concepts
- Why the tangent is horizontal at a stationary point
- The geometric meaning of concavity and how it links to the second derivative
- Why $f''(x) = 0$ is inconclusive and a sign table is needed
Skills
- Find stationary points by solving $f'(x) = 0$
- Classify stationary points using the second derivative test or a sign table
- Sketch the shape of a curve using stationary point information
At a stationary point, the tangent to the curve is horizontal — the gradient equals zero. To find stationary points, solve $f'(x) = 0$. There are three possible types:
Three types of stationary point: local maximum (red), local minimum (teal), horizontal inflection (gold).
The second derivative test
Once you have solved $f'(x) = 0$, substitute each $x$-value into $f''(x)$:
- $f''(x) < 0$: the curve is concave down at this point → local maximum
- $f''(x) > 0$: the curve is concave up at this point → local minimum
- $f''(x) = 0$: the test is inconclusive → use a sign table for $f'(x)$
To find stationary points: solve $f'(x) = 0$, then substitute into $f(x)$ for the $y$-coordinate; Second derivative test: $f''(x) < 0 \Rightarrow$ max, $f''(x) > 0 \Rightarrow$ min, $f''(x) = 0 \Rightarrow$ test further
Pause — copy the stationary point procedure (solve $f'(x) = 0$, substitute into $f(x)$) and the second derivative test ($f'' < 0$ = max, $f'' > 0$ = min, $f'' = 0$ = test further) into your book.
Quick check: True or false — if $f'(a) = 0$ and $f''(a) = 0$, then the point $(a, f(a))$ must be a horizontal inflection.
Worked examples · 3 in a row, reveal as you go
Find and classify the stationary points of $f(x) = x^3 - 3x^2$.
Classify the stationary points of $f(x) = x^3 - 3x^2$ using the second derivative test.
Find and classify the stationary points of $f(x) = x^4 - 4x^3$.
Quick check: For $f(x) = x^3 - 3x$, which of the following correctly identifies a stationary point and its nature?
Common errors · the 3 traps that cost marks
Think & type: Explain why the second derivative test is inconclusive when $f''(a) = 0$, and what you should do instead.
Quick-fire practice · 5 problems
Find the stationary points of $f(x) = x^2 - 4x + 3$.
Find and classify the stationary points of $f(x) = x^3 - 6x^2 + 9x + 1$.
Does $f(x) = e^x$ have any stationary points? Explain.
For $f(x) = x^4 - 2x^2$, find all stationary points and classify them.
Sketch the curve $y = x^3 - 3x$ showing stationary points, labelling coordinates.
Fill the blanks: drag each token into the matching blank.
A stationary point occurs where $f'(x)$ equals ___. If $f''(x) < 0$, the point is a local ___. If $f''(x) > 0$, the point is a local ___. If $f''(x) = 0$, use a ___ to determine the nature.
Match each condition to the correct conclusion.
- $f'(a) = 0$ and $f''(a) < 0$
- $f'(a) = 0$ and $f''(a) > 0$
- $f'(a) = 0$ and $f''(a) = 0$
- inconclusive, use sign table
- local minimum at $x = a$
- local maximum at $x = a$
Earlier you were asked about where curves pause. Stationary points are exactly these pause points — where the gradient is momentarily zero and the tangent is horizontal. The first derivative finds them; the second derivative (or a sign table) tells us what kind of pause it is: a peak, a trough, or a level crossing.
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 and classify the stationary points of $f(x) = x^3 - 3x^2 + 4$. Show all working. 4 MARKS
View comprehensive answer
$f'(x) = 3x^2 - 6x = 3x(x - 2)$ [0.5]. $f'(x) = 0$ when $x = 0$ or $x = 2$ [0.5]. $f(0) = 4$, $f(2) = 0$ [0.5]. $f''(x) = 6x - 6$. $f''(0) = -6 < 0$ (local max at $(0, 4)$). $f''(2) = 6 > 0$ (local min at $(2, 0)$) [2.5].
Q2. Show that $f(x) = x^3$ has a horizontal inflection at $x = 0$. 3 MARKS
View comprehensive answer
$f'(x) = 3x^2$, so $f'(0) = 0$ (stationary point) [1]. $f''(x) = 6x$, so $f''(0) = 0$ (inconclusive) [0.5]. For $x < 0$: $f'(x) = 3x^2 > 0$; for $x > 0$: $f'(x) = 3x^2 > 0$. Since $f'(x)$ does not change sign, it is a horizontal inflection [1.5].
Q3. The curve $y = ax^3 + bx^2 + cx + d$ has a local maximum at $(1, 4)$ and a local minimum at $(3, 0)$. Find the values of $a$, $b$, $c$ and $d$. 5 MARKS
View comprehensive answer
$y' = 3ax^2 + 2bx + c$. At stationary points $y' = 0$: $3a + 2b + c = 0$ and $27a + 6b + c = 0$ [1]. Also $a + b + c + d = 4$ and $27a + 9b + 3c + d = 0$ [1]. Subtracting: $24a + 4b = 0 \Rightarrow b = -6a$ [1]. Then $c = 9a$ and $d = 4 - 4a$. Using the second point: $d = 0$, hence $a = 1$, $b = -6$, $c = 9$, $d = 0$ [2].
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 stationary point questions. Lighter alternative to the boss.
Mark lesson as complete
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