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 confidencethat 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.
Climb platforms by answering stationary point questions. Lighter alternative to the boss.
Mark lesson as complete
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