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.
Today's hook, A ball thrown upward has height $h(t) = 20t - 5t^2$. At the very top of its arc, is it moving upward, downward, or neither? What does the derivative equal at that exact moment, and how does that help us find exactly when (and how high) the ball peaks?
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Orient and recall
Meet the destination, bring back what you already know, and gather the terms and formulas this lesson leans on.
Worksheets
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?
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The two moves
Work through the core explanation before applying it.
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The two moves
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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.
Solve $f'(x) = 0$ for $x$, then substitute into $f(x)$ for $y$. Both coordinates are needed for a point.
Second derivative test
$f''(x) < 0$: concave down, local max. $f''(x) > 0$: concave up, local min. $f''(x) = 0$: inconclusive.
Sign table backup
When $f''(x) = 0$, test the sign of $f'(x)$ on either side. If it doesn't change sign, it's a horizontal inflection.
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What you will master
Meet the destination, bring back what you already know, and gather the terms and formulas this lesson leans on.
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What you will master
Know
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
Understand
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
Can do
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
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Key terms
Stationary pointA point on a curve where the derivative is zero: $f'(x) = 0$. The tangent is horizontal.
Local maximumA stationary point where the function changes from increasing to decreasing. $f''(x) < 0$.
Local minimumA stationary point where the function changes from decreasing to increasing. $f''(x) > 0$.
Horizontal inflectionA stationary point where the function does not change from increasing to decreasing or vice versa.
Second derivative testUsing $f''(x)$ to classify the nature of a stationary point.
Sign tableA table showing the sign of $f'(x)$ on either side of a stationary point to determine its nature.
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Finding and classifying stationary points
Work through the core explanation before applying it.
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Finding and classifying stationary points
core concept
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).
Beyond the syllabus. Everything above this point is Core. MAV-11-06 requires you to find stationary points by solving $f'(x) = 0$ and to say where a function is increasing or decreasing. The second derivative test below is not a Year 11 method. It is shown because it is quick and because you will meet it in Year 12, but the Year 11 way to classify a stationary point is a sign table for $f'(x)$ either side of it, and that is the method you will be examined on. Learn the sign table first and treat the $f''$ shortcut as a check, not as the answer.
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 $f'(x)$
Real-world example. The height of a ball thrown upward is $h(t) = 20t - 5t^2$. We have $h'(t) = 20 - 10t$. Setting $h'(t) = 0$ gives $t = 2$ s. Since $h''(t) = -10 < 0$ everywhere, the stationary point is a local maximum. The ball peaks at $h(2) = 40 - 20 = 20$ m.
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
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Work examples end to end
Follow the reasoning through complete worked solutions.
PROBLEM 1 · FIND STATIONARY POINTS
Find and classify the stationary points of $f(x) = x^3 - 3x^2$.
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$f'(x) = 3x^2 - 6x = 3x(x - 2)$
Differentiate and factorise.
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$f'(x) = 0$ when $x = 0$ or $x = 2$
Solve for stationary points.
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$f(0) = 0$ and $f(2) = 8 - 12 = -4$. Stationary points at $(0, 0)$ and $(2, -4)$.
Find $y$-coordinates by substituting back into $f(x)$.
PROBLEM 2 · SECOND DERIVATIVE CLASSIFICATION
Classify the stationary points of $f(x) = x^3 - 3x^2$ using the second derivative test.
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$f''(x) = 6x - 6$
Find the second derivative.
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$f''(0) = -6 < 0$, so $(0, 0)$ is a local maximum.
Negative second derivative indicates concave down: local maximum.
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$f''(2) = 12 - 6 = 6 > 0$, so $(2, -4)$ is a local minimum.
Positive second derivative indicates concave up: local minimum.
PROBLEM 3 · INCONCLUSIVE SECOND DERIVATIVE
Find and classify the stationary points of $f(x) = x^4 - 4x^3$.
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$f'(x) = 4x^3 - 12x^2 = 4x^2(x - 3)$
Differentiate and factorise.
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$f'(x) = 0$ when $x = 0$ or $x = 3$. $f(0) = 0$ and $f(3) = 81 - 108 = -27$.
Find stationary points and their $y$-coordinates.
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$f''(x) = 12x^2 - 24x$. $f''(0) = 0$ (inconclusive → use sign table: $f'(x) \geq 0$ on both sides of $x=0$, so horizontal inflection). $f''(3) = 36 > 0$, so $(3, -27)$ is a local minimum.
At $x = 0$, $f''(x) = 0$ so the second derivative test fails. The sign table shows $f'(x)$ does not change sign across $x = 0$.
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
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Dodge the traps, then apply
Meet the mistakes that cost marks, then do it yourself.
Trap 01
Forgetting to find the $y$-coordinate
Stationary points are points $(x, y)$, not just $x$-values. Always substitute back into $f(x)$ to find the $y$-coordinate. Missing this loses marks every time.
Trap 02
Assuming $f''(x) = 0$ means inflection without testing
When $f''(x) = 0$, the second derivative test is inconclusive. You must use a sign table or examine the behaviour of $f'(x)$ on either side.
Trap 03
Confusing local and global maxima/minima
A local maximum is only the highest point in its neighbourhood, not necessarily the highest point on the entire curve. The HSC usually asks for local maxima/minima unless specified otherwise.
Think & type: Explain why the second derivative test is inconclusive when $f''(a) = 0$, and what you should do instead.
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Quick-fire practice · 5 problems
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Drill it, then lock it in
Run the quick drill and copy the summary into your book.
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Find the stationary points of $f(x) = x^2 - 4x + 3$.
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Find and classify the stationary points of $f(x) = x^3 - 6x^2 + 9x + 1$.
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Does $f(x) = e^x$ have any stationary points? Explain.
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For $f(x) = x^4 - 2x^2$, find all stationary points and classify them.
Fill the blanks: drag each token into the matching blank.
zeromaximumminimumsign table
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 $x = a$
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Revisit your thinking
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.
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Multiple choice
Answer the drill bank and rate your confidence.
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Multiple choice
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Short answer
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Short answer
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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].
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ApplyBand 4
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].
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AnalyseBand 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].
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Review and finish
Take the module quiz if you are ready, then mark the lesson complete.
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Take the full module quiz
quiz
A full module quiz covering every lesson in this module, not just this one. Set aside a decent block of time and treat it like a real assessment.
