Solving $f'(x) = 0$ finds the points where the curve levels off. It does not tell you what happens there. The sign of $f'$ on either side does, and it separates a maximum from a minimum from the point that is neither.
Commit to a prediction about what a zero gradient means.
A curve has $f'(2) = 0$. Without looking ahead list every possibility for what the curve is doing at $x = 2$, and say what extra information you would need to decide between them.
Define a stationary point and list the three kinds.
Three kinds of stationary point
We just saw that a derivative exists only where the curve is smooth. That raises a question: where it does exist, the places it equals zero are special, but special how? This card answers it → zero gradient means the tangent is horizontal, and that alone allows three quite different behaviours.
A stationary point is defined by the tangent being horizontal, nothing more. Whether the curve turns there is a separate question, answered by what $f'$ does on either side.
The first two are turning points: the curve reverses direction. The third is not, and calling it one is the error this lesson is built to prevent. "Local" means compared with nearby points only — a local maximum can be lower than points far away on the same curve.
A stationary point is where $f'(c) = 0$: the tangent is horizontal. Nothing more is implied.; Turning points are the max and min cases. A horizontal point of inflection is stationary but NOT a turning point.
Pause, copy the definition of a stationary point and the three classifications with their sign patterns $+0-$, $-0+$ and same-sign, into your book.
True or false: Every stationary point of a function is a turning point.
Build a sign table and read the classification from it.
We just saw that the three cases are distinguished by the sign of $f'$ either side. That raises a question: how do you organise that so nothing is missed and the working is markable? This card answers it → a sign table, built once and read off for every stationary point at the same time.
The method, in four steps:
Worked through. $f(x) = x^3 - 3x$, so $f'(x) = 3x^2 - 3 = 3(x - 1)(x + 1)$, giving stationary points at $x = -1$ and $x = 1$.
| $x$ | $-2$ | $-1$ | $0$ | $1$ | $2$ |
|---|---|---|---|---|---|
| $f'(x)$ | $9$ | $0$ | $-3$ | $0$ | $9$ |
| sign | $+$ | $0$ | $-$ | $0$ | $+$ |
| shape | ↗ | — | ↘ | — | ↗ |
Reading the table: at $x = -1$ the sign goes $+$ to $-$, a local maximum. At $x = 1$ it goes $-$ to $+$, a local minimum.
The $y$-values come from $f$: $f(-1) = -1 + 3 = 2$ and $f(1) = 1 - 3 = -2$. So the local maximum is $(-1, 2)$ and the local minimum is $(1, -2)$.
Sign test: differentiate, factorise, solve $f'(x) = 0$, test one easy value in each interval, classify from the sign pattern.; Get $y$-coordinates from $f$, never from $f'$ (which is 0 at every stationary point).
Pause, copy the four-step sign-test method and the warning that $y$-coordinates come from $f$ rather than $f'$, into your book.
Quick check: At a stationary point $x = c$, the derivative is negative just to the left and negative just to the right. What is the point?
Use the explorer. Press Build your own f′ and set $p = 1$, $q = 3$ and $m = 2$, so that $f'(x) = (x - 1)^2 (x - 3)$. Scrub across both stationary points until the table fills. Type whole numbers.
The stationary point that is not a turning point is at $x =$
The local minimum is at $x =$
Use the explorer. With $f(x) = x^3 - 3x$ loaded and the table complete, move the $C$ slider to $3$. True or false: the classification of at least one stationary point changed.
Work three examples, including one that does not turn.
Worked examples · 3 in a row, reveal as you go
Find and classify the stationary points of $f(x) = 2x^3 - 9x^2 + 12x - 3$.
Find and classify the stationary points of $f(x) = x^4 - 4x^3$.
$f(x) = x^3 + ax^2 + bx$ has a stationary point at $(1, -2)$. Find $a$ and $b$, then classify the other stationary point.
Fill the blanks: $f(x) = x^3 - 12x$. Give the $x$-values in increasing order.
smaller stationary $x =$
larger stationary $x =$
$f$ at the larger one $=$
Read the traps, work the quick-fire set, then revisit.
Common errors · traps that cost marks
Quick-fire practice · find, test, classify
Find and classify the stationary points of $f(x) = x^2 - 6x + 5$.
Find and classify the stationary points of $f(x) = x^3 - 3x^2 + 3x$.
$f(x) = x^4 - 2x^2$. Find all stationary points and classify each.
Explain why the sign of $f'$ cannot change between two consecutive stationary points, and why that makes one test value per interval enough.
$f(x) = x^3 + px + q$ has a stationary point at $(2, 5)$. Find $p$ and $q$.
Odd one out: Three of these functions have a local minimum at $x = 0$. Which one does not?
Match each sign pattern of $f'$ to the kind of stationary point:
Earlier you were asked what a curve with $f'(2) = 0$ could be doing there. There are exactly three possibilities: a local maximum (gradient $+$ then $-$), a local minimum (gradient $-$ then $+$), or a horizontal point of inflection (gradient the same sign on both sides). The extra information you need is the sign of $f'$ just either side of $x = 2$ — the value $f'(2) = 0$ on its own cannot distinguish them, which is exactly why the sign test exists.
The sign test always works, and it is sometimes slow: it needs a test value in every interval and careful bookkeeping. The next two lessons build the second derivative, which measures how the gradient itself is changing. That gives a faster classification for most stationary points, and it also answers a question the sign test cannot: which way the curve is bending between them.
Apply the lesson methods, then compare each response with its comprehensive answer.
Use the visible short-answer practice below during this lesson. After completing the focus area, use Checkpoint 1 for the checkpoint question bank.
Q1. Consider $f(x) = 2x^3 + 3x^2 - 12x + 1$.
(a) Find $f'(x)$ and solve $f'(x) = 0$. (b) Construct a sign table for $f'$ and classify each stationary point. (c) State the coordinates of both stationary points. (3 marks)
Q2. A curve has $f'(x) = (x + 2)^2(x - 4)$.
(a) State the $x$-values of all stationary points. (b) Classify each one using a sign table, and justify why the repeated factor matters. (c) You are told $f(-2) = 7$. Explain what more you would need in order to sketch the curve, and what you can already say about its shape near $x = -2$. (3 marks)
Q3. A student is asked to find the maximum value of $f(x) = 3x - x^3$ on the interval $-3 \leq x \leq 3$. They compute $f'(x) = 3 - 3x^2$, solve to get $x = \pm 1$, identify $x = 1$ as a local maximum with $f(1) = 2$, and answer "the maximum value is 2".
(a) Verify that the student's local maximum is correct. (b) Explain why their final answer is nevertheless wrong, computing the value they should have given. (c) State the general principle their method omits, and describe when omitting it is safe. (3 marks)
Drill 1: $f'(x) = 2x - 6 = 0$ gives $x = 3$. $f'(2) = -2 < 0$ and $f'(4) = 2 > 0$, so the sign runs $-$ to $+$: a local minimum. $f(3) = 9 - 18 + 5 = -4$, so the point is $(3, -4)$.
Drill 2: $f'(x) = 3x^2 - 6x + 3 = 3(x-1)^2$, which is zero only at $x = 1$. Since $(x-1)^2 \geq 0$, $f'$ is positive on both sides, so the sign does not change: $(1, 1)$ is a horizontal point of inflection, not a turning point. ($f(1) = 1 - 3 + 3 = 1$.)
Drill 3: $f'(x) = 4x^3 - 4x = 4x(x-1)(x+1)$, zero at $x = -1, 0, 1$. Testing: $f'(-2) = -24 < 0$, $f'(-0.5) = 1.5 > 0$, $f'(0.5) = -1.5 < 0$, $f'(2) = 24 > 0$. So $(-1, -1)$ is a local minimum, $(0, 0)$ is a local maximum, and $(1, -1)$ is a local minimum. The curve is a "W".
Drill 4: $f'$ is a polynomial, so it is continuous; a continuous function cannot pass from positive to negative without taking the value 0 somewhere in between. Every zero of $f'$ is a stationary point and has already been found, so between two consecutive ones there is no zero and therefore no sign change. One test value per interval is therefore enough to determine the sign throughout it.
Drill 5: On the curve: $2^3 + 2p + q = 5$, so $2p + q = -3$. Stationary: $f'(x) = 3x^2 + p$, so $f'(2) = 12 + p = 0$ and $p = -12$. Substituting: $-24 + q = -3$, so $q = 21$.
Q1 (3 marks): (a) $f'(x) = 6x^2 + 6x - 12 = 6(x^2 + x - 2) = 6(x + 2)(x - 1)$, so $x = -2$ or $x = 1$ [1]. (b) Test $f'(-3) = 6(-1)(-4) = 24 > 0$; $f'(0) = 6(2)(-1) = -12 < 0$; $f'(2) = 6(4)(1) = 24 > 0$. So at $x = -2$ the sign runs $+$ to $-$, a local maximum; at $x = 1$ it runs $-$ to $+$, a local minimum [1]. (c) $f(-2) = 2(-8) + 3(4) + 24 + 1 = -16 + 12 + 24 + 1 = 21$ and $f(1) = 2 + 3 - 12 + 1 = -6$. So the stationary points are $(-2, 21)$ and $(1, -6)$ [1].
Q2 (3 marks): (a) $f'(x) = 0$ when $(x+2)^2 = 0$ or $x - 4 = 0$, so $x = -2$ and $x = 4$ [0.5]. (b) $(x+2)^2$ is positive for every $x \neq -2$, so it never affects the sign; the sign of $f'$ is decided entirely by $(x - 4)$. Testing: $f'(-3) = (1)(-7) < 0$, $f'(0) = (4)(-4) < 0$, $f'(5) = (49)(1) > 0$. At $x = -2$ the sign is negative on both sides, so it is a horizontal point of inflection; at $x = 4$ the sign runs $-$ to $+$, a local minimum. The repeated factor is exactly what makes the first one fail to turn [1.5]. (c) To sketch the curve I would also need $f$ itself, or at least $f(4)$, since $f'$ alone fixes the shape but not the vertical position, and any constant added to $f$ gives the same $f'$. Near $x = -2$ I can already say the curve is decreasing on both sides, flattening momentarily at the point $(-2, 7)$ before continuing downward [1].
Q3 (3 marks): (a) $f'(x) = 3 - 3x^2 = 3(1-x)(1+x)$, so $f'(1) = 0$ ✓. Testing either side: $f'(0) = 3 > 0$ and $f'(2) = -9 < 0$, so the sign runs $+$ to $-$ and $x = 1$ is indeed a local maximum, with $f(1) = 3 - 1 = 2$ ✓ [1]. (b) The student never examined the endpoints of the interval. $f(-3) = -9 + 27 = 18$ and $f(3) = 9 - 27 = -18$. The largest value of $f$ on $[-3, 3]$ is therefore $18$, attained at the endpoint $x = -3$, which is nine times the value they gave [1]. (c) The principle: a local maximum is not automatically a global maximum. On a closed interval the greatest value must be found by comparing every local maximum with the values at both endpoints (and at any point where $f'$ fails to exist). Omitting the endpoint check is safe only when the domain has no endpoints to check — for instance a function on all of $\mathbb{R}$ whose behaviour at infinity is known to rule them out — or when the question explicitly asks for a local maximum rather than the maximum value [1].
Retrieve the central ideas, then mark the lesson complete or continue to the module quiz.
Answer from memory first, then return to the matching Learn checkpoint to check and correct your response.
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.
Start the module quiz →Tick when you've finished the practice and review.
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