Stationary points

Find them and justify their nature

In 11 of 20 recent exams
2% of the marks
Paper 1 or 2
3 to 6 marks

How it is set in the exam

Paper 1 (no calculator) and Paper 2 (calculator). Very common (11 of 20 exam versions), usually as two parts of a longer question: the coordinates (3–6 marks), then the nature with a reason (2–3 marks). On Paper 2 (calculator) the GDC gives the coordinates, for about 2 marks. A sketch that rewards the turning points is worth 3 marks.

Spot it in the exam

Exam phrases to spot (6)

“find the coordinates of the stationary points”
“the local maximum point”
“the turning points of the graph”
“determine the nature of each stationary point”
“justify your answer”
“find the least value of $f$ for $0 \le x \le 4$ ”

In disguise (7)

wordings that mean $f'(x) = 0$ with the $x$ unknown (Gradient zero: which type?): "the tangent at $P$ is horizontal", "… parallel to the x-axis"
"the line $y = k$ is a tangent to the curve", $k$ unknown: the point of contact is a stationary point, and $k$ is its $y$ -value
"the graph of $y = x^3 - 12x + c$ touches the x-axis", $c$ unknown and added on (not in $f'$ ): a stationary point with $y = 0$
"find the range of $f$ " on a restricted domain: the turning points and the ends
a context, with the model already given as a formula: "the highest temperature", "the smallest number of …"
a given model in an angle $\theta$ , where the question already says which angle gives the least (or greatest) value: no test, and most of the work is the trig equation $f'(\theta) = 0$ , often $a\sin\theta = b\cos\theta$ , so $\tan\theta = \frac{b}{a}$ (Solve a trig equation on an interval), then exact values on
the greatest difference between two given models, with the answer as a calendar year: the difference $D(t)$ has $D'(t) = 0$ where the two rates are equal

Variants (3)

the greatest or least value on a given interval: compare the turning points with the ends
" $f(x) = k$ has exactly two solutions" (one, three): the horizontal line $y = k$ against the graph
"sketch the graph of $f$ ": the turning points, the intercepts and the ends each earn a mark

Looks like it, but isn't (9)

the $x$ is given and a constant is unknown (" $f$ has a local minimum at $x = 2$ ; find $a$ "): Find unknown constants
a quadratic, or a horizontal line against a parabola: Find the vertex is quicker
"the rate of change is greatest": the maximum of $f'$ , in Concavity and points of inflexion
the graph of $f'$ is given, not its formula: Graphs of f, f′ and f″
two quantities linked by a constraint, to be set up first: Optimisation
a sketch on , on a given domain, with no derivative found or used: the features come from the GDC, Sketch a function on a given domain with the GDC
the maximum or minimum is given, and only a range, an intercept, an asymptote or a sign of an exponential or log function is asked: Exponential and log graphs
a sinusoid $a\sin\left(b(t - c)\right) + d$ with no derivative asked for: its greatest and least values are $d \pm \lvert a \rvert$ , and their times come from the period (Read a sinusoid's features)
an exponential or log function with no turning point, whose range on a domain comes from an end and its asymptote: Exponential and log graphs

Worked example

long Section B question: a cubic, its nature, then f(x) = k

Paper 1 (no calculator)

Question

Let $f(x) = 2x^3 - 3x^2 - 12x + 5$ .
(a) Find $f'(x)$ . [2]
(b) Find the coordinates of the stationary points of the graph of $f$ . [4]
(c) Use the second derivative to determine the nature of each stationary point. Justify your answers. [5]
(d) Find the values of $k$ for which the equation $f(x) = k$ has three distinct real solutions. [3]

Try it yourself first, then compare.

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Write your working on paper or on screen. Paper Zero marks it step by step, like the worked example, and shows where each mark went. Free, with no account.

Try one like it

See it move: Where the tangent is flat

At $x = -2.00$ : $f'(x) = 24.00$ , so $f$ is increasing here.
$f'(x) = 6(x + 1)(x - 2)$ : the tangent is flat at $x = -1$ and $x = 2$ , where the lower graph crosses 0.
$f''(-1) = -18 < 0$ : $(-1, 12)$ is a local maximum, and $f'$ goes from positive to negative there.
$f''(2) = 18 > 0$ : $(2, -15)$ is a local minimum, and $f'$ goes from negative to positive there.
Add the line $y = k$ to see how many times $f(x) = k$ is solved.

Which method?

Situation	Test for the nature
$f''$ is quick to find (polynomials, most questions) and $f''(a) \ne 0$	second derivative: $f''(a) < 0$ maximum, $f''(a) > 0$ minimum
$f''$ would be long (a quotient, a product with $e^x$ ), or $f''(a) = 0$	the sign of $f'$ on each side, from the factors of $f'$ (a sign table labelled $f'$ )
, an awkward function	the GDC finds the point (maximum or minimum); for the nature, give the sign of $f'$ on each side (gradient at a point), e.g. in Example 2.4 $d'(3.2) = 0.0635 > 0$ and $d'(3.4) = -0.0611 < 0$ , or the sign of $f''$ at the point (its value from the GDC, or a sketch of the graph of $f''$ showing its sign there). The calculator's label is not a reason, and a sketch of $f$ with the turning point marked is not a safe reason (markschemes have accepted it only for a quadratic).
the question states the nature ("the local maximum $Q$ ")	no test needed

Steps

Differentiate and factorise
$f'(x)$ .
Solve $f'(x) = 0$ .
Keep the solutions in the domain, and reject the others with a reason (for example, $e^x = -3$ has no solution). On : the maximum or minimum of the graph of $f$ , or the zeros of $f'$ (zeros and intersections).
Find $y$ from $f$ ,
not from $f'$ : the point is $(a, f(a))$ .
Nature,
when it is asked for: the second-derivative test or the sign test (Testing the nature). Write the value, its sign and the conclusion together.
Greatest or least value on a given interval:
add the ends, $f(a)$ and $f(b)$ , to the values at the stationary points, and compare (Greatest and least values).
$f(x) = k$ has $n$ solutions:
sketch $y = f(x)$ from its turning points (and the ends of the domain), slide the horizontal line $y = k$ up and down, and count the meetings. The boundary values of $k$ are the $y$ -values of the turning points, of the ends of the domain, and of any horizontal asymptote (for $y = xe^{-x}$ , $k = 0$ is one).
Sketch:
plot the turning points, the intercepts and the ends of the domain, label each with its coordinates, and join them with the right shape, including how the curve behaves far out (for example, flattening towards the x-axis).

Common mistakes

The $y$ -coordinate from $f'$ ,
or no $y$ at all: a stationary point is $(a, f(a))$ .
A nature with no reason,
or " $f''(2) = 18$ " with no " $> 0$ ": the reason mark needs the sign and the conclusion together.
Treating $f''(a) = 0$ as "a point of inflexion".
The second-derivative test has failed; use the sign test.

More mistakes (4)

Leaving in a rejected solution,
such as $e^x = -2 - \sqrt{7}$ , or an $x$ outside the domain: , even if the right answer is there too.
Taking a local maximum as the greatest value on an interval
without checking the ends (Example 2.4).
Counting the solutions of $f(x) = k$ at a turning value
at $k = 12$ in Example 2.1 the line touches the graph at $(-1, 12)$ and crosses it once more: two solutions, not three.
Unlabelled or unfinished sketches
each turning point, intercept and end is a separate mark, and the curve must stop at the ends of a given domain.

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