Stationary points and the second derivative

AS · P1 · 17 min

At the top of a hill or the bottom of a valley a curve is momentarily flat: its gradient is zero. These points are called stationary points, and finding them and deciding whether each is a maximum or a minimum is one of the most frequent tasks on Paper 1. It appears as a standalone question, inside curve sketching, and as the engine of every practical maximum and minimum problem. The syllabus asks you to locate stationary points, determine their nature (including by the second derivative test) and use them in sketching graphs.

What a stationary point is

Picture walking left to right along the curve $y = 2x^3 - 3x^2 - 12x + 5$ . You climb, reach a peak, descend into a dip, then climb again. At the peak and the bottom of the dip the tangent is horizontal, so the gradient is $0$ .

y = 2x^3 - 3x^2 - 12x + 5 y = 12 y = -15 (-1, 12) (2, -15)

The horizontal tangents touch the curve at $(-1, 12)$ and $(2, -15)$ . These are the stationary points of this curve.

Definition

A stationary point of the curve $y = f(x)$ is a point where $\dfrac{dy}{dx} = 0$ . The tangent there is horizontal.

A maximum point (local maximum) is a stationary point where the curve changes from increasing to decreasing. A minimum point (local minimum) is a stationary point where the curve changes from decreasing to increasing. Maximum and minimum points are together called turning points.

The word "local" matters. A maximum point is the highest point in its neighbourhood, not necessarily the highest point on the whole curve. The cubic above goes off to $+\infty$ on the right, so $(-1, 12)$ is not the greatest value of $y$ overall; it is just the top of a hill. Example 2 below has a maximum point that is lower than its minimum point.

Finding stationary points

The condition is a single equation, so finding stationary points is mostly algebra.

Finding stationary points

Rewrite the equation as powers of $x$ if needed, then find $\dfrac{dy}{dx}$ (using the chain rule for brackets raised to a power).
Set $\dfrac{dy}{dx} = 0$ and solve for $x$ . Factorise wherever you can.
Reject any solutions outside the domain given in the question.
Substitute each $x$ into the equation of the curve to find the $y$ -coordinate.
Write each stationary point as a coordinate pair.

For $y = 2x^3 - 3x^2 - 12x + 5$ :

\frac{dy}{dx} = 6x^2 - 6x - 12 = 6(x^2 - x - 2) = 6(x - 2)(x + 1)

so $\dfrac{dy}{dx} = 0$ at $x = 2$ and $x = -1$ , giving $y = -15$ and $y = 12$ .

Deciding the nature: the second derivative test

The second derivative $\dfrac{d^2y}{dx^2}$ is the rate of change of the gradient. It tells you which way the gradient is moving as you pass through the stationary point.

At a minimum, the gradient goes from negative, through zero, to positive. The gradient is increasing, so $\dfrac{d^2y}{dx^2} > 0$ . The curve is shaped like a cup, $\cup$ .
At a maximum, the gradient goes from positive, through zero, to negative. The gradient is decreasing, so $\dfrac{d^2y}{dx^2} < 0$ . The curve is shaped like a cap, $\cap$ .

A memory aid: a positive second derivative is a smile (minimum); a negative one is a frown (maximum).

Second derivative test

At a stationary point, where $\dfrac{dy}{dx} = 0$ :

$\dfrac{d^2y}{dx^2}$	Nature
$> 0$	minimum point
$< 0$	maximum point
$= 0$	test fails: use the first derivative test

For the cubic, $\dfrac{d^2y}{dx^2} = 12x - 6$ . At $x = 2$ it is $18 > 0$ , so $(2, -15)$ is a minimum. At $x = -1$ it is $-18 < 0$ , so $(-1, 12)$ is a maximum. This agrees with the graph.

The first derivative test

When the second derivative is zero, or is awkward to find, look at the sign of $\dfrac{dy}{dx}$ just either side of the stationary point. A sign change from $+$ to $-$ means the curve rises then falls: a maximum. From $-$ to $+$ it falls then rises: a minimum. This uses the same idea as increasing and decreasing functions.

First derivative test

Evaluate $\dfrac{dy}{dx}$ at a value of $x$ slightly less than and slightly greater than the stationary value (with no other stationary point in between).

Sign of $\dfrac{dy}{dx}$ : left, then right	Shape	Nature
$+$ then $-$	$\nearrow\ \searrow$	maximum point
$-$ then $+$	$\searrow\ \nearrow$	minimum point
same sign both sides	$\nearrow\ \nearrow$ or $\searrow\ \searrow$	neither

The third row is the case $y = x^3$ at the origin: the gradient is zero there, but positive on both sides, so the curve flattens for an instant and carries on rising. Such a point is neither a maximum nor a minimum.

Beyond the syllabus

A stationary point that is neither a maximum nor a minimum is called a stationary point of inflexion. The syllabus states that knowledge of points of inflexion is not included in Paper 1, so you will not be asked to find them. You must still not misclassify one: if the second derivative is zero, you cannot conclude anything until you check the signs either side.

A second derivative of zero does not mean "neither". The curve $y = (x - 2)^4 + 3$ has $\dfrac{d^2y}{dx^2} = 12(x - 2)^2 = 0$ at its stationary point $x = 2$ , yet the point $(2, 3)$ is clearly a minimum, since $(x - 2)^4 \ge 0$ . Only the first derivative test settles it.

Using stationary points to sketch a curve

Stationary points, intercepts and end behaviour together pin down the shape of a graph. For a cubic or a quartic, the stationary points tell you exactly where the hills and dips are, and the discriminant of a quadratic $\dfrac{dy}{dx}$ tells you how many there are.

Sketching from stationary points

Find and classify the stationary points.
Find the $y$ -intercept (put $x = 0$ ) and, if they factorise easily, the $x$ -intercepts.
Decide the behaviour for large positive and large negative $x$ from the leading term.
Plot the stationary points, join them smoothly with the correct turning shapes, and label every key point with its coordinates.

A cubic $y = ax^3 + bx^2 + cx + d$ has $\dfrac{dy}{dx} = 3ax^2 + 2bx + c$ , a quadratic. It has two stationary points when this quadratic has discriminant $> 0$ , one stationary point (neither max nor min) when the discriminant is $0$ , and none when it is $< 0$ . Example 5 uses this.

Worked examples

Locating and classifying

Find the coordinates of the stationary points on the curve $y = x^3 - 12x + 1$ and determine their nature.

Solution

\frac{dy}{dx} = 3x^2 - 12 = 3(x - 2)(x + 2)

Setting $\dfrac{dy}{dx} = 0$ gives $x = 2$ or $x = -2$ .

x = 2:\ y = 8 - 24 + 1 = -15, \qquad x = -2:\ y = -8 + 24 + 1 = 17

Second derivative: $\dfrac{d^2y}{dx^2} = 6x$ .

At $x = 2$ : $\dfrac{d^2y}{dx^2} = 12 > 0$ , so $(2, -15)$ is a minimum point.

At $x = -2$ : $\dfrac{d^2y}{dx^2} = -12 < 0$ , so $(-2, 17)$ is a maximum point.

A chain rule curve

A curve has equation $y = x + \dfrac{9}{x + 2}$ for $x \neq -2$ . Find the coordinates of the stationary points and determine their nature.

Solution

Write $y = x + 9(x + 2)^{-1}$ . By the chain rule,

\frac{dy}{dx} = 1 - 9(x + 2)^{-2} = 1 - \frac{9}{(x + 2)^2}

Set this equal to zero:

\frac{9}{(x + 2)^2} = 1 \quad\Rightarrow\quad (x + 2)^2 = 9 \quad\Rightarrow\quad x + 2 = \pm 3

so $x = 1$ or $x = -5$ .

x = 1:\ y = 1 + \frac{9}{3} = 4, \qquad x = -5:\ y = -5 + \frac{9}{-3} = -8

Differentiate again: $\dfrac{d^2y}{dx^2} = 18(x + 2)^{-3} = \dfrac{18}{(x + 2)^3}$ .

At $x = 1$ : $\dfrac{18}{27} = \dfrac{2}{3} > 0$ , so $(1, 4)$ is a minimum point.

At $x = -5$ : $\dfrac{18}{-27} = -\dfrac{2}{3} < 0$ , so $(-5, -8)$ is a maximum point.

The maximum point has a smaller $y$ -coordinate than the minimum point. That is no contradiction: they are on different branches of the curve, separated by the asymptote $x = -2$ , and each is only a local maximum or minimum.

y = x + 9/(x + 2) x = -2 (1, 4) (-5, -8)

When the second derivative is zero

Find the stationary points of $y = x^4 - 4x^3$ and determine their nature.

Solution

\frac{dy}{dx} = 4x^3 - 12x^2 = 4x^2(x - 3)

So $x = 0$ or $x = 3$ , with $y = 0$ and $y = 81 - 108 = -27$ .

$\dfrac{d^2y}{dx^2} = 12x^2 - 24x$ .

At $x = 3$ : $108 - 72 = 36 > 0$ , so $(3, -27)$ is a minimum point.

At $x = 0$ : $\dfrac{d^2y}{dx^2} = 0$ , so the test fails. Use the first derivative either side:

$x$	$-0.5$	$0$	$0.5$
$\dfrac{dy}{dx} = 4x^2(x - 3)$	$-3.5$	$0$	$-2.5$
sign	$-$	$0$	$-$

The gradient is negative on both sides, so $(0, 0)$ is neither a maximum nor a minimum. (The factor $4x^2$ is never negative, so the sign of $\dfrac{dy}{dx}$ near $0$ is the sign of $x - 3$ , which is negative.)

y = x^4 - 4x^3 (0, 0) (3, -27)

Finding constants from a stationary point

The curve $y = x^3 + ax^2 + bx + 4$ has a stationary point at $(2, -8)$ .

(a) Find the values of $a$ and $b$ .

(b) Find the coordinates of the other stationary point and determine the nature of both.

Solution

(a) Two facts give two equations. The point is on the curve:

8 + 4a + 2b + 4 = -8 \quad\Rightarrow\quad 2a + b = -10 \qquad (1)

The gradient is zero at $x = 2$ . Since $\dfrac{dy}{dx} = 3x^2 + 2ax + b$ ,

12 + 4a + b = 0 \quad\Rightarrow\quad 4a + b = -12 \qquad (2)

Subtract (1) from (2): $2a = -2$ , so $a = -1$ and $b = -8$ .

(b) Now $y = x^3 - x^2 - 8x + 4$ and

\frac{dy}{dx} = 3x^2 - 2x - 8 = (x - 2)(3x + 4)

The factor $x - 2$ must be there, which is a useful check. The other root is $x = -\tfrac{4}{3}$ :

y = -\frac{64}{27} - \frac{16}{9} + \frac{32}{3} + 4 = \frac{-64 - 48 + 288 + 108}{27} = \frac{284}{27}

$\dfrac{d^2y}{dx^2} = 6x - 2$ . At $x = 2$ it is $10 > 0$ , so $(2, -8)$ is a minimum. At $x = -\tfrac{4}{3}$ it is $-10 < 0$ , so $\left(-\tfrac{4}{3}, \tfrac{284}{27}\right)$ is a maximum.

How many stationary points

Find the set of values of the constant $k$ for which the curve $y = x^3 - 3kx^2 + 12x$ has two stationary points.

Solution

\frac{dy}{dx} = 3x^2 - 6kx + 12

Two stationary points means $3x^2 - 6kx + 12 = 0$ has two distinct real roots, so its discriminant is positive:

(-6k)^2 - 4(3)(12) > 0 \quad\Rightarrow\quad 36k^2 > 144 \quad\Rightarrow\quad k^2 > 4

So $k < -2$ or $k > 2$ .

At $k = \pm 2$ the derivative is $3(x \mp 2)^2$ , which has a repeated root: one stationary point, which is neither a maximum nor a minimum. For $-2 < k < 2$ there are none.

Watch out

Using $\dfrac{dy}{dx}$ to find the $y$ -coordinate. Once you have $x$ , substitute into the equation of the curve. Substituting into $\dfrac{dy}{dx}$ just gives $0$ .

Reading $\dfrac{d^2y}{dx^2} = 0$ as "neither". It means only that the test has failed. Use the signs of $\dfrac{dy}{dx}$ either side, as with $y = (x - 2)^4 + 3$ , which has a minimum.

Swapping the signs. Positive second derivative means minimum. Students who think "positive means top" lose the classification mark.

Losing a root when dividing. In $4x^3 - 12x^2 = 0$ , dividing by $x^2$ throws away $x = 0$ . Factorise instead: $4x^2(x - 3) = 0$ .

Forgetting the negative root. $(x + 2)^2 = 9$ gives $x + 2 = \pm 3$ , two stationary points, not one.

Ignoring the domain. If the question says $x > 0$ , discard any stationary value with $x \le 0$ and say so.

Exam tip

"Find the coordinates" needs both $x$ and $y$ . Leaving only $x$ costs the final accuracy mark.
"Determine the nature" needs a reason. Write the value of $\dfrac{d^2y}{dx^2}$ (or its sign) and the conclusion: " $\dfrac{d^2y}{dx^2} = 18 > 0$ , so minimum." A bare "minimum" with no evidence usually scores nothing.
Use the first derivative test if the question allows "any method" and the second derivative is messy. Show the actual values of $\dfrac{dy}{dx}$ at your chosen points, not just signs.
Exact values: give $\tfrac{284}{27}$ , not $10.5$ , unless decimals are asked for.
Check by factor: if you know one stationary value (as in Example 4), the derivative must have that factor. If it does not factorise with it, recheck your constants.
"Hence sketch" expects the stationary points, intercepts and correct end behaviour, each labelled. You do not need to plot to scale.

Summary

A stationary point is where $\dfrac{dy}{dx} = 0$ ; the tangent there is horizontal.
Solve $\dfrac{dy}{dx} = 0$ for $x$ by factorising, then use the curve for $y$ .
Second derivative test: $\dfrac{d^2y}{dx^2} > 0$ gives a minimum, $< 0$ a maximum, $= 0$ is inconclusive.
First derivative test: $+$ to $-$ is a maximum, $-$ to $+$ a minimum, no change is neither.
Maximum and minimum points are local: a maximum can be lower than a minimum on another branch.
A cubic's derivative is a quadratic, so its discriminant decides whether there are two, one or no stationary points.
Unknown constants: "stationary point at $(p, q)$ " gives two equations, $y(p) = q$ and $y'(p) = 0$ .
Knowledge of points of inflexion is not required, but never classify from $\dfrac{d^2y}{dx^2} = 0$ alone.

Practice questions

Question

Find the stationary points of $y = 2x^3 - 3x^2 - 12x - 7$ and determine their nature.
Find the coordinates of the stationary point on the curve $y = 2x + \dfrac{8}{x^2}$ and determine its nature.
The curve $y = 4\sqrt{x} - x$ is defined for $x > 0$ . Find the stationary point and show that it is a maximum.
Find the stationary point of $y = (x - 2)^4 + 3$ and determine its nature. Explain why the second derivative test does not help.
Find the stationary points of $y = (2x - 1)^3 - 24x$ and determine their nature.
The curve $y = x^2 + \dfrac{a}{x}$ has a stationary point where $x = 3$ . Find $a$ , the $y$ -coordinate of the stationary point and its nature.
The curve $y = 2x^3 + ax^2 + bx - 7$ has stationary points where $x = -1$ and $x = 2$ . Find $a$ and $b$ , the coordinates of both stationary points, and their nature.
(a) Find the stationary points of $y = (x^2 - 3)^2$ and determine their nature. (b) Sketch the curve. (c) Hence find the set of values of $k$ for which the line $y = k$ meets the curve at exactly four points.
The curve $y = k\sqrt{x} - x$ , where $k$ is a positive constant, has a stationary point with $y$ -coordinate $9$ . Find $k$ , and show that the stationary point is a maximum.

Answers

$\dfrac{dy}{dx} = 6x^2 - 6x - 12 = 6(x - 2)(x + 1)$ , so $x = 2$ or $x = -1$ . At $x = 2$ : $y = 16 - 12 - 24 - 7 = -27$ ; at $x = -1$ : $y = -2 - 3 + 12 - 7 = 0$ . $\dfrac{d^2y}{dx^2} = 12x - 6$ : at $x = 2$ it is $18 > 0$ , so $(2, -27)$ is a minimum; at $x = -1$ it is $-18 < 0$ , so $(-1, 0)$ is a maximum.
$y = 2x + 8x^{-2}$ , $\dfrac{dy}{dx} = 2 - 16x^{-3} = 0$ gives $x^3 = 8$ , $x = 2$ , $y = 4 + 2 = 6$ . $\dfrac{d^2y}{dx^2} = 48x^{-4} = 3 > 0$ at $x = 2$ , so $(2, 6)$ is a minimum.
$y = 4x^{\frac{1}{2}} - x$ , $\dfrac{dy}{dx} = 2x^{-\frac{1}{2}} - 1 = 0$ gives $\sqrt{x} = 2$ , $x = 4$ , $y = 8 - 4 = 4$ . $\dfrac{d^2y}{dx^2} = -x^{-\frac{3}{2}} = -\tfrac{1}{8} < 0$ , so $(4, 4)$ is a maximum.
$\dfrac{dy}{dx} = 4(x - 2)^3 = 0$ only at $x = 2$ , giving $(2, 3)$ . $\dfrac{d^2y}{dx^2} = 12(x - 2)^2 = 0$ at $x = 2$ , so the test is inconclusive. At $x = 1.9$ , $\dfrac{dy}{dx} = 4(-0.1)^3 < 0$ ; at $x = 2.1$ , $\dfrac{dy}{dx} = 4(0.1)^3 > 0$ . The sign goes $-$ to $+$ , so $(2, 3)$ is a minimum.
$\dfrac{dy}{dx} = 3(2x - 1)^2 \times 2 - 24 = 6(2x - 1)^2 - 24 = 0$ , so $(2x - 1)^2 = 4$ , $2x - 1 = \pm 2$ , $x = \tfrac{3}{2}$ or $x = -\tfrac{1}{2}$ . At $x = \tfrac{3}{2}$ : $y = 8 - 36 = -28$ . At $x = -\tfrac{1}{2}$ : $y = -8 + 12 = 4$ . $\dfrac{d^2y}{dx^2} = 24(2x - 1)$ : at $x = \tfrac{3}{2}$ it is $48 > 0$ , minimum $\left(\tfrac{3}{2}, -28\right)$ ; at $x = -\tfrac{1}{2}$ it is $-48 < 0$ , maximum $\left(-\tfrac{1}{2}, 4\right)$ .
$\dfrac{dy}{dx} = 2x - ax^{-2}$ . At $x = 3$ : $6 - \dfrac{a}{9} = 0$ , so $a = 54$ . Then $y = 9 + \dfrac{54}{3} = 27$ . $\dfrac{d^2y}{dx^2} = 2 + 2ax^{-3} = 2 + \dfrac{108}{27} = 6 > 0$ , so $(3, 27)$ is a minimum.
$\dfrac{dy}{dx} = 6x^2 + 2ax + b$ has roots $-1$ and $2$ , so it equals $6(x + 1)(x - 2) = 6x^2 - 6x - 12$ . Comparing, $2a = -6$ and $b = -12$ , so $a = -3$ , $b = -12$ . The curve is $y = 2x^3 - 3x^2 - 12x - 7$ , so (as in question 1) $(-1, 0)$ is a maximum and $(2, -27)$ is a minimum.
(a) By the chain rule $\dfrac{dy}{dx} = 2(x^2 - 3)(2x) = 4x(x^2 - 3)$ , which is zero at $x = 0$ and $x = \pm\sqrt{3}$ . Points: $(0, 9)$ , $(\sqrt{3}, 0)$ , $(-\sqrt{3}, 0)$ . Expanding, $y = x^4 - 6x^2 + 9$ , so $\dfrac{d^2y}{dx^2} = 12x^2 - 12$ . At $x = 0$ : $-12 < 0$ , maximum. At $x = \pm\sqrt{3}$ : $36 - 12 = 24 > 0$ , minima. (b) A "W" shape, symmetrical about the $y$ -axis, touching the $x$ -axis at $(\pm\sqrt{3}, 0)$ with a maximum at $(0, 9)$ , rising steeply for large $|x|$ . (c) A horizontal line cuts the W four times when it lies strictly between the minimum value and the maximum value: $0 < k < 9$ .
$\dfrac{dy}{dx} = \dfrac{k}{2\sqrt{x}} - 1 = 0$ gives $\sqrt{x} = \dfrac{k}{2}$ , so $x = \dfrac{k^2}{4}$ . Then $y = k \cdot \dfrac{k}{2} - \dfrac{k^2}{4} = \dfrac{k^2}{4} = 9$ , so $k^2 = 36$ and $k = 6$ (as $k > 0$ ). The point is $(9, 9)$ . $\dfrac{d^2y}{dx^2} = -\dfrac{k}{4}x^{-\frac{3}{2}} = -\dfrac{6}{4 \times 27} = -\dfrac{1}{18} < 0$ , so it is a maximum.

What a stationary point is

Finding stationary points

Deciding the nature: the second derivative test

The first derivative test

Using stationary points to sketch a curve

Worked examples

Practice questions

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