5.8 Stationary points, optimisation and inflexion

The derivative $f'(x)$ gives the gradient of a curve at every point. Where the gradient is zero the curve is momentarily flat: it may be at the top of a hill, at the bottom of a valley, or levelling off before it continues in the same direction. Finding and classifying these points is the main use of differentiation in sketching graphs, and it answers practical questions of the form "what is the greatest volume?", "what is the least cost?" or "when is the temperature highest?".

The second derivative adds a second layer of information: how the gradient itself is changing. Where the curve switches from bending one way to bending the other there is a point of inflexion, which is often the place where a quantity is growing or falling fastest.

Stationary points

What a stationary point is

On a smooth curve, the tangent at each point has a gradient equal to the value of the derivative there. At most points the tangent slopes up or down. At a few points it is horizontal, and these are the points where the function stops increasing or decreasing, at least for an instant.

The word "stationary" comes from motion: if $f(x)$ gives the position of an object, then $f'(x)$ is its velocity, and at a stationary point the object is not moving for that instant.

The three types of stationary point

A stationary point is always one of three types. The type depends on what the gradient does on either side of the point.

• A local maximum: the curve rises to the point and falls after it, so the gradient changes from positive to negative. The point is higher than every nearby point.

• A local minimum: the curve falls to the point and rises after it, so the gradient changes from negative to positive. The point is lower than every nearby point.

• A stationary point of inflexion: the curve flattens out and then continues in the same direction, so the gradient has the same sign on both sides. The point is neither a maximum nor a minimum.

A local maximum or local minimum is also called a turning point, because the curve turns from going up to going down or the reverse. A stationary point of inflexion is a stationary point but not a turning point.

The curve $y = 3x^5 - 5x^3$ has all three types. Its derivative is $\frac{dy}{dx} = 15x^4 - 15x^2 = 15x^2(x^2 - 1)$, which is zero at $x = -1$, $x = 0$ and $x = 1$. The table summarises how the sign of the gradient identifies each type.

Finding stationary points

Stationary points are found by solving an equation, then using the original function to find the $y$-coordinates.

1. Differentiate to find $f'(x)$.

2. Solve $f'(x) = 0$. Each solution is the $x$-coordinate of a stationary point.

3. Substitute each solution into $f(x)$, not $f'(x)$, to find the $y$-coordinate.

A second error is to solve $f(x) = 0$ instead of $f'(x) = 0$. The solutions of $f(x) = 0$ are where the curve crosses the $x$-axis; the solutions of $f'(x) = 0$ are where it is flat. These are different points except by coincidence.

Classifying stationary points

Solving $f'(x) = 0$ finds the stationary points but does not say what type they are. Determining the type, also called the nature of the point, needs a test.

The first derivative test

The first derivative test looks at the sign of $f'(x)$ at values of $x$ a little to the left and a little to the right of the stationary point. The sign shows whether the function is increasing or decreasing there, which is exactly what separates the three types.

The test values must lie between neighbouring stationary points. If a test value jumps past another stationary point, the sign it gives may belong to a different part of the curve. A sign table with one column for each interval keeps this organised.

The first derivative test always works for a function whose derivative is defined near the point, which makes it the safest choice when a stationary point might be a point of inflexion.

The second derivative test

The second derivative $f''(x)$ measures how the gradient is changing. At a local maximum the gradient goes from positive, through zero, to negative, so it is decreasing; this usually means $f''(a) < 0$ and the curve is concave down. At a local minimum the gradient is increasing, so usually $f''(a) > 0$ and the curve is concave up.

A helpful picture is a cup. A curve that is concave up holds water, like the bottom of a valley, so a flat point there is a minimum. A curve that is concave down spills water, like the top of a hill, so a flat point there is a maximum.

The second derivative test is often quicker than a sign table, especially for functions that are awkward to evaluate at test points. It also works well with functions other than polynomials.

This example shows why the word "local" matters. The local maximum value $-4$ is smaller than the local minimum value $4$. A local maximum is only higher than the points close to it, not necessarily higher than every point on the curve.

The same method applies to exponential and trigonometric functions, using the rules of 5.6.

When the second derivative test fails

If $f''(a) = 0$, the second derivative test gives no information. The three functions $y = x^4$, $y = -x^4$ and $y = x^3$ show why. All three have $f'(0) = 0$ and $f''(0) = 0$, yet the origin is a minimum on the first, a maximum on the second and a stationary point of inflexion on the third.

Choosing a test

Both tests are valid, and each has its strengths.

Stationary points and unknown constants

Some questions give information about a stationary point and ask for constants in the function. Each piece of information gives an equation. A stationary point at $(p, q)$ gives two equations: $f'(p) = 0$ because the point is stationary, and $f(p) = q$ because the point lies on the curve.

The number of stationary points of a cubic depends on the quadratic $f'(x)$. Its discriminant decides whether $f'(x) = 0$ has two, one or no real solutions, which gives a cubic with two turning points, one stationary point of inflexion, or no stationary points at all.

Greatest and least values on an interval

A local maximum is the highest point in its own neighbourhood. In many problems the question is different: what is the greatest value the function takes anywhere on a given interval? This is the global maximum on that interval, and the least value is the global minimum.

Endpoints matter

On a closed interval $a \le x \le b$, the greatest or least value of a continuous function occurs either at a stationary point inside the interval or at one of the endpoints $x = a$ and $x = b$. At an endpoint the gradient need not be zero, so solving $f'(x) = 0$ alone can miss the answer.

To find the global maximum and minimum on $a \le x \le b$:

1. Find the stationary points that lie inside the interval.

2. Evaluate $f$ at each of these stationary points and at both endpoints.

3. The largest of these values is the global maximum; the smallest is the global minimum.

Points of inflexion

Concavity and where it changes

A curve is concave up on an interval where $f''(x) > 0$, so the gradient increases, and concave down where $f''(x) < 0$, so the gradient decreases (see 5.7). A point where the curve changes from one to the other is a point of inflexion.

At a point of inflexion the gradient reaches a local maximum or local minimum value: it stops increasing and starts decreasing, or the reverse. That is why points of inflexion often mark the moment of fastest growth or fastest decline in a model, for example the point at which a spreading infection is gaining new cases most quickly.

Zero and non-zero gradient

The gradient at a point of inflexion can be zero or non-zero.

• If $f'(a) = 0$ as well, the point is a stationary point of inflexion, as at the origin on $y = x^3$. The tangent is horizontal.

• If $f'(a) \neq 0$, the point is a non-stationary point of inflexion. The tangent slopes, and it crosses the curve at the point.

For $y = x^3 - 3x$, the second derivative is $\frac{d^2y}{dx^2} = 6x$, which is negative for $x < 0$ and positive for $x > 0$. The curve changes from concave down to concave up at the origin, where the gradient is $3(0)^2 - 3 = -3$. This point is not stationary, so the methods for stationary points do not find it: solving $f'(x) = 0$ gives only $x = \pm 1$.

Finding points of inflexion

1. Find $f''(x)$.

2. Solve $f''(x) = 0$ to find the candidates.

3. Check that $f''(x)$ changes sign at each candidate, using test values on either side or the factors of $f''(x)$.

4. Find the $y$-coordinate from $f(x)$. If a question asks, find the gradient $f'(a)$ to decide whether the point is stationary.

Why $f''(x) = 0$ is not enough

A zero second derivative only identifies candidates. On $y = x^4$, $f''(x) = 12x^2$ is zero at $x = 0$, but $12x^2 \ge 0$ on both sides, so the curve is concave up throughout and the origin is a minimum, not a point of inflexion.

The function $f(x) = x^4 - 4x^3$, whose stationary points are $(0, 0)$ and $(3, -27)$, shows both kinds of point of inflexion on one curve.

Reading points of inflexion from the graph of $f'$

Since $f''$ is the derivative of $f'$, the points of inflexion of $f$ are where $f'$ has a local maximum or local minimum. The stationary points of $f$ are where the graph of $f'$ meets the $x$-axis. When only the graph of $f'$ is given, both kinds of point can still be found from it.

The graph of $y = f'(x)$ for $f'(x) = x^3 - 3x + 2 = (x + 2)(x - 1)^2$ crosses the $x$-axis at $x = -2$, touches it at $x = 1$, and has a local maximum at $x = -1$.

Optimisation

The method

Optimisation means finding the best value of a quantity: the largest volume, the least cost, the shortest time. Calculus solves these problems once the quantity is written as a function of a single variable, because the best value is then a maximum or minimum of that function.

Most problems involve two or more variables linked by a fixed condition, such as a given volume or a given length of material. This condition is the constraint. It is used to eliminate all but one variable.

1. Draw a diagram if the situation is geometric, and name the variables.

2. Write an expression for the quantity to be optimised.

3. Use the constraint to write that quantity in terms of one variable.

4. State the domain: the values of the variable that make sense in context.

5. Differentiate, and solve derivative $= 0$.

6. Verify that the value gives a maximum or minimum (second derivative test or first derivative test), and check the endpoints of the domain.

7. Answer the question in context, with units.

Maximising the volume of an open box

An open box is made from a rectangular sheet of card $24$ cm by $15$ cm by cutting a square of side $x$ cm from each corner and folding up the sides.

A small cut gives a wide, shallow box; a large cut gives a deep box with a tiny base. Both have small volumes, so somewhere in between the volume is greatest.

Because $V$ is close to $0$ at both ends of the domain and has only one stationary point inside it, the local maximum at $x = 3$ is also the global maximum. It is good practice to say this when a question asks for "the maximum".

Minimising the surface area of a can

A drinks can is a closed cylinder. A manufacturer wants to hold $330\ \text{cm}^3$ while using as little metal as possible, so the surface area should be a minimum for the fixed volume.

The answer has a neat form. From $r^3 = \frac{165}{\pi}$, $\pi r^3 = 165$, so

The can with the least surface area has its height equal to its diameter, for any fixed volume. Real drinks cans are taller and narrower than this. A model that counts only area leaves out other factors: the top and base are made of thicker metal than the side, and a narrow can is easier to hold. Optimisation gives the best answer to the question the model asks, so the model must include whatever matters in reality.

Shortest distance to a curve

The distance between two points involves a square root, which makes differentiation messy. Since distance is never negative, the distance $D$ is least exactly when $D^2$ is least, so it is simpler to minimise $D^2$.

When the optimum is at an endpoint

Not every optimisation problem has its answer at a stationary point. If the function has no maximum inside the domain, or the stationary point is a minimum when a maximum is wanted, the answer lies at an endpoint.

Every optimisation problem rests on a mathematical model. The function is a simplified description of reality, the domain records what is physically possible, and the derivative locates the best choice within those limits. A box made from real card loses some volume to the thickness of the material, and the can model ignored the cost of thicker lids, so an optimum found from a model should be checked against the situation it describes.

Exam-style questions on stationary points and optimisation

The first question is typical of a non-calculator paper. It combines a geometric "show that" with trigonometric differentiation and a justification.

The second question allows a GDC. It mixes stationary points, a point of inflexion and the global behaviour of a model on a closed interval.

Self-check

• To do:

  I can find the coordinates of the stationary points of a function by solving $f'(x) = 0$ and substituting into $f(x)$.

• To do:

  I can classify a stationary point as a local maximum, local minimum or stationary point of inflexion with the first derivative test.

• To do:

  I can use the second derivative test, and I know it is inconclusive when $f''(a) = 0$.

• To do:

  I can find constants in a function from information about its stationary points.

• To do:

  I can find the greatest and least values of a function on a closed interval, including values at the endpoints.

• To do:

  I can find points of inflexion by solving $f''(x) = 0$ and checking that $f''(x)$ changes sign, and decide whether each is stationary.

• To do:

  I can read stationary points and points of inflexion of $f$ from the graph of $f'$.

• To do:

  I can set up an optimisation problem using a constraint and a domain, solve it, and justify that my answer is a maximum or minimum.

• To do:

  I can interpret the result of an optimisation in context, with units, and comment on the limitations of the model.

1. (a)

   Find $f'(x)$.

   [2]2 marks

   Markscheme (a)

   $f'(x) = 3x^2 - 12x + 9$ A1A1

   Award A1 for each two correct terms.

2. (b)

   Find the coordinates of the two stationary points of the graph of $f$.

   [4]4 marks

   Markscheme (b)

   Setting $f'(x) = 0$: $3(x-1)(x-3) = 0$

3. (c)

   Find the equation of the tangent to the graph of $f$ at the point where $x = 4$.

   [3]3 marks

   Markscheme (c)

   $f'(4) = 48 - 48 + 9 = 9$

So $x = 2$ or $x = -1$. Substitute each value into $f(x)$.

The stationary points are $(-1, 12)$ and $(2, -15)$.

The gradient changes from positive to negative at $x = -1$, so $(-1, 12)$ is a local maximum. It changes from negative to positive at $x = 2$, so $(2, -15)$ is a local minimum.

Solve $f'(x) = 0$.

The $y$-coordinates are $f(2) = 2 + 2 = 4$ and $f(-2) = -2 - 2 = -4$.

The second derivative is

$f''(2) = 1 > 0$, so $(2, 4)$ is a local minimum. $f''(-2) = -1 < 0$, so $(-2, -4)$ is a local maximum.

Since $e^{-x} > 0$ for every $x$, $f'(x) = 0$ only when $x = 1$, and $f(1) = e^{-1}$. The stationary point is $\left(1, \frac{1}{e}\right)$.

Differentiate again, using the product rule on $e^{-x}(1 - x)$.

$f''(1) = e^{-1}(-1) = -\frac{1}{e} < 0$, so the point is a maximum.

(b) $g'(x) = 1 - 2\cos x$. Solve $g'(x) = 0$.

The second derivative is $g''(x) = 2\sin x$.

$g''\left(\frac{\pi}{3}\right) = \sqrt{3} > 0$, so $\left(\frac{\pi}{3}, \frac{\pi}{3} - \sqrt{3}\right)$ is a local minimum.

$g''\left(\frac{5\pi}{3}\right) = -\sqrt{3} < 0$, so $\left(\frac{5\pi}{3}, \frac{5\pi}{3} + \sqrt{3}\right)$ is a local maximum.

The point is stationary, so the derivative is zero at $x = 1$.

The point lies on the curve, so $y = -4$ when $x = 1$.

Equation (1) gives $2a + b = -3$. Subtracting (2) from this gives $a = 2$, and then $b = -7$.

(b) Now $\frac{dy}{dx} = 3x^2 + 4x - 7 = (3x + 7)(x - 1)$, so the other stationary point is at $x = -\frac{7}{3}$.

The second derivative is $\frac{d^2y}{dx^2} = 6x + 4$.

At $x = 1$: $\frac{d^2y}{dx^2} = 10 > 0$, so $(1, -4)$ is a local minimum.

At $x = -\frac{7}{3}$: $\frac{d^2y}{dx^2} = -14 + 4 = -10 < 0$, so this point is a local maximum. Its $y$-coordinate is $\left(-\frac{7}{3}\right)^3 + 2\left(-\frac{7}{3}\right)^2 - 7\left(-\frac{7}{3}\right) = \frac{392}{27}$.

(b) When $k = 3$, $f'(x) = 3x^2 - 6x + 3 = 3(x - 1)^2$. This is zero only at $x = 1$, so there is exactly one stationary point, at $(1, f(1)) = (1, 1)$.

Since $3(x - 1)^2 > 0$ for every $x \neq 1$, the gradient is positive on both sides of $x = 1$. The point $(1, 1)$ is a stationary point of inflexion.

The second derivative test would not help here: $f''(x) = 6x - 6$ gives $f''(1) = 0$.

Evaluate $f$ at the stationary points and the endpoints.

The greatest value is $56$, at the endpoint $x = 6$. The least value is $-25$, at the local minimum $x = 3$. The range is $-25 \le f(x) \le 56$.

• y = x⁴ − 6x² + 2

(b) The only solution of $f''(x) = 0$ is $x = 1$. For every $x \neq 1$, $(x - 1)^2 > 0$, so $f''(x) > 0$ on both sides of $x = 1$. The second derivative does not change sign, so the curve is concave up on both sides and $x = 1$ is not a point of inflexion. Since there are no other candidates, the graph has no point of inflexion.

The point $(1, 1)$ is not stationary either, because $f'(1) = 1 \neq 0$.

At $x = -2$, the graph of $f'$ crosses from negative to positive, so $f$ has a local minimum.

At $x = 1$, the graph of $f'$ touches the axis and stays positive on both sides, so $f$ has a stationary point of inflexion.

(b) Points of inflexion of $f$ are where $f'$ has turning points, that is, where $f''(x) = 0$ and changes sign.

This is zero at $x = -1$ (the local maximum of $f'$) and $x = 1$ (the local minimum of $f'$), and changes sign at both. The points of inflexion of $f$ are at $x = -1$ and $x = 1$. The one at $x = 1$ is the stationary point of inflexion found in (a).

(b) The cut must have positive length, and the shorter side of the base, $15 - 2x$, must be positive. So $0 < x < 7.5$.

(c) Differentiate and solve $\frac{dV}{dx} = 0$.

So $x = 3$ or $x = 10$. The value $x = 10$ is outside the domain $0 < x < 7.5$ (it would need a cut wider than half of the 15 cm side), so it is rejected.

Check the nature with the second derivative.

So $x = 3$ gives a maximum.

(d) $V = 3(24 - 6)(15 - 6) = 3 \times 18 \times 9 = 486$. The maximum volume is $486\ \text{cm}^3$, when $3$ cm squares are cut from the corners.

(b) The domain is $r > 0$. Write $A = 2\pi r^2 + 660r^{-1}$ and differentiate.

Set this equal to zero.

The second derivative is

which is positive for every $r > 0$. So $r = 3.74$ cm (3 s.f.) gives a minimum.

(c) $h = \frac{330}{\pi (3.7449\ldots)^2} = 7.4898\ldots$, so $h = 7.49$ cm. The minimum area is

The least surface area is $264\ \text{cm}^2$ (3 s.f.), for a can of radius $3.74$ cm and height $7.49$ cm.

Adding the two areas gives $A(x)$. The domain includes $0$ and $20$, where all the wire goes into one shape.

(b) Differentiate, using the chain rule on $(20 - x)^2$.

Set $A'(x) = 0$ and multiply by $8\pi$.

$A''(x) = \frac{1}{8} + \frac{1}{2\pi} > 0$, so this is a minimum. The minimum area is

when $11.2$ cm is used for the square.

(c) The only stationary point is a minimum, so the maximum must be at an endpoint.

The maximum total area is $\frac{100}{\pi} \approx 31.8\ \text{cm}^2$, when the wire is not cut and all of it forms a circle.

(b) Expand first: $A = 100\sin\theta + 100\sin\theta\cos\theta$. Use the product rule on $\sin\theta\cos\theta$.

Since $\cos^2\theta - \sin^2\theta = \cos 2\theta$,

(c) Set the derivative equal to zero and use $\cos 2\theta = 2\cos^2\theta - 1$.

So $\cos\theta = \frac{1}{2}$ or $\cos\theta = -1$. In the domain $0 < \theta \le \frac{\pi}{2}$, $\cos\theta = -1$ has no solution, so $\theta = \frac{\pi}{3}$.

Justify with the first derivative test, using exact values on each side.

At $\theta = \frac{\pi}{4}$: $\frac{dA}{d\theta} = 100\left(\frac{\sqrt{2}}{2} + \cos\frac{\pi}{2}\right) = 50\sqrt{2} > 0$.

At $\theta = \frac{\pi}{2}$: $\frac{dA}{d\theta} = 100(0 + \cos\pi) = -100 < 0$.

The gradient changes from positive to negative, so $\theta = \frac{\pi}{3}$ gives a maximum.

(d)

The maximum cross-sectional area is $75\sqrt{3}\ \text{cm}^2$ (about $130\ \text{cm}^2$).

$h(20) = -8 + 36 - 48 + 30 = 10$ and $h(40) = -64 + 144 - 96 + 30 = 14$.

$h''(x) = -0.006x + 0.18$. Then $h''(20) = 0.06 > 0$, so $(20, 10)$ is a local minimum, and $h''(40) = -0.06 < 0$, so $(40, 14)$ is a local maximum.