Stationary Points: Maxima and Minima

You are cycling up a long hill. The bottom is brutally steep, and every few seconds it eases off a little, until one turn of the pedals where your legs stop working for a living. The road has gone flat. That is the top. Two completely different questions live on that same hill. One asks where the road is flat. The other asks where the road is climbing at exactly one in ten, the sort of number printed on the warning sign at the bottom. One piece of maths answers both, and the only thing that changes between them is a single number on one side of an equals sign. Mix the two up and every line you write afterwards is correct working aimed at a question nobody asked you.

So the first job is telling those two questions apart on paper, before any algebra starts. Worth knowing where this turns up: calculus is examined on the Edexcel International GCSE Higher papers and on the Cambridge IGCSE Extended papers. On a specification without calculus, a curve's turning point is found by completing the square, and the video on completing the square covers that method. When you differentiate a curve you get d y by d x, and that is not just an answer, it is a gradient machine. Feed it a value of x and it hands back how steep the curve is at that exact place. Building that machine is the job of the video called Differentiating Powers of x, so from here it is assumed done. The machine runs in two directions. Give it an x, and a gradient comes out. Or fix the gradient at the number you want, and solve the equation to find the x that produces it. That is why questions on this come in three shapes, and they are easy to tell apart once you know there are only three. Find the gradient at x equals two. Find where the gradient equals eight. Find the stationary point.

Those three are on screen now. Only one of them is asking you to set d y by d x equal to zero. Which is it: the first, the second, or the third? Have a think. I'll wait. The answer is the third one. A stationary point is a place where the curve has levelled off, so the gradient there is zero, and zero is what the derivative gets set equal to. The first one hands you the x, so nothing is being set equal to anything. You substitute two into the derivative and read off the gradient that comes back. The second hands you the gradient instead, so you write the derivative equals eight and solve that equation for x. Same machine, run the other way round.

Which gives you a line to carry into the exam hall: zero only when it says flat. If the question has not told you the curve is level, stationary, turning, at a maximum or at a minimum, then zero has no business sitting on the right of your equals sign. The left-hand side of that equation is always the derivative. It is the right-hand side that the question chooses for you.

Time to run the harder of the two, the one where the question names a gradient that is not zero. A cinema models its brightness level b at h hours after opening as b equals h cubed, minus twenty seven h, plus five. You are asked to find d b by d h, and then to find the value, or values, of h where the rate of change of brightness equals six. Differentiating term by term gives d b by d h equals three h squared minus twenty seven. The letters have changed because this curve is b against h rather than y against x, and the derivative just takes their names.

That derivative is the gradient machine, and the question has already told you what it wants the gradient to be. So take it from there. Set three h squared minus twenty seven equal to six, solve for h, and then decide which of the answers can actually be a time. Have a go at this one. I'll wait. Here is the working. Three h squared minus twenty seven equals six. Add twenty seven to both sides and three h squared equals thirty three. Divide by three and h squared equals eleven, so h is the square root of eleven, or minus the square root of eleven. The context does the last piece of work. h counts hours after the cinema opened, so a negative time is not on offer here, and the answer is h equals the square root of eleven. Leave it as root eleven unless the question asks for a decimal, because root eleven is exact and a decimal is only a rounded copy. Root eleven is about three point three two, so a little over three and a quarter hours after opening.

Now look at what the reflex answer would have been. Set that same derivative to zero instead of six, and you get h squared equals nine, so h equals three. Three hours. It is a tidy whole number, it sits right beside the correct answer, and nothing about it looks wrong on the page. The danger is that the mistake does not produce nonsense, it produces a believable number for a question you were never asked.

On to the other kind, where the question wants the flat spot itself. A company models its weekly profit p as minus two t squared, plus sixteen t, minus ten, where t is the number of weeks since a product launched. Find the stationary point of this curve, and state whether it is a maximum or a minimum. Differentiating gives d p by d t equals minus four t plus sixteen, and that expression is the gradient of the profit curve in any week t. The words stationary point are the question telling you the gradient. At the very top of a curve, or the very bottom, the curve has stopped rising and has not yet started falling, so for that instant it is level. Level is a gradient of zero. So set minus four t plus sixteen equal to zero. That gives sixteen equals four t, and t equals four, four weeks after the launch. But t equals four is only half an answer, because a point needs two coordinates. Put four back into the original profit equation, not into the derivative, and p comes out as twenty two. The stationary point is four, twenty two. That leaves the last word, maximum or minimum. Test the gradient on either side of week four: put t equals three into the derivative, then put t equals five in, and watch which way the signs go.

Take your time. I'll wait right here. At t equals three the derivative gives four, a positive gradient, so the curve is still climbing. At t equals five it gives minus four, a negative gradient, so the curve is falling. Climbing, then level, then falling, is a maximum.

That sign test works on any curve you are ever handed, not only on this one, which is why it is worth doing even when you think you can picture the shape. So the answer is the point four, twenty two, and it is a maximum. Write the coordinates and the word maximum down together, because the question asked for both.

Here is what that mix-up looks like from the marking side of a real paper. This line comes from an examiner report, on a question where a cubic had to be differentiated and the gradient then set equal to a given value. Quote: Some students set d y by d x equal to zero and therefore could not gain this third method mark. Notice where in the working it went wrong. It was the third method mark, not the first, so the differentiating had already gone in correctly. The whole thing turned on what those students wrote after the equals sign. So here is the habit that stops it. Before you touch the algebra, find the gradient word in the question and underline it. If that word is a number, the number goes on the right. If the words are stationary, turning, maximum or minimum, zero goes on the right. Two more things the wording hands you on the same read-through. Value or values, with the s in brackets, is a quiet warning that the equation may have more than one solution, which is exactly what root eleven and its negative were. And when a question says find the stationary point, the word point is asking for a pair of coordinates. Stopping at the x value leaves the answer half written.

Zero only when it says flat. That is the chant, and here is the method it belongs to, from the top. Differentiate first, whatever the question turns out to be asking, because that is what gives you the gradient machine for the curve. If the question hands you an x, substitute it into the derivative and the gradient drops straight out. If the question hands you a gradient, put that number on the right and solve for x. A stationary point is the same job with zero as the number, because flat is a gradient like any other. Then finish the answer properly. Substitute back into the original equation for the second coordinate, throw out any root the context forbids, and test the gradient either side to name it a maximum or a minimum. Get that one decision right at the start, and everything after it is differentiating you can already do.

Next in the chapter: Applying Calculus to Kinematics Problems, where the curve becomes distance against time, and a gradient of zero becomes the moment something is momentarily at rest.

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