Lesson 04 · Unit 12 · Differentiation
Stationary points
Lesson 03 ended on a curve whose tangent came out flat. That was not an accident. The places where a curve goes momentarily level are where its peaks and valleys are — and finding them is what the rest of this topic is built on.
A stationary point is a point on a curve where \(\dfrac{dy}{dx} = 0\). The tangent there is horizontal. For an instant the curve is neither climbing nor falling.
There are two kinds you will meet. At a maximum the curve climbs, levels off and falls. At a minimum it falls, levels off and climbs.
- the curve \(y=\tfrac14x^{3}-3x\)
- the tangent at P, and negative gradient on the strip
- the stationary points, and positive gradient on the strip
Slide P from left to right. The gradient starts positive, passes through zero at \((-2,\,4)\), goes negative, passes through zero again at \((2,\,-4)\), and turns positive. The strip along the bottom records that sign as you go.
A stationary point is a point, so it has two coordinates. A question that says “find the stationary points” wants \((-2,\,4)\) and \((2,\,-4)\), not \(x = -2\) and \(x = 2\). Half-answers like that are among the easiest marks to throw away in the whole paper.
Points of inflexion are not on the 0606 syllabus, so you will not be asked to classify one. But a stationary point that is neither a maximum nor a minimum can still turn up — \(y = x^{3}\) has one at the origin. If a test ever gives you nothing, tab 4 explains what to do.
The words questions use
- stationary point — \(\dfrac{dy}{dx} = 0\)
- turning point — a stationary point where the curve actually turns
- maximum / minimum — which way it turns
- nature of the stationary point — an instruction to say which it is
- greatest / least value — the \(y\)-coordinate there, not the \(x\)
The method
- Differentiate.
- Set \(\dfrac{dy}{dx} = 0\).
- Solve for \(x\) — usually by factorising.
- Put each \(x\) back into the curve to get \(y\).
Step 4 uses the original equation, exactly as it did for tangents.
- \(\dfrac{dy}{dx} = 6x^{2} - 18x + 12\)
- \(6x^{2} - 18x + 12 = 0\)stationary means zero gradient
- \(6(x^{2} - 3x + 2) = 0\)take the 6 out first — it makes the factorising easy
- \(6(x-1)(x-2) = 0 \ \Rightarrow\ x = 1 \text{ or } x = 2\)
- \(x=1:\ y = 2 - 9 + 12 - 3 = 2\)into the curve
- \(x=2:\ y = 16 - 36 + 24 - 3 = 1\)
- Stationary points: \((1,\,2)\) and \((2,\,1)\)coordinates, as pairs
Factor out the common number before factorising. \(6x^{2}-18x+12\) looks unpleasant; \(6(x-1)(x-2)\) takes five seconds. On a no-calculator paper this habit saves real time.
Find the stationary point of \(y = x + \dfrac{9}{x}\) for \(x > 0\).
- \(y = x + 9x^{-1}\)rewrite first
- \(\dfrac{dy}{dx} = 1 - 9x^{-2} = 1 - \dfrac{9}{x^{2}}\)
- \(1 - \dfrac{9}{x^{2}} = 0 \ \Rightarrow\ x^{2} = 9\)multiply through by \(x^{2}\)
- \(x = 3\), since \(x > 0\)the condition rules out \(x=-3\)
- \(y = 3 + 3 = 6\), so \((3,\,6)\)
When a question states a restriction such as \(x > 0\), it is there to be used. Giving both roots when one has been excluded loses a mark even though your algebra was right.
Differentiating gives \(3x^{2} - 6x + k\). What is \(k\)?
Setting it to zero and factorising gives two values of \(x\). What is the smaller one?
And the larger one?
Find \(y\) when \(x = -1\).
Find \(y\) when \(x = 3\).
- \(y = x^{3} - 3x^{2} + 4\). Find the smaller of the two values of \(x\) at which the curve is stationary.
- The same curve. Find the larger one.
- The same curve. Find \(y\) at that larger value of \(x\).
- \(y = x^{3} + 3x^{2} - 24x\). Find the positive value of \(x\) at which the curve is stationary.
- The same curve. Find the negative one.
- The same curve. Find \(y\) at \(x = 2\).
- \(y = 2x^{3} - 3x^{2} - 12x + 1\). Find the smaller of its two stationary values of \(x\).
- The same curve. Find \(y\) there.
- \(y = x^{3} - 12x\). Find the positive stationary value of \(x\).
- The same curve. Find \(y\) at \(x = -2\).
- \(y = x^{3} + 6x^{2} + 9x + 2\). Find the larger stationary value of \(x\).
- \(y = x^{3} - 3x + 5\). Find \(y\) at the stationary point where \(x = 1\).
- \(y = x^{3} + 3x + 1\). How many stationary points does this curve have? (Differentiate first, then look at the discriminant.)
- \(y = x^{3} - 3x^{2} + 3x - 1\). This curve has exactly one stationary point. Find its \(x\)-coordinate.
- \(y = x^{3} + ax^{2}\) is stationary at \(x = 2\) as well as at \(x = 0\). Find \(a\).
- \(y = 2x^{3} + 9x^{2} + 12x + 4\). Find the smaller stationary value of \(x\).
Two of the last four questions break the pattern of the first twelve: one of those curves has no stationary point at all, and another has exactly one. A cubic is not obliged to have two — that depends entirely on whether \(\dfrac{dy}{dx} = 0\) has two roots, one, or none.
Knowing where a curve is flat is not the same as knowing whether it is a peak or a valley. There are two ways to decide. This is the first, and it always works.
The first derivative test
Look at the sign of \(\dfrac{dy}{dx}\) just before and just after the point.
- \(+\) then \(-\) : the curve climbs then falls — a maximum
- \(-\) then \(+\) : the curve falls then climbs — a minimum
Go back to the model in tab 1 and read the strip along the bottom. That strip is this test.
For \(y = 2x^{3} - 9x^{2} + 12x - 3\) we found \(\dfrac{dy}{dx} = 6(x-1)(x-2)\). The factorised form makes the signs easy to read without a calculator.
| \(x\) | 0.9 | 1 | 1.1 |
|---|---|---|---|
| \(6(x-1)(x-2)\) | \(6(-0.1)(-1.1)\) | 0 | \(6(0.1)(-0.9)\) |
| sign | \(+\) | 0 | \(-\) |
\(+\) then \(-\), so \((1,\,2)\) is a maximum.
| \(x\) | 1.9 | 2 | 2.1 |
|---|---|---|---|
| \(6(x-1)(x-2)\) | \(6(0.9)(-0.1)\) | 0 | \(6(1.1)(0.1)\) |
| sign | \(-\) | 0 | \(+\) |
\(-\) then \(+\), so \((2,\,1)\) is a minimum.
You only need the sign, never the value. Writing \(6(-0.1)(-1.1)\) and saying “positive” is a complete answer. Multiplying it out to \(0.66\) is extra arithmetic and one more place to slip.
Choose test values close to the point, and never past the other stationary point. If a curve is stationary at \(x=1\) and at \(x=2\), then testing \(x=0\) and \(x=3\) steps straight over the second point. Both come out with the same sign, and you would wrongly conclude that \(x=1\) is not a turning point at all.
One curve, both of its stationary points, four test values and two verdicts. For steps 1 to 4 enter \(1\) for a positive gradient and \(-1\) for a negative one; steps 5 and 6 have their own code, given in the question.
Sign of \(\dfrac{dy}{dx}\) at \(x = -1.1\)?
Sign at \(x = -0.9\)?
Sign at \(x = 2.9\)?
Sign at \(x = 3.1\)?
Enter 1 if \(x=-1\) is a maximum, or 2 if it is a minimum.
Enter 1 if \(x=3\) is a maximum, or 2 if it is a minimum.
There is a faster way, and in most questions it is the one to use.
The second derivative test
At a stationary point, find \(\dfrac{d^{2}y}{dx^{2}}\).
- \(\dfrac{d^{2}y}{dx^{2}} < 0\) : maximum
- \(\dfrac{d^{2}y}{dx^{2}} > 0\) : minimum
- \(\dfrac{d^{2}y}{dx^{2}} = 0\) : the test tells you nothing — go back to the first derivative test
The second derivative measures how the gradient itself is changing. At a peak the gradient is falling from positive to negative, so the second derivative is negative. At a valley it is rising, so the second derivative is positive.
A way to remember which is which
Negative curves downwards, like the top of a hill: \(\cap\). Positive curves upwards, like the bottom of a valley: \(\cup\). The sign is literally the shape.
- \(\dfrac{dy}{dx} = 6x^{2} - 18x + 12\)from before
- \(\dfrac{d^{2}y}{dx^{2}} = 12x - 18\)differentiate again
- At \(x=1:\ 12 - 18 = -6 < 0\), so \((1,\,2)\) is a maximum
- At \(x=2:\ 24 - 18 = 6 > 0\), so \((2,\,1)\) is a minimum
The same conclusions as the sign tables in the last tab, in a quarter of the space.
Write the inequality, not just the number. “\(-6\)” on its own is an incomplete argument; “\(-6 < 0\), therefore maximum” is the answer the mark scheme is looking for. The reasoning is what earns the mark, not the arithmetic.
If the second derivative comes out as zero, you have learned nothing — it does not mean the point is neither. Go back and test the sign of \(\dfrac{dy}{dx}\) on each side. Knowing that the test can fail, and what to do about it, is itself examinable.
- Conclude it is neither a maximum nor a minimum
- Go back and test the sign of \(\dfrac{dy}{dx}\) on each side
- Conclude it is a maximum
- Conclude it is a minimum
Setting \(\dfrac{dy}{dx} = 0\) gives two values. What is the smaller?
And the larger?
Find \(\dfrac{d^{2}y}{dx^{2}}\) at \(x = 0\).
And at \(x = 4\)?
Enter 1 if \(x=0\) is a maximum, or 2 if it is a minimum.
- \(y = x^{3} - 3x^{2} + 1\) is stationary at \(x = 2\). Find \(\dfrac{d^{2}y}{dx^{2}}\) there.
- The same curve is also stationary at \(x = 0\). Find \(\dfrac{d^{2}y}{dx^{2}}\) there.
- So which of those two is the maximum? Give its \(x\)-coordinate.
- \(y = x^{3} + 3x^{2} - 9x\) is stationary at \(x = 1\). Find \(\dfrac{d^{2}y}{dx^{2}}\) there.
- The same curve is stationary at \(x = -3\). Find \(\dfrac{d^{2}y}{dx^{2}}\) there.
- \(y = x + \dfrac{4}{x}\) is stationary at \(x = 2\). Find \(\dfrac{d^{2}y}{dx^{2}}\) there.
- The same curve is also stationary at \(x = -2\). Find \(\dfrac{d^{2}y}{dx^{2}}\) there.
- So the curve \(y = x + \dfrac{4}{x}\) has a local minimum at \(x = 2\) and a local maximum at \(x = -2\), where \(y = -4\). Enter the value of \(y\) at the minimum.
- \(y = x^{2} + \dfrac{16}{x}\) is stationary at \(x = 2\). Find \(\dfrac{d^{2}y}{dx^{2}}\) there.
- \(y = 2x^{3} - 3x^{2} - 36x\) is stationary at \(x = 3\). Find \(\dfrac{d^{2}y}{dx^{2}}\) there.
- The same curve is stationary at \(x = -2\). Find \(\dfrac{d^{2}y}{dx^{2}}\) there.
- \(y = x^{4} - 2x^{2}\) is stationary at \(x = 0\). Find \(\dfrac{d^{2}y}{dx^{2}}\) there.
- The same curve is stationary at \(x = 1\). Find \(\dfrac{d^{2}y}{dx^{2}}\) there.
- \(y = x^{4}\) is stationary at \(x = 0\). Find \(\dfrac{d^{2}y}{dx^{2}}\) there — and notice what the test then tells you.
The last question is the honest limit of the test. \(y = x^{4}\) has a clear minimum at the origin, and yet the second derivative there does not carry the sign the test needs. Whenever it comes out zero, the test has told you nothing and you must fall back on the sign of \(\dfrac{dy}{dx}\) on either side.
Question. Find the nature of the stationary point of \(y = x^{2} - 8x + 3\).
- \(\dfrac{dy}{dx} = 2x - 8\)
- \(2x - 8 = 0 \ \Rightarrow\ x = 4\)
- \(\dfrac{d^{2}y}{dx^{2}} = 2\)
- \(2 > 0\), so the point is a maximum
- \(y = 16 - 32 + 3 = -13\)
- Maximum at \((4,\,-13)\)
A curve has equation \(y = x^{3} - 12x + 5\).
- (a)Find the coordinates of the two stationary points.[4]
- (b)Determine the nature of each.[3]
- (a) \(\dfrac{dy}{dx} = 3x^{2} - 12\)
- \(3x^{2} - 12 = 0 \ \Rightarrow\ x^{2} = 4 \ \Rightarrow\ x = \pm 2\)both roots
- \(x=2:\ y = 8 - 24 + 5 = -11\)
- \(x=-2:\ y = -8 + 24 + 5 = 21\)
- \((2,\,-11)\) and \((-2,\,21)\)
- (b) \(\dfrac{d^{2}y}{dx^{2}} = 6x\)
- At \(x=2:\ 12 > 0\), minimum
- At \(x=-2:\ -12 < 0\), maximum
\(x^{2} = 4\) has two solutions. Writing only \(x = 2\) loses half of part (a) and a mark in part (b) — three marks gone from a missing sign.
A curve has equation \(y = x + \dfrac{25}{x}\) for \(x > 0\).
- (a)Show that the curve has a stationary point at \(x = 5\).[3]
- (b)Show that this point is a minimum, and write down the least value of \(y\).[3]
- (a) \(y = x + 25x^{-1}\ \Rightarrow\ \dfrac{dy}{dx} = 1 - 25x^{-2}\)
- \(1 - \dfrac{25}{x^{2}} = 0 \ \Rightarrow\ x^{2} = 25\)
- \(x = 5\) since \(x > 0\)the restriction rules out \(-5\)
- (b) \(\dfrac{d^{2}y}{dx^{2}} = 50x^{-3} = \dfrac{50}{x^{3}}\)differentiate \(-25x^{-2}\)
- At \(x=5:\ \dfrac{50}{125} = 0.4 > 0\), so minimum
- least value \(y = 5 + 5 = 10\)the \(y\)-coordinate, not the \(x\)
“Least value” means the value of \(y\). Answering “\(x = 5\)” to that part is a very common and very expensive misreading.
A curve has equation \(y = x^{2}e^{-x}\).
- (a)Show that \(\dfrac{dy}{dx} = xe^{-x}(2 - x)\).[3]
- (b)Find the \(x\)-coordinates of the two stationary points.[2]
- (c)Find the \(y\)-coordinate of the stationary point at \(x=2\), correct to 3 significant figures.[2]
- (a) \(u = x^{2},\ v = e^{-x}\)product rule
- \(\dfrac{du}{dx} = 2x,\quad \dfrac{dv}{dx} = -e^{-x}\)chain rule on \(v\): the inside derivative is \(-1\)
- \(\dfrac{dy}{dx} = -x^{2}e^{-x} + 2xe^{-x} = xe^{-x}(2-x)\)as required
- (b) \(e^{-x}\) is never zero, so \(x = 0\) or \(x = 2\)that observation is worth stating
- (c) \(y = 4e^{-2} = 0.541\)3 s.f.
\(e^{\text{anything}}\) is never zero. Saying so explicitly is what lets you divide it out and keep only \(x(2-x) = 0\). Candidates who try to “solve” \(e^{-x} = 0\) waste time on an equation with no solutions.
A curve has equation \(y = 2\sin x + x\) for \(0 \le x \le 2\pi\).
- (a)Find \(\dfrac{dy}{dx}\).[1]
- (b)Find the exact values of \(x\) at the stationary points.[3]
- (c)Determine the nature of each.[3]
- (a) \(\dfrac{dy}{dx} = 2\cos x + 1\)
- (b) \(2\cos x + 1 = 0 \ \Rightarrow\ \cos x = -\tfrac12\)
- \(x = \dfrac{2\pi}{3}\) and \(x = \dfrac{4\pi}{3}\)both solutions in \(0 \le x \le 2\pi\), in radians
- (c) \(\dfrac{d^{2}y}{dx^{2}} = -2\sin x\)
- At \(\tfrac{2\pi}{3}:\ -2\left(\tfrac{\sqrt3}{2}\right) = -\sqrt3 < 0\), maximum
- At \(\tfrac{4\pi}{3}:\ -2\left(-\tfrac{\sqrt3}{2}\right) = \sqrt3 > 0\), minimum
A trigonometric equation over a stated interval usually has more than one solution. Sketch \(\cos x\) mentally and count how many times it hits \(-\tfrac12\) between \(0\) and \(2\pi\) — twice — before you decide you have finished.
You should now be able to
- Explain what \(\dfrac{dy}{dx} = 0\) means about a curve
- Find the coordinates of stationary points, as pairs
- Use the first derivative test with a sign table
- Use the second derivative test, and state the inequality
- Say what to do when the second derivative test gives zero
- Handle stationary points of expressions involving \(e^{x}\) and trigonometric functions
- Answer the question actually asked — greatest value, least value, or coordinates
Lesson 05 takes this machinery outside and points it at boxes, tanks and fences.