In secondary school, you may have been taught that to find the maximum and minimum points of f , simply follow this procedure:
The Incorrect Recipe for Finding Maximum and Minimum Points.
Given a differentiable function f ∶ D → R,
1. Compute f′(x). Find the points x at which f′(x) = 0.
2. These points are also the maximum and minimum points. (If we also want to know which are maximum and which are minimum points, then simply employ some method like sketch-the-graph or the Second Derivative Test.)
Unfortunately, the above procedure (let’s call it the Incorrect Recipe) may sometimes fail. It rests on the false belief that “f′(x) = 0 ⇐⇒ x is an extremum”. This is false because
1. The IET does NOT say, “f′(x) = 0 Ô⇒ x is an extremum.” It is perfectly possible that f′(x) = 0 without x being an extremum.
2. The IET does NOT say, “x is an extremum Ô⇒ f′(x) = 0 .” Instead, it says, “x is an extremum AND an interior point Ô⇒ f′(x) = 0.” Thus, it is perfectly possible that x is an extremum without f′(x) = 0.
Here is an example to illustrate these two failings of the Incorrect Recipe.
Example 144 (revisited). Graphed below is the function f ∶ [−1.5, 0.5] → R defined by x↦ x5+ 2x4+ x3. Five points are labelled.
According to the Incorrect Recipe,
1. Compute f′(x) = 5x4 + 8x3 + 3x2 = x2(5x2+ 8x + 3) = x2(5x + 3)(x + 1). We see that f′(x) = 0 ⇐⇒ x = −3
5,−1, 0.
2. So −3
5,−1, 0 are the maximum and minimum points of f.
The Incorrect Recipe does correctly identify the points B = (−1, f(−1)) and C = (−3
5, f(−3
5)) as maximum and minimum points, respectively. But it makes two mistakes.
Mistake #1: D = (0, 0) is neither a maximum nor a minimum point, contrary to the Incorrect Recipe.
Mistake #2: A and E are respectively a minimum and a maximum point, but neither is detected by the Incorrect Recipe.
B D
f (x) = x
5+ 2x
4+ x
3x y
A
C
E
The Correct Recipe for Finding Maximum and Minimum Points.
Given a differentiable function f ∶ D → R,
1. Identify all the stationary points (i.e. x where f′(x) = 0).
2. Identify all the non-interior points.
3. Check to see if each stationary point and each non-interior point is a maximum point, a minimum point, or neither. (To do so, employ some method like sketch-the-graph or the Second Derivative Test.)
The Correct Recipe rectifies the Incorrect Recipe in two ways:
1. The Correct Recipe demands that you also check the non-interior points, which may possibly be maximum or minimum points, but may not be detected by the Incorrect Recipe.
2. The Correct Recipe does not assume that every single one of our shortlist of points (the stationary points and the non-interior points) is either a maximum point or a minimum point. It allows for the possibility that some of these points could be neither.
By the way, the condition that f is differentiable is very important. If f is not differentiable, then the above Correct Recipe might not work. But not to worry, since most functions on the A-levels are usually differentiable.
Example 124. Consider f ∶ [−1, 1] → R defined by x ↦ x3. Let’s apply the Correct Recipe.
1. Identify all the stationary points (i.e. x where f′(x) = 0).
f′(x) = 3x2. So f′(x) = 0 ⇐⇒ x = 0. The only stationary point is x = 0.
2. Identify all the non-interior points.
Every point x∈ (−1, 1) is in the interior of [−1, 1]. The only non-interior points are −1 and 1.
3. Check if each of these points is a maximum point, a minimum point, or neither.
From a sketch of the graph, we see that the stationary point x = 0 is neither a maximum nor a minimum point. The non-interior point −1 is a minimum point. The non-interior point 1 is a maximum point.
Altogether, we conclude that −1 is the only minimum point and 1 is the only maximum point.
Exercise 68. For each of the following functions, find all the maximum and minimum points using the Correct Recipe. (Answer on p. 1029.)
(a) f ∶ R → R defined by x ↦ x.
(b) g∶ [0, 1] → R defined by x ↦ x.
(c) h ∶ R → R defined by x ↦ x4 − 2x2. Identify also the global minimum point(s) of h (if any exist).
12 Concavity, Inflexion Points, and the 2DT
• A function is concave downwards (or simply concave) on an interval if the line segment connecting any two points of the graph in this interval is below the graph.
• A function is concave upwards (or simply convex) on an interval if the line segment connecting any two points of the graph in this interval is above the graph.
• An inflexion point is any point where the concavity of the function changes, either from downwards to upwards, or upwards to downwards.22
Example 125. Graphed below is f ∶ R → R defined by x ↦ x3.
f is concave downwards on R−0 because there, the line segment connecting any two points on f is below the graph of f .
In contrast, f is concave upwards on R+0 because there, the line segment connecting any two points on f is above the graph of f .
0 is an inflexion point because this is where the function f changes from being concave downwards to being concave upwards.
A test for whether a point is an inflexion point is this: Draw the tangent line to the graph at that point. The point is an inflexion point ⇐⇒ The line is above the graph on one side of the point and below the graph on the other side (see Fact 95 in the Appendices).
The tangent line to the graph at the point 0 is drawn in green (it coincides with the horizontal axis). We indeed see that the line is above the graph on the left side of the point and below the graph on the right side of the point. Therefore, 0 is an inflexion point.
22These are informal definitions. For the formal definitions, see p. 964in the Appendices (optional).
For a graph to be concave downwards, its slope must be decreasing. Conversely, to be concave upwards, its slope must be increasing. Altogether then, the following proposition is intuitively plausible.
Proposition 2. Let f ∶ D → R be a twice-differentiable function.
(a) f is concave downwards on an interval ⇐⇒ f′′(x) ≤ 0 for every x in this interval.
(b) f is convex upwards on an interval ⇐⇒ f′′(x) ≥ 0 for every x in this interval.
(c) x is an inflexion point Ô⇒ f′′(x) = 0.
Proof. Optional, see p. 967 in the Appendices.
Example 151 (revisited). Consider f ∶ R → R defined by x ↦ x3. f is concave downwards on R−0, concave upwards on R+0, and has an inflexion point at x= 0.
We can verify that, as per the above proposition:
f′(x) = 3x2⎧⎪⎪⎪⎪⎪⎪
⎨⎪⎪⎪⎪⎪⎪⎩
< 0, for x ∈ R−0,
= 0, for x = 0,
> 0, for x ∈ R+0.
It is very tempting to believe that the converse of part (c) of the above proposition is true.
That is, it is very tempting to believe that
“f′′(x) = 0 Ô⇒ x is an inflexion point.”
But this is wrong! It is perfectly possible that f′′(x) = 0 without x being an inflexion point! Here’s an example:
Example 126. Consider g ∶ R → R defined by x ↦ x4. We have g′(x) = 4x3 and g′′(x) = 12x2, so that g′′(x) = 0 ⇐⇒ x = 0.
We might thus be tempted to conclude that 0 is an inflexion point. However, this is not the case. Although g′′(0) = 0, we have g′′(x) > 0 for x > 0 and we also have g′′(x) > 0 for x< 0, and so the concavity of g does not change at the point 0.
To qualify as an inflexion point, the concavity of the function must change. At 0, the concacivty of g does not change. Therefore, 0 is NOT an inflexion point.
y
x g is concave upwards everywhere
.
However, is not an
inflexion point of .
Inflexion points may be further sub-divided into stationary points of inflexion and non-stationary points of inflexion.
Definition 46. A stationary point of inflexion is simply any point that is both an inflexion point and a stationary point.
A non-stationary point of inflexion is simply any point that is an inflexion point, but not a stationary point.
The A-level syllabuses explicitly exclude non-stationary points of inflexion. Nonetheless, there is the temptation to believe that “every inflexion point must also be a stationary point”. Here’s a quick counter-example that dispels this belief:
Example 127. The graph below is for the function f ∶ R → R defined by x ↦ x3 + x.
We have f′(x) = 3x2 + 1 and f′′(x) = 6x. The point 0 is not a stationary point because f′(0) = 1 ≠ 0.
However, 0 is an inflexion point, because to the left of 0, f is concave downwards; and to the right, f is concave upwards. So 0 is a point of inflexion. Indeed, it is a non-stationary point of inflexion.
Also illustrated is the tangent line at y = x (whose slope is indeed non-zero). Observe that indeed, to the left of 0, the tangent line is above the graph; while to the right of 0, the tangent line is below the graph. This serves as a second way to verify that 0 is a point of inflexion.
y
x Concave upwards on
Concave downwards on
Tangent line at 0
12.1 The Second Derivative Test (2DT)
y
x f must be concave upwards
around the minimum turning point 0.
So for all x near 0.
y x
f must be concave downwards around the maximum
turning point 0.
So for all x near 0.
From graphs, it looks like around a maximum turning point a, f must be concave down-wards, i.e. f′′(a) < 0. Similarly, around a minimum turning point b, f must be concave upwards, i.e. f′′(b) > 0. The next proposition is thus intuitively plausible.
Proposition 3. (Second Derivative Test [2DT].) Let f be a twice-differentiable func-tion. Let a be a stationary point (i.e. f′(a) = 0).
1. If f′′(a) < 0, then a is a maximum point.
2. If f′′(a) > 0, then a is a minimum point.
3. If f′′(a) = 0, then the 2DT is uninformative. That is, a could be a maximum point, a minimum point, an inflexion point, or something else altogether!
Proof. Optional, see p. 968 in the Appendices.
The third part of the above Proposition must be heavily emphasised: If f′(a) = 0 and f′′(a) = 0, then the 2DT tells us absolutely nothing about a! a could be a maximum point, a minimum point, an inflexion point, or something else altogether!
We previously gave the Correct Recipe for finding maximum and minimum points. Let’s now add the 2DT to this recipe:
The Enriched Recipe for Finding Maximum and Minimum Points.
Given a twice-differentiable function f ∶ D → R,
1. Identify all the stationary points (i.e. a where f′(a) = 0).
(a) Evaluate f′′ at each of these points.
(b) f′′(a) < 0 Ô⇒ a is a maximum point. Conversely, f′′(a) > 0 Ô⇒ x is a minimum point. If f′′(a) = 0, then we need to determine the nature of a using some other method (e.g. sketch-the-graph).
2. Identify all the non-interior points.
(a) Check if each of these points is a maximum point, a minimum point, or neither.
If f is not twice-differentiable, then the Enriched Recipe may not work. Fortu-nately, most functions in A-levels are twice-differentiable.
Example 121 (revisited). Consider f ∶ [−1.5, 0.5] → R defined by x ↦ x5+ 2x4+ x3. 1. Identify all the stationary points. f′(x) = 5x4 + 8x3 + 3x2 = x2(5x2 + 8x + 3) = 0 ⇐⇒
x= 0 or x = −1, −0.6 (quadratic formula).
(a) f′′(x) = 20x3+ 24x2+ 6x = 2x(10x2+ 12x + 3).
(b) f′′(−0.6) > 0 Ô⇒ −0.6 is a minimum point. f′′(−1) < 0 Ô⇒ −1 is a maximum point. But f′′(0) = 0, so the 2DT tells us nothing. By sketching the graph (non-rigorous method that will suffice for the A-levels), we see that 0 is an inflexion point.
2. The only two non-interior points are −1.5 and 0.5. Again by sketching the graph, we see that −1.5 is a minimum point and 0.5 is a maximum point.
Altogether, we conclude that there are two maximum points — −1 and 0.5 — and two minimum points — −0.6 and −1.5.
Exercise 69. Use the Enriched Recipe to find the maximum and minimum points of each of the following functions. (Answer on p. 1031.)