Let us start out with two examples. In Figure 3.9 you see the graph of the function
q(x) = x2− 2x + 3.
Consider two points on the graph, say (−1, 6) and (1, 2), and connect them by a line segment. As you see, the line segment lies above the graph. The same is true, if we take any two points on the graph. This property of the function will be called being concave up.
In contrast, if you consider the the graph of the function (see Figure 3.10)
3.5. THE SECOND DERIVATIVE AND CONCAVITY 157 -2 -1 1 2 2 4 6 8 10 Figure 3.9: q(x) = x2− 2x + 3 -2 -1 1 2 3 4 -8 -6 -4 -2 2 4 Figure 3.10: g(x) =−x2+ 5x− 1
and take any two points on its graph, say (0,−1) and (3, 5), then the line segment which connects the two points lies below the graph. This property will be called being concave down.
We may use the monotonicity of the first derivative or information about the second derivative of a function to find criteria which tell us that a func- tion is concave up or down on an interval. We also study the corresponding notion at a point.
Concavity on an Interval
First of all, let us define the concept of being concave up or down on an interval. Let (a, f (a)) and (b, f (b)) be two distinct points on the graph of a function f . The two point formula for a line provides us with an expression for the line through these two point. For any x∈ (−∞, ∞)
l(x) = f (a) +f (b)− f(a)
b− a (x− a).
The following definition expresses in mathematical notation that a function is concave up (down) if every line segment connecting two points on its graph lies above (below) the graph.
Definition 3.28. Let f be a function which is defined on an interval I. We say that f is concave up on I if
for all a, b, c∈ I with a < c < b. Here l(x) is the line through (a, f(a)) and
(b, f (b)). We say that f is concave down on I if
f (c) > l(c) for all a, b, c∈ I with a < c < b.
In Figures 3.9 and 3.10 you saw the graph of a function which is concave up and of a function which is concave down. We state a theorem which provides you with assumptions under which a function is concave up or down. We will not provide a proof of the theorem.
Theorem 3.29. Let f be a function which is defined on an interval I. 1. Suppose that f (x) is differentiable on I. If f0(x) is increasing on I,
then f (x) is concave up on I. If f0(x) is decreasing on I, then f (x) is
concave down on I.
2. Suppose that f (x) is twice differentiable8 on I. If f00(x) > 0 for all x
in I, then f (x) is concave up on I. If f00(x) < 0 for all x in I, then
f (x) is concave down on I.
3. More generally, the conclusions in (2) still hold if in each finite interval J ⊂ I there are only finitely many points at which the assumption f00(x) > 0, resp. f00(x) < 0, is not satisfied.
Let us apply this theorem in a few examples.
Example 3.30. Confirm the statements which we made in the prolog to
this section.
Solution: The second derivative of the function (see Figure 3.9) q(x) = x2− 2x + 3 is q00(x) = 2,
and this function is positive on the entire real line. Theorem 3.29 (2) says that q is concave up on (−∞, ∞).
In comparison, the second derivative of the function
g(x) =−x2+ 5x− 1 is g00(x) =−2,
and this function is negative on the entire real line. The theorem says that
g is concave down on (−∞, ∞). ♦
8Strictly speaking, so far we can consider being ‘twice differentiable’ only for functions
which are defined on open intervals. More generally, we proceed as in Section 3.1. We say that f (x) is twice differentiable on I, if f (x) extends to a function F (x) which is defined on an open interval J which contains I, and F (x) is twice differentiable on J. The second derivative will be unique at all points in I if I does not consist of exactly one point.
3.5. THE SECOND DERIVATIVE AND CONCAVITY 159
Example 3.31. Show that the function h(x) = ln x
is concave down on the interval (0,∞).
Solution: The first derivative of h(x) is h0(x) = 1/x, and its second derivative is (see Table 2.6)
h00(x) =−1/x2.
Apparently h00(x) < 0 for all positive numbers x, so that we may conclude from Theorem 3.29 (2) that ln x is concave down on its domain, (0,∞).
Another Solution: Our calculation shows that h0(x) is decreasing on the interval (0,∞), because the first derivative h00(x) of h0(x) is negative on this interval. Theorem 3.29 (1) implies that h(x) = ln x is concave down on its domain, (0,∞). ♦
Example 3.32. Study the concavity properties of the exponential function f (x) = ex.
Solution: We asserted the existence and differentiability of the exponen-
tial function in Theorems 1.12 on page 20 and 2.12 on page 52. Theorem 2.12 also says that f (x) = f0(x). Applying the theorem twice we conclude that
f00(x) = ex.
By definition, f (x) = f00(x) > 0 for all real numbers x, see Theorem 1.12. Theorem 3.29 (2) implies that the exponential function is concave up on its entire domain, the interval (−∞, ∞). ♦
Remark 16. Let f be a function which is defined on an interval I. Suppose
that its image is an interval J , and that f has an inverse, which we call g. If
f is concave up, then g is concave down, and vice versa. This follows from
a geometric argument. You should convince yourself that if a secant line is above the graph, and you reflect the picture at the diagonal (that is how to get the graph of the inverse function) then, in the reflected picture, the secant line will be below the graph. We could have used this argument to exploit the statement that the logarithm function is concave down to deduce that the exponential function is concave up.
Exercise 109. In each of the Figures 3.11 and 3.12 you see the graphs of
1. decide which graph belongs belongs to the function and which one to its second derivative.
2. determine (approximately) intervals on which the second derivative is positive, resp., negative.
3. determine (approximately) intervals on which the function is concave up, resp., concave down.
0.5 1 1.5 2 2.5 3
-4 -2 2 4
Figure 3.11: Graphs of f and f00.
0.6 0.8 1.2 1.4 1.6 1.8 2
0.5 1 1.5
Figure 3.12: Graphs of f and f00.
Exercise 110. Discuss the concavity properties of the functions
(1) f (x) = x2− 5x + 8 and (2) g(x) =√3x− 1.
So far we applied Theorem 3.29 (2) to obtain conclusions. Let us look at some examples where we apply condition (3).
Example 3.33. Study the concavity properties of the function p(x) = x3− 3x2− 9x + 3.
Solution: You find the graph of this function in Figure 3.7, and we
discussed its monotonicity properties in Example 3.22 on page 150. An easy calculation provides us with the second derivative of this function:
3.5. THE SECOND DERIVATIVE AND CONCAVITY 161
We see that p00(x) > 0 for x∈ (1, ∞), and p00(x) < 0 for x∈ (−∞, 1). This means that p00(x) > 0 for all x ∈ [1, ∞) with only one exception, x = 1. Theorem 3.29 (3) tells us that p(x) is concave up on the interval [1,∞). Similarly, p00(x) < 0 for x∈ [−∞, 1) with only one exception, x = 1. One deduces that f (x) is concave down on the interval (−∞, 1]. ♦
Example 3.34. Study the monotonicity and concavity properties of the
tangent function tan x on the interval (−π/2, π/2).
Solution: You find the first and second derivative of the function tan x
in Table 2.6 on page 136:
tan0x = sec2x & tan00x = 2 sec2x tan x.
By definition, sec2x > 0 on the interval (−π/2, π/2). This means that tan x
is increasing on (−π/2, π/2), see Theorem 3.12 on page 144.
In addition, tan x < 0 for x∈ (−π/2, 0) and tan x > 0 for x ∈ (0, π/2). This means that tan00x < 0 for x ∈ (−π/2, 0) and tan00x > 0 for x ∈
(0, π/2). Theorem 3.29 (3) implies that tan x is concave down on (−π/2, 0] and concave up on [0, π/2). Compare Figures 5.11 and 5.12 to confirm our conclusions visually. ♦
Example 3.35. Find intervals on which the function f (x) = sin x is con-
cave up or concave down.
Solution: The sine function is defined and twice differentiable on the
interval (−∞, ∞). Its second derivative is (see Table 3.1)
f00(x) =− sin x.
You may use the graph shown in Figure 5.9, or the geometry of the unit circle, to conclude that sin x > 0 on intervals of the form (2nπ, (2n + 1)π) and sin x < 0 on intervals of the form ((2n+1)π, 2nπ). Here n is an arbitrary integer (whole number). We conclude that sin00(x) < 0 on all intervals of the form (2nπ, (2n + 1)π) and that sin x is concave down on the intervals [2nπ, (2n + 1)π]. Similarly, sin x is concave up on all intervals of the form [(2n + 1)π, 2nπ]. ♦
Remark 17. Let us discuss the statement about the poverty rate which we
quoted in the beginning of the chapter:“the rate at which the rate of poverty
is increasing is decreasing.” You may view it politically. The speech writer
carefully designed a sentence which was not untrue, and which ended on a positive note. Something about the poverty rate was decreasing. We have encounter functions, like ln, which are increasing and concave down (the rate
at which they increase decreases). These functions are not even bounded, so arbitrarily large values are obtained as we wait long enough. That would be rather bad news, in case the function described the poverty rate. The question is whether the relevant political decisions and current social and economical conditions effect the first or the second derivative of the function
P (t), the poverty rate as a function of time. If P00(t)≤ −A for some positive number A for a sufficiently long time, then P0(t) will continue to decrease at least at rate A and eventually become negative, so that the poverty rate itself would start decreasing. Maybe a good policy will be designed to have an effect on the second derivative of a quantity which needs change. It may not bring immediate relief, but eventually lasting improvement. A change in the first derivative, without control of the second one, may bring temporary relief without solving the long term problem.
Exercise 111. Find intervals on which the following functions are concave
up, resp., concave down. 1. f (x) = x3− 4x2+ 8x− 7 2. g(x) = x4+ 2x3− 3x2+ 5x− 2 3. h(x) = x + 1/x 4. i(x) = 2x4− x2 5. j(x) = x/(x2− 1) 6. k(x) = 2 cos2x− x2 for x∈ [0, 2π]. Concavity at a Point
The notion of being concave up or down was defined for functions which are defined on intervals. Still, we got a picture how the function has to look like near a point, and this is the behavior which we like to capture in a definition.
Definition 3.36. Let f be a function and c an interior point 9 of its do- main. We say that f is concave up, resp., concave down, at c if there exists an open interval I and a line10 l such that l(c) = f (c) and
f (x) > l(x), resp., f (x) < l(x),
9The idea of an interior point was defined in Definition 2.1 on page 42.
10A line l, as required in this definition, is called a support line. In general, there could
be more than one such line, but if the function is differentiable at c, then the support line is unique.
3.5. THE SECOND DERIVATIVE AND CONCAVITY 163
for all x∈ I with x 6= c.
In other words, we are asking for a line l(x), which agrees with f at c, and on some open interval around c the function is larger (resp., smaller) that l(x). The inequality is required to be strict for x 6= c. You see this situation illustrated in two generic pictures in Figures 3.13 and 3.14. One shows a function which is concave up at the indicated point, one shows a function which is concave down.
0.5 1 1.5 2 1 2 3 4 Figure 3.13: Concave up at• 1 2 3 4 5 -0.5 0.5 1 1.5 2
Figure 3.14: Concave down at •
There are two obvious questions. How can we detect whether a function is concave up or down at a point? What is the slope of the line l referred to in the definition? The answer to both question is given in our next theorem.
Theorem 3.37. Let f be a function and c an interior point of its domain. 1. If f0 is increasing at c or if f00(c) > 0, then f is concave up at c.
2. If f0 is decreasing at c or if f00(c) < 0, then f is concave down at c.
3. If f is differentiable and concave up or down at c, then there is only one line which plays the role of l(x) in Definition 3.36, and this line is the tangent line to the graph of f at c.
We can use the theorem to check the concavity of a function at a point. Just calculate the second derivative of the function at the point in question, and see whether it is positive or negative. If this second derivative at the point should turn out to be 0, then the theorem is inconclusive. It does not tell us anything.
Remark 18. Item (3) in Theorem 3.37 describes the situation alluded to
in Example 2 on page 40. In this case we can find the tangent line to a graph by holding a ruler against it.
Example 3.38. Check the concavity of
f (x) = x5− 7x4+ 2x3+ 2x2− 5x + 4 at x = 2.
Solution: We calculate the second derivative of f : f00(x) = 20x3− 84x2+ 12x + 4.
Evaluated at x = 2 we find f00(2) = −148 < 0. So the function is concave down at x = 2 ♦
Exercise 112. Decide at which points on the real line the following func-
tions are concave up, resp., concave down: (a) f (x) = x3− 2x2+ 5x− 3.
(b) f (x) = x4+ x3− 3x2+ 6x + 1.
To relate concavity properties on an interval to those at each point in the interval we state, without proof, the following theorem.
Theorem 3.39. Let f be a function which is defined on an open interval
(a, b). Then f is concave up (resp., down) on (a, b) if and only if f is concave
up (resp., down) at each point in (a, b).