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7 Inflection Points, Concavity, and Convexity

# 7 Inflection Points, Concavity, and Convexity

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312

CHAPTER 8

/

SINGLE-VARIABLE OPTIMIZATION

INFLECTION POINTS

If the function f is twice differentiable, the point c is called an inﬂection point

for f if there exists an interval (a, b) about c such that:

(a) f (x) ≥ 0 in (a, c) and f (x) ≤ 0 in (c, b); or

(b) f (x) ≤ 0 in (a, c) and f (x) ≥ 0 in (c, b).

Brieﬂy, x = c is an inﬂection point if f (x) changes sign at x = c, and we refer to the

point (c, f (c)) as an inﬂection point on the graph. Figure 8.7.1 gives an abstract example

from mathematics, while Fig. 8.7.2 gives a sporting example: it shows the proﬁle of a ski

jump. The point P, where the slope is steepest, is an inﬂection point.

y

P

P

f (x)

0

f (x)

0

c

Figure 8.7.1 Point P is an inﬂection point on

the graph; x = c is an inﬂection point for

the function

Figure 8.7.2 The point P, where the slope is steepest,

is an inﬂection point

When looking for possible inﬂection points of a function, we usually use part (ii) in the

following theorem:

THEOREM 8.7.1

(TEST FOR INFLECTION POINTS)

Let f be a function with a continuous second derivative in an interval I, and let

c be an interior point of I.

(i) If c is an inﬂection point for f , then f (c) = 0.

(ii) If f (c) = 0 and f changes sign at c, then c is an inﬂection point for f .

The proof of this theorem is rather simple:

(i) Because f (x) ≤ 0 on one side of c and f (x) ≥ 0 on the other, and because f is continuous, it

must be true that f (c) = 0.

(ii) If f changes sign at c, then c is an inﬂection point for f , by deﬁnition.

SECTION 8.7

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INFLECTION POINTS, CONCAVITY, AND CONVEXITY

313

This theorem implies that f (c) = 0 is a necessary condition for c to be an inﬂection

point. It is not a sufficient condition, however, because f (c) = 0 does not imply that f

changes sign at x = c. A typical case is given in the next example.

Show that the function f (x) = x4 does not have an inﬂection point at x = 0, even

though f (0) = 0.

E X A M P L E 8.7.1

Solution: Here f (x) = 4x3 and f (x) = 12x2 , so that f (0) = 0. But f (x) > 0 for all x = 0,

and so f does not change sign at x = 0. Hence, x = 0 is not an inﬂection point—in fact, it

is a global minimum, as shown in Fig. 8.6.3.

E X A M P L E 8.7.2

Find possible inﬂection points for f (x) = 19 x3 − 16 x2 − 23 x + 1.

Ð

Solution: From Example 8.6.1, we have f (x) = 23 x − 13 = 23 x − 12 . Hence, f (x) ≤ 0

for x ≤ 1/2, whereas f (1/2) = 0 and f (x) ≥ 0 for x > 1/2. According to part (ii) in

Theorem 8.7.1, x = 1/2 is an inﬂection point for f .

E X A M P L E 8.7.3

Find possible inﬂection points for f (x) = x6 − 10x4 .

Solution: In this case f (x) = 6x5 − 40x3 and

f (x) = 30x4 − 120x2 = 30x2 (x2 − 4) = 30x2 (x − 2)(x + 2)

A sign diagram for f is as follows:

22

21

0

1

2

30x2

x 22

x12

f ʺ(x)

f(x)

From the sign diagram we see that f changes sign at x = −2 and at x = 2, so these are

inﬂection points. Since f does not change sign at x = 0, it is not an inﬂection point, even

though f (0) = 0.

Economic models often involve functions having inﬂection points. The cost function in

Fig. 4.7.2 is a typical example. Here is another.

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CHAPTER 8

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SINGLE-VARIABLE OPTIMIZATION

A ﬁrm produces a commodity using only one input. Let y = f (x), for x ≥ 0, be the

output obtained when x units of the input are used. Then f is called a production function.

Its ﬁrst derivative measures the increase in output that is obtained by increasing the input

used inﬁnitesimally; this derivative is called the ﬁrm’s marginal product. It is often assumed

that the graph of a production function is “S-shaped”. That is, the marginal product, f (x),

is increasing up to a certain production level c, and then decreasing. Such a production

function is indicated in Fig. 8.7.1. If f is twice differentiable, then f (x) ≥ 0 in [0, c], and

f (x) ≤ 0 in [c, ∞). Hence, f is ﬁrst convex and then concave, with c as an inﬂection point.

Note that at x = c a unit increase in input gives the maximum increase in output.

E X A M P L E 8.7.4

More General Deﬁnitions of Concave and Convex Functions

So far the convexity and concavity properties of functions have been deﬁned by looking

at the sign of the second derivative. An alternative geometric characterization of convexity

and concavity suggests a more general deﬁnition that is valid even for functions that are not

differentiable.

CONCAVE AND CONVEX FUNCTIONS

A function f is called concave if the line segment joining any two points on the

graph is below the graph, or on it. It is called convex if any such line segment

lies above, or on the graph.

y

y

f

f

x

x

Figure 8.7.3 f is concave

Figure 8.7.4 f is convex

These deﬁnitions are illustrated in Figs 8.7.3 and 8.7.4. Because the graph has a “kink”

in Fig. 8.7.3, this function is not even differentiable, let alone twice differentiable. For twice

differentiable functions, one can prove that this general deﬁnition is equivalent to the deﬁnitions in (6.9.3) and (6.9.4). Now, in order to use the deﬁnition to examine the convexity

or concavity of a given function, we need an algebraic formulation. This will be discussed

in FMEA.

SECTION 8.7

/

315

INFLECTION POINTS, CONCAVITY, AND CONVEXITY

Strictly Concave and Strictly Convex Functions

A function f is called strictly concave if the line segment joining any two points on the

graph is strictly below the graph, except at the end points of the segment; it is called strictly

convex if any such segment lies strictly above the graph, again except at the end points of

the segment. For instance, the function whose graph is shown in Fig. 8.7.3 has two linear

pieces, on which line segments joining two points coincide with part of the graph. Thus this

function is concave, but not strictly concave. By contrast, the function graphed in Fig. 8.7.4

is strictly convex.

Fairly obvious sufficient conditions for strict concavity/convexity are the following,

which will be further discussed in FMEA:

f (x) < 0 for all x ∈ (a, b) =⇒ f is strictly concave in (a, b)

(8.7.1)

f (x) > 0 for all x ∈ (a, b) =⇒ f is strictly convex in (a, b)

(8.7.2)

The reverse implications are not correct. For instance, one can prove that f (x) = x4 is

strictly convex in the interval (−∞, ∞), but f (x) is not > 0 everywhere, because f (0) =

0—see Fig. 8.6.3.

For twice differentiable functions, it is usually much easier to check concavity/convexity

by considering the sign of the second derivative than by using the deﬁnitions of the properties. However, in theoretical arguments the deﬁnitions are often very useful, especially

because they generalize easily to functions of several variables. (See FMEA.)

EXERCISES FOR SECTION 8.7

1. Let f be deﬁned for all x by f (x) = x3 + 32 x2 − 6x + 10.

(a) Find the critical points of f and determine the intervals where f increases.

(b) Find the inﬂection point for f .

2. Decide where the following functions are convex and determine possible inﬂection points:

(a) f (x) =

SM

x

1 + x2

(b) g(x) =

1−x

1+x

(c) h(x) = xex

3. Find local extreme points and inﬂection points for the functions deﬁned by the following formulas:

(a) y = (x + 2)e−x

ln x

(d) y = 2

x

(b) y = ln x + 1/x

(c) y = x3 e−x

(e) y = e2x − 2ex

(f) y = (x2 + 2x)e−x

4. A competitive ﬁrm receives a price p for each unit of its output, and pays a price w for each unit

of its only variable

√ input. It also incurs set up costs of F. Its output from using x units of variable

input is f (x) = x.

(a) Determine the ﬁrm’s revenue, cost, and proﬁt functions.

(b) Write the ﬁrst-order condition for proﬁt maximization, and give it an economic interpretation.

Check whether proﬁt really is maximized at a point satisfying the ﬁrst-order condition.

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SINGLE-VARIABLE OPTIMIZATION

5. Find the extreme points and the inﬂection points of the function f whose graph is shown in

Fig. 8.7.5.

y

4

y

f (x)

3

4

2

1

3

2

1

0

1

2

5

6

x

Figure 8.7.5 Exercise 5

6. Find numbers a and b such that the graph of f (x) = ax3 + bx2 passes through (−1, 1) and has an

inﬂection point at x = 1/2.

7. Consider the following cubic cost function, deﬁned for x ≥ 0: C(x) = ax3 + bx2 + cx + d, where

a > 0, b < 0, c > 0, and d > 0. Find the intervals where the function is convex and where it is

concave. Find also the unique inﬂection point.

8. Use the same coordinate system to draw the graphs of two concave functions f and g, both deﬁned

for all x. Let the function h be deﬁned by h(x) = min{f (x), g(x)}—that is, for each given x, the

number h(x) is the smaller of f (x) and g(x). Draw the graph of h and explain why it is also concave.

REVIEW EXERCISES

x2

.

x2 + 2

(a) Compute f (x) and determine where f (x) is increasing/decreasing.

1. Let f (x) =

(b) Find possible inﬂection points.

(c) Determine the limit of f (x) as x → ±∞, and sketch the graph of f (x).

2. A ﬁrm’s production function is Q(L) = 12L2 −

with L ∈ [0, 200].

1 3

20 L ,

where L denotes the number of workers,

(a) What size of the work force maximizes output Q(L)?

(b) What size of the work force maximizes output per worker, Q(L)/L? Letting L∗ denote such

size, note that Q (L∗ ) = Q(L∗ )/L∗ . Is this a coincidence?

3. A farmer has one thousand metres of fence wire to make a rectangular enclosure, as in

Exercise 4.6.7, but this time no fencing is needed on one side of the enclosure that is a straight

canal bank. What should be the dimensions of the enclosure in order to maximize area?

4. By producing and selling Q units of some commodity a ﬁrm earns a total revenue

R(Q) = −0.0016Q2 + 44Q and incurs a cost of C(Q) = 0.0004Q2 + 8Q + 64 000.

(a) What production level maximizes proﬁts?

(b) The elasticity ElQ C(Q) ≈ 0.12 for Q = 1000. Interpret this result.

CHAPTER 8

/

REVIEW EXERCISES

317

5. The unit price P obtained by a ﬁrm in producing and selling Q ≥ 0 units is P(Q) = a − bQ2 ,

and the cost of producing and selling Q units is C(Q) = α − βQ. All constants are positive.

Find the level of production that maximizes proﬁts.

6. Let g(x) = x − 2 ln(x + 1).

(a) Where is g deﬁned?

(b) Find g (x) and g (x).

(c) Find possible extreme points and inﬂection points, and sketch the graph.

7. Let f (x) = ln(x + 1) − x + 12 x2 − 16 x3 .

(a) Find the domain of the function and prove that, for x in the domain:

f (x) =

x 2 − x3

2(x + 1)

(b) Find possible extreme points and inﬂection points.

(c) Check f (x) as x → (−1)+ , and sketch the graph on the interval (−1, 2].

SM

8. Consider the function deﬁned, for all x, by h(x) = ex /(2 + e2x ).

(a) Where is h increasing/decreasing? Find possible maximum and minimum points for h.

(b) If one restricts the domain of h to (−∞, 0], it has an inverse. Why? Find an expression for

the inverse function.

9. Let f (x) =

e2x

+

Ð2

4e−x .

(a) Find f (x) and f (x).

(b) Determine where f is increasing/decreasing, and show that f is convex.

(c) Find possible global extreme points for f .

SM

10. [HARDER] Letting a > 0, consider the function

f (x) = √

3

x

x2

−a

(a) Find the domain Df of f and the intervals where f (x) is positive. Show that the graph of f is

(b) Where is f increasing and where is it decreasing? Find possible local extreme points.

(c) Find possible inﬂection points for f .

SM

11. [HARDER] Classify the critical points of

f (x) =

x4

6x3

+ x2 + 2

by using the ﬁrst-derivative test. Sketch the graph of f .

9

INTEGRATION

Is it right I ask; is it even prudence; to bore thyself and bore the students?

—Mephistopheles to Faust, in Johann Wolfgang von Goethe’s Faust

T

he main topic of the preceding three chapters was differentiation, which can be directly

applied to many interesting economic problems. Economists, however, especially when

doing statistics, often face the mathematical problem of ﬁnding a function from information

about its derivative. This process of reconstructing a function from its derivative can be regarded

as the “inverse” of differentiation. Mathematicians call this process integration.

There are simple formulas that have been known since ancient times for calculating the area

of any triangle, and so of any polygon that, by deﬁnition, is entirely bounded by straight lines.

Over 4000 years ago, however, the Babylonians were concerned with accurately measuring the

area of plane surfaces, like circles, that are not bounded by straight lines. Solving this kind of

area problem is intimately related to integration, as will be explained in Section 9.2.

Apart from providing an introduction to integration, this chapter will also discuss some

important applications of integrals that economists are expected to know. A brief introduction

to some simple differential equations concludes the chapter.

9.1 Indeﬁnite Integrals

Suppose we do not know the function F, but we do know that its derivative is x2 , so that

F (x) = x2 . What is F? Since the derivative of x3 is 3x2 , we see that 13 x3 has x2 as its

derivative. But so does 13 x3 + C where C is an arbitrary constant, since additive constants

disappear with differentiation.

In fact, let G(x) denote an arbitrary function having x2 as its derivative. Then the derivative of G(x) − 13 x3 is equal to 0 for all x. But, by (6.3.3), a function that has derivative equal

to 0 for all x must be constant. This shows that

F (x) = x2 ⇔ F(x) = 13 x3 + C

with C as an arbitrary constant.

320

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INTEGRATION

Assume that the marginal cost function of a ﬁrm is C (Q) = 2Q2 + 2Q + 5, and that

the ﬁxed costs are 100. Find the cost function C(Q).

E X A M P L E 9.1.1

Solution: Considering separately each of the three terms in the expression for C (Q),

we realize that the cost function must have the form C(Q) = 23 Q3 + Q2 + 5Q + c,

because if we differentiate this function we obtain precisely 2Q2 + 2Q + 5. But the

ﬁxed costs are 100, which means that C(0) = 100. Inserting Q = 0 into the proposed formula for C(Q) yields c = 100. Hence, the required cost function must be

C(Q) = 23 Q3 + Q2 + 5x + 100.

Suppose f (x) and F(x) are two functions of x having the property that f (x) = F (x) for all

x in some interval I. We pass from F to f by taking the derivative, so the reverse process of

passing from f to F could appropriately be called taking the antiderivative. But following

usual mathematical

practice, we call F an indeﬁnite integral of f over the interval I, and

R

denote it by f (x) dx. Two functions having the same derivative throughout an interval

must differ by a constant, so:

THE INDEFINITE INTEGRAL

If F (x) = f (x), then

Z

f (x) dx = F(x) + C

(9.1.1)

where C is an arbitrary constant.

For instance, the solution to Example 9.1.1 implies that

Z

2

(2x2 + 2x + 5) dx = x3 + x2 + 5x + C

3

R

The symbol is the integral sign, and the function f (x) appearing in (9.1.1) is the integrand.

Then we write dx to indicate that x is the variable of integration. Finally, C is a constant

of integration. We read (9.1.1) this way: The indeﬁnite integral of f (x) w.r.t. x is F(x) plus

a constant. We call it an indeﬁnite integral because F(x) + C is not to be regarded as one

deﬁnite function, but as a whole class of functions, all having the same derivative f .

Differentiating each side of (9.1.1) shows directly that

Z

d

f (x) dx = f (x)

(9.1.2)

dx

namely, that the derivative of an indeﬁnite integral equals the integrand. Also, (9.1.1) can

obviously be rewritten as

Z

F (x) dx = F(x) + C

Thus, integration and differentiation cancel each other out.

(9.1.3)

SECTION 9.1

/

INDEFINITE INTEGRALS

321

Some Important Integrals

There are some important integration formulas which follow immediately from the corresponding rules for differentiation. Let a be a ﬁxed number, different from −1. Because the

derivative of xa+1 /(a + 1) is xa , one has

If a = −1, then

Z

xa dx =

1 a+1

+C

x

a+1

(9.1.4)

This very important result states that the indeﬁnite integral of any power of x, except

x , is obtained by increasing the exponent of x by 1, then dividing by the new exponent,

and ﬁnally adding a constant of integration. Here are three prominent examples.

−1

E X A M P L E 9.1.2

R

R

1 1+1

1

+ C = x2 + C

x

1+1

2

R 1

R −3

1

1

−3+1

(b) x3 dx = x dx =

+C =− 2 +C

x

−3 + 1

2x

R√

R 1/2

1 +1

1

2 3/2

(c)

x dx = x dx = 1

x2 + C = x + C

3

2 +1

(a)

x dx =

x1 dx =

When a = −1, the formula in (9.1.4) is not valid, because the right-hand side involves

division by zero and so becomes meaningless. The integrand is then 1/x, and the problem

is thus to ﬁnd a function having 1/x as its derivative. Now F(x) = ln x has this property, but

it is only deﬁned for x > 0. Note, however, that ln(−x) is deﬁned for x < 0, and according

to the chain rule, its derivative is [1/(−x)] (−1) = 1/x. Recall too that |x| = x when x ≥ 0

and |x| = −x when x < 0. Thus, whether x > 0 or x < 0, we have:

Z

1

dx = ln |x| + C

x

Consider next the exponential function. The derivative of ex is ex . Thus

x

e + C. More generally, since the derivative of (1/a)eax is eax , we have:

If a = 0, then

Z

eax dx =

1 ax

e +C

a

(9.1.5)

R

ex dx =

(9.1.6)

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/

INTEGRATION

For a > 0 we can write ax = e(ln a)x . As a special case of (9.1.6), when ln a = 0 because

a = 1, we obtain:

When a > 0 and a = 1,

Z

ax dx =

1 x

a +C

ln a

(9.1.7)

The above were examples of how knowing the derivative of a function given by a formula automatically gives us a corresponding indeﬁnite integral. Indeed, suppose it were

possible to construct a complete table with every formula that we knew how to differentiate

in the ﬁrst column, and the corresponding derivative in the second column. For example,

corresponding to the entry y = x2 ex in the ﬁrst column, there would be y = 2xex + x2 ex

in the second column. Because integration

is the reverse of differentiation, we infer the

R

x

corresponding integration result that (2xe + x2 ex ) dx = x2 ex + C for a constant C.

2

Even after this superhuman effort, you would look in vain for e−x in the second col2

umn of this table. The reason is that there is no “elementary” function that has e−x as its

2

derivative. Indeed, the integral of e−x is used in the deﬁnition of a new very special “error

function” that plays a prominent role in statistics because of its relationship to the “normal

distribution” — see Exercises 4.9.5 and 9.7.12. A list of a few such impossible “integrals”

is given in (9.3.9).

Using the proper rules systematically allows us to differentiate very complicated functions. On the other hand, ﬁnding the indeﬁnite integral of even quite simple functions can

be very difficult, or even impossible. Where it is possible, mathematicians have developed a

number of integration methods to help in the task. Some of these methods will be explained

in the rest of this chapter.

It is usually quite easy, however, to check whether a proposed indeﬁnite integral is correct. We simply differentiate the proposed function to see if its derivative really is equal to

the integrand.

E X A M P L E 9.1.3

Verify that, for x > 0,

R

ln x dx = x ln x − x + C.

Solution: We put F(x) = x ln x − x + C. Then

F (x) = 1 · ln x + x · (1/x) − 1 = ln x + 1 − 1 = ln x

which shows that the integral formula is correct.

Some General Rules

The two differentiation rules (6.7.1) and (6.7.2) immediately imply that (aF(x)) = aF (x)

and (F(x) + G(x)) = F (x) + G (x). These equalities then imply the following:

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