What can be learnt from the sign of the second derivative? Recall how the sign of the first derivative determines whether a function is increasing or decreasing:
fix) > 0 on (a, b) <*==> fix) is increasing on ia, b) [1]
fix) < 0 on (a, b) •<==>■ f(x) is decreasing on (a. b) [2]
The second derivative fix) is the derivative of fix). Hence:
fix) > 0 on ia, b) fix) is increasing on ia, b) [3]
fix) < 0 on (a, b) fix) is decreasing on ia, b) [4]
The equivalence in [3] is illustrated in Fig. 9.14. The slope of the tangent, fix), is increasing as x increases. On the other hand, the slope of the tangent to the graph in Fig. 9.15 is decreasing as x increases. (Place a ruler as a tangent to the graph of the function. As the ruler slides along the curve from left to right, the tangent rotates counterclockwise in Fig. 9.14, clockwise in Fig. 9.15.)
FIGURE 9.14 The slope of the tangent increases as x increases. f'{x) is increasing.
FIGURE 9.14 The slope of the tangent increases as x increases. f'{x) is increasing.
We introduce the following definitions, assuming that / is continuous in the interval 7 and twice differentiable in the interior of 7, denoted by 7°:
/ is convex on / <*= |
=► /"(*) > 0 for all x in 7° |
/ is concave on I < |
=► fix) < 0 for all x in 7° |
The distinction between convexity and concavity of a function is absolutely crucial in many economic models. Study carefully the cases illustrated in Fig. 9.16.
Example 9.10
Check the convexity/concavity of the following:
(a) f(x) =x2-2x+2 and (b) fix) = ax2 + bx+c Solution
(a) Here fix) = 2x-2so fix) = 2. Because fix) > 0 for all x, / is convex.
(b) Here fix) - 2ax + b, so fix) =2a. If a = 0, then / is linear and / is convex as well as concave. If a > 0, then fix) > 0, so / is convex. If a < 0, then fix) < 0, so / is concave. Compare with the graphs in Fig. 3.1 in Section 3.1.
Some Typical Examples
We consider two typical examples of convex and concave functions. In Fig. 9.17 we have drawn roughly the graph of function P, where
P(t) = world population (in 1000 millions) in year t
It appears from the figure that not only is P (t) increasing, but the rate of increase increases. (Each year the increase becomes larger.) So P(t) is convex.
The graph in Fig. 9.18 shows the crop of wheat Y(N) when N pounds of fertilizer per acre are used, based on fertilizer experiments in Iowa during 1952 (see Example 9.2 in Section 9.2). The function has a maximum at N = No 172. Increasing the amount of fertilizer beyond No will cause wheat production to decline. Moreover, Y(N) is concave. If N < No, increasing N by one unit will lead to less increase in Y(N) the larger is N. On the other hand, if N > No, increasing N by one unit will lead to a larger decrease in Y(N) the larger is N.
Example 9.11
Examine the concavity/convexity of the production function
Y = AKa (A >0, 0 < a < 1) defined for all K > 0.
Solution Differentiating Y twice with respect to K yields
Because a € (0, 1), coefficient Aa(a - 1) < 0, so that Y" < 0 for all K > 0. Hence, the function is concave. The graph of Y = AKa. for 0 < a < 1, is y y
shown in in Fis. 9.19. If a > 1, then Y" > 0 and Y is a convex function of K, as shown in Fig. 9.20.
Suppose that functions U and g are both increasing and concave, so that U' > 0, U" < 0, g' > 0, and g" < 0. Prove that the composite function f(x) = g{U(:c))
is also increasing and concave. Solution Using the chain rule yields
Because g' and U' are both > 0, so f'(x) > 0. Hence, / is increasing. (An increasing transformation of an increasing function is increasing.)
In order to compute f"(x), we must differentiate the product of the two functions g'(U(x)) and U'(x). According to the chain rule, the derivative of g'(U(x)) is equal to g"(U(x)) ■ U'(x). Hence,
Because g" < 0, g' > 0. and U" < 0. it follows that /"(je) < 0. (An increasing concave transformation of a concave function is concave.)
Functions we study in economics are often convex in some parts of the domain but concave in others. Points at which a function changes from being convex to being concave, or vice versa, are called inflection points.
f"(x) = g"[U(x)] • {U'(x)f + g'(U(x)) ■ U"{x)
Point c is an inflection point for a twice differentiable function / if there is an interval (a, b) containing c such that either of the following two conditions holds:
(a) /"(» > 0 if a < x < c and f"(x) < 0 if c <x <b or
(b) f"(x) <0 if a<x<c and f"(x) > 0 if c <x <b
Briefly, x = c is an inflection point if f"{x) changes sign at c. We also refer to the point (c, /(c)) as an inflection point on the graph. An example is given in Fig. 9.21. Figure 9.22 shows the profile of a ski jump. Point P, where the hill is steepest, is an inflection point.
Theorem 93 (Test for Inflection Points) Let / be a function with a continuous second derivative in an interval /, and suppose that c is an interior point of I.
(a) If c is an inflection point for /, then /"(c) = 0.
(b) If /"(c) = 0 and f" changes sign at c, then c is an inflection point for /.
FIGURE 9.21 Point P is an inflection point on the graph (x = c is an inflection point for the function).
FIGURE 9.22 Point P, where the slope is steepest, is an inflection point.
Proof
(a) Because fix) < 0 on one side of c and fix) > 0 on the other, f"(c) = 0.
(b) If f" changes sign about point c, then c is an inflection point for / according to [9.11].
According to Theorem 9.3 (a), the condition /"(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.
Example 9.13
Show that f(x) = x4 does not have an inflection point at x = 0, even though /"(0) = 0.
Solution Here f'(x) = 4x3 and fix) = 12x2, so that /"(0) = 0. But fix) > 0 for all x ^ 0, and hence f" does not change sign at x = 0. Hence, x = 0 is not an inflection point. (In fact, it is a global minimum, of course, as shown in Fig. 9.10 in Section 9.4.)
Find possible inflection points for f(x) = |x3 — \x2 — |x + 1. Solution We find the first and second derivatives to be f(x) = ¡X2 -\x-\ and f(x) = \x - \ = f (x - i)
Hence, /"(*) < 0 for x < 1/2, whereas /"(1/2) = 0 and fix) > 0 for x > 1/2. According to Theorem 9.3(b), x = 1/2 is an inflection point for /.
Example 9.15
Find possible inflection points for fix) = x2ex. Draw its graph. (See Example 8.2, Section 8.1.)
Solution The first derivative of / is fix) = 2xex +x2ex, so the second derivative is fix) = 2ex + 2xex + 2xex + xV = exix2 + 4x + 2) = exix - x,)(x - x2)
where x\ = -2 - as -3.41 and x2 = -2 + -Jl % -0.59 are the two roots of the quadratic equation x2 + 4x + 2 = 0. The sign diagram associated with fix) is shown below. From this diagram we see that / has inflection points at x = X| and at x = X2- The graph is convex in the intervals (—oo,xj] and [xj, oo), and it is concave in [X],X2]. See
Fig. 9.23 in which we have also taken advantage of the results of Example 8.2.
Fig. 9.23 in which we have also taken advantage of the results of Example 8.2.
A firm produces a commodity using only one input. Let x = f(v), v > 0, be the maximum production obtainable when v units of the input are used. Then / is called a production function. It is often assumed that the marginal product f(v) is increasing up to a certain production level i>o, and then decreasing. Such a production function is indicated in Fig. 9.24. If / is FIGURE 9.24 x f is a production function. v0 is an inflection point. twice differentiate, then f"(y) is > 0 in [0, uo) and < 0 in (uq, oo). Hence, / is first convex and then concave, with vo as an inflection point. An example of such a function is given in Problem 9. Suppose that f"(x) < 0 for all x in an interval I. Then f'(x) is decreasing in I. So if f'{c) = 0 for an interior point c in I, then f'(x) must be > 0 to the left of c, whereas f'{x) < 0 to the right of c. This implies that the function itself is increasing to the left of c, and decreasing to the right of c. We conclude that * = c is a maximum point for / in I. This important observation is illustrated in Fig. 9.25. We have a corresponding result for the minimum of a convex function. Theorem 9.4 (Maximum/Minimum for Concave/Convex Functions) Suppose / is a concave (convex) function in an interval I. If c is a stationary point for / in the interior of /, then c is a maximum point (minimum point) for / in I. Briefly stated, when c is an interior point of I, then A Useful Result f"(x) < 0 for all X € /, and f(c) = 0 =» x = c is a maximum point for / in I Example 9.17 Let the total cost of producing Q units of a commodity be FIGURE 9.25 f is concave, f'{c) = 0, and c is a maximum point. >• = m c where a, b, and c are positive constants. Prove that the average cost function A(Q) = aQ + b + c/Q has a minimum at Q* = -Jcja. (See also Problem 8 in Section 9.3.) Solution The first-order derivative of A(Q) is and the only stationary point is Q* = -JcJa. Because A"(Q) = 2c/Q3 > 0 for all Q > 0, A(Q) is convex, and by Theorem 9.4, Q* = y/c/a is the minimum point. 1. Determine the concavity/convexity of fix) = — hx2 -I- 8* — 3. 2. Let / be defined for all x by fix) = x3 + \x2 - 6x + 10. b. Find the stationary points of / and the intervals where f is increasing. c. Find the inflection points of / and the intervals of concavity/convexity. 3. A competitive firm receives a price p for each unit of its output, pays a price xv for each unit of its only variable input, and incurs fixed costs of F. Its output from using x units of variable input is fix) = ~Jx. a. Write the firm's revenue, cost, and profit functions. b. Write the first-order condition for profit maximization, and give it an economic interpretation. c. Check whether profit really is maximized at a point satisfying the first-order condition. d. Explain how your answers would change if fix) = x2. 4. What are the extreme points and the inflection points of function / whose graph is given in Fig. 9.26? 5. Decide where the following functions are convex and determine possible inflection points: FIGURE 9.26 FIGURE 9.26 6. Find numbers a and b such that the graph of passes through (—1. 1) and has an inflection point at x = 1/2. 7. Find the intervals where the following cubic cost function is convex and where it is concave, and find the unique inflection point: C(Q) = aQ3 + bQ2 + cQ +d, (a > 0. b< 0, c > 0, d > 0) 8. With reference to Example 9.5, let R(Q) = PQ and C(Q) = aQh+c, where P, a, b, and c are positive constants with b > 1. Find the value of Q that maximizes profits jr(fi) = PQ - (aQb -f c). (Use Theorem 9.4.) Harder Problems 9. With reference to Example 9.16, let f(v) = (v - 1)1/3 -r 1 for v > 0. a. Show that / is an increasing function of v and that f"(v) > 0 in [0, 1), f'iy) < 0 in (1, oo). Draw the graph of /. b. Suppose that the price per unit of the commodity is 1 and that the price the firm must pay per unit of the input is p. The profit is then tt(v) = f(v) — pv. Suppose that vm > 0 maximizes tz(v) for the given value of p > 0. Find vm expressed in terms of p. c. Draw the graph of it for the case p = 1. Use the same diagram as in part (a). d. Find the nonnegative roots of the equation n(v) = 0. For which values of p are there three real roots? e. For all values of p, find the solution of the problem maximize it(v) subject to v > 0 |
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