The main theorems used so far in this chapter to locate extreme points require the function to be steadily increasing on one side of the point and steadily decreasing on the other side.
Many functions with a derivative whose sign varies in a more complicated way may still have a maximum or minimum. This section shows how to locate possible extreme points for an important class of such functions.
Examples such as those illustrated in Figs. 2–4 of Section 8.1 show how not all functions have extreme points. The following theorem gives important sufficient conditions for their existence.
T H E O R E M 8 . 4 . 1 ( E X T R E M E V A L U E T H E O R E M )
Suppose thatf is a continuous function over a closed and bounded interval [a, b].
Then there exist a pointdin [a, b] wherefhas a minimum, and a pointcin [a, b]
wheref has a maximum, so that
f (d)≤f (x)≤f (c) for allxin [a, b]
NOTE 1 One of the most common misunderstandings of the extreme value theorem is illustrated by the following statement from a student’s exam paper: “The function is continuous, but since it is not defined on a closed, bounded interval, the extreme value theorem shows that there is no maximum.” The misunderstanding here is that, although the conditions of the theorem are sufficient, they certainly are notnecessaryfor the existence of an extreme point. In Problem 9, you are asked to study a function defined in an interval that is neither closed nor bounded, and moreover the function is not even continuous. Even so, it has both a maximum and a minimum.
The proof of the extreme value theorem is surprisingly difficult. Yet the result is not hard to believe. Imagine, for example, a cyclist going for a ride along some mountain roads. Since roads avoid going over cliffs, the height of the road above sea level is a continuous function of the distance travelled, as illustrated in Fig. 1. As that figure also shows, the trip must take the cyclist over some highest pointP, as well as through a lowest pointQ. (These points could also be at the start or finish of the ride.)
x
y P
Q
Figure 1
Searching for Maxima/Minima
Suppose we know that a functionf has a maximum and/or a minimum in some bounded intervalI. The optimum must occur either at an interior point ofI, or else at one of the end
S E C T I O N 8 . 4 / T H E E X T R E M E V A L U E T H E O R E M 271 points. If it occurs at an interior point (inside the intervalI) and iff is differentiable, then the derivativefis zero at that point. In addition, there is the possibility that the optimum occurs at a point wheref is not differentiable. Hence, every extreme point must belong to one of the following three different sets:
(a) Interior points inI wheref(x)=0 (b) End points ofI (if included inI)
(c) Interior points inI wherefdoes not exist
Points satisfying any one of these three conditions will be calledcandidate extreme points.
Whether they are actual extreme points depends on a careful comparison of function values, as explained below.
A typical example showing that a minimum can occur at a point of type (c) is shown in Fig. 8.1.3. However, most functions that economists study are differentiable everywhere.
The following recipe, therefore, covers most problems of interest.
Problem: Find the maximum and minimum values of a differentiable function f defined on a closed, bounded interval [a, b].
Solution:
(I) Find all stationary points off in(a, b)—that is, find all pointsx in(a, b) that satisfy the equationf(x)=0.
(II) Evaluatef at the end pointsaandbof the interval and also at all stationary points.
(III) The largest function value found in (II) is the maximum value, and the smallest function value is the minimum value off in [a, b].
(1)
A differentiable function is continuous, so the extreme value theorem assures us that max- imum and minimum points do exist. Following the procedure just given, we can, in principle, find these extreme points.
E X A M P L E 1 Find the maximum and minimum values for
f (x)=3x2−6x+5, x ∈[0,3]
Solution: The function is differentiable everywhere, andf(x) = 6x −6 = 6(x−1).
Hencex =1 is the only stationary point. The candidate extreme points are the end points 0 and 3, as well asx=1. We calculate the value off at these three points. The results are f (0)=5,f (3)=14, andf (1)=2. We conclude that the maximum value is 14, obtained atx =3, and the minimum value is 2 atx=1.
E X A M P L E 2 Find the maximum and minimum values of
f (x)= 14x4−56x3+12x2−1, x ∈[−1,3]
Solution: The function is differentiable everywhere, and f(x)=x3−52x2+x=x
x2−52x+1
Solving the quadratic equationx2− 52x +1 =0, we get the rootsx = 1/2 andx = 2.
Thusf(x)=0 forx=0, 1/2, and 2. These three points, together with the two end points
−1 and 3 of the interval [−1,3], constitute the five candidate extreme points. We find that f (−1)=7/12,f (0)= −1,f (1/2)= −185/192,f (2)= −5/3, andf (3)=5/4. Thus the maximum value off is 5/4 atx=3. The minimum value is−5/3 atx =2.
Note that it is unnecessary to study the sign variation off(x)or to use other tests such as second-order conditions in order to verify that we have found the maximum and minimum values.
In the two previous examples we had no trouble in finding the solutions to the equation f(x)=0. However, in some cases, finding all the solutions tof(x)=0 might constitute a formidable or even insuperable problem. For instance,
f (x)=x26−32x23−11x5−2x3−x+28, x ∈[−1,5]
is a continuous function, so it does have a maximum and a minimum in [−1,5]. Yet it is impossible to find any exact solution to the equationf(x)=0.
Difficulties of this kind are often encountered in practical optimization problems. In fact, only in very special cases can the equationf(x)=0 be solved exactly. Fortunately, there are standard numerical methods for use on a computer that in most cases will find points arbitrarily close to the actual solutions of such equations—see, for example, Newton’s method discussed in Section 7.10.
The Mean Value Theorem
This optional section deals with the mean value theorem, which is a principal tool for the precise demonstration of results in calculus.
Consider a functionf defined on an interval [a, b], and suppose that the graph off is connected and lacks corners, as illustrated in Fig. 2. Because the graph off joinsAtoBby a connected curve having a tangent at each point, it is geometrically plausible that for at least one value ofxbetween aandb, the tangent to the graph atxshould be parallel to the lineAB. In Fig. 2,x∗appears to be such a value ofx. The lineABhas slope [f (b)−f (a)]/(b−a). So the condition for the tangent line at(x∗, f (x∗))to be parallel to the lineABis thatf(x∗)=[f (b)−f (a)]/(b−a). In fact,x∗ can be chosen so that the vertical distance between the graph off andABis as large as possible.
The proof that follows is based on this fact.
S E C T I O N 8 . 4 / T H E E X T R E M E V A L U E T H E O R E M 273
y ⫽ f(x)
b ⫺ a
f(b) ⫺ f(a) y
a x A
B
b x*
Figure 2
T H E O R E M 8 . 4 . 2 ( T H E M E A N V A L U E T H E O R E M )
Iff is continuous in the closed bounded interval [a, b], and differentiable in the open interval(a, b), then there exists at least one interior pointx∗in(a, b)such that
f(x∗)= f (b)−f (a) b−a
(2)
Proof: According to the point–point formula, the straight line throughAandBin Fig. 2 has the equation
y−f (a)=f (b)−f (a) b−a (x−a) The function
g(x)=f (x)−f (a)−f (b)−f (a) b−a (x−a)
therefore measures the vertical distance between the graph off and the lineAB. Note that g(x)=f(x)−f (b)−f (a)
b−a (∗)
Obviously,g(a)=g(b)=0. The functiong(x)inherits fromf the properties of being continuous in [a, b] and differentiable in(a, b). By the extreme value theorem,g(x) has a maximum and a minimum over [a, b]. Becauseg(a) = g(b), at least one of these extreme pointsx∗must lie in (a, b). Theorem 8.1.1 tells us thatg(x∗)=0, and the conclusion follows from(∗).
E X A M P L E 3 Test the mean value theorem onf (x)=x3−xin [0,2].
Solution: We find that [f (2)−f (0)]/(2−0)=3 andf(x)=3x2−1. The equationf(x)=3 has two solutions,x= ±2√
3/3. Because the positive rootx∗=2√
3/3∈(0,2), we have f(x∗)=f (2)−f (0)
2−0 Thus, the mean value theorem is confirmed in this case.
NOTE 2 (Increasing and decreasing functions) Over any intervalI, iff (x2)≥f (x1)whenever x2 > x1, then in Section 6.3 we called the functionf increasinginI. Using the definition of the derivative, we see easily that iff (x)is increasing and differentiable, thenf(x) ≥ 0. The mean value theorem can be used to make this precise and to prove the converse. Letf be a function which is continuous in the intervalIand differentiable in the interior ofI(that is, at points other than the end points). Supposef(x)≥0 for allxin the interior ofI. Letx2> x1be two arbitrary numbers inI. According to the mean value theorem, there exists a numberx∗in(x1, x2)such that
f (x2)−f (x1)=f(x∗)(x2−x1) (∗) Because x2 > x1 andf(x∗) ≥ 0, it follows thatf (x2) ≥ f (x1), sof (x) is increasing. This proves (6.3.1). The equivalence in (6.3.2) can be proved by considering the condition for−f to be increasing. Finally, (6.3.3) involves bothfand−f being increasing. Alternatively it follows easily by using equation(∗).
NOTE 3 (Proof of Lagrange’s remainder formula (7.6.2)) We start by proving that the formula is correct forn=1. This means that we want to prove formula (7.6.4). Forx=0, define the function S(x)implicitly by the equation
f (x)=f (0)+f(0)x+12S(x)x2 (∗) If we can prove that there exists a numbercbetween 0 andxsuch thatS(x)=f(c), then (7.6.4) is established. Keepxfixed and define the functiongfor alltbetween 0 andxby
g(t )=f (x)−[f (t )+f(t )(x−t )+21S(x)(x−t )2] (∗∗) Then(∗)and(∗∗)imply thatg(0) =f (x)−[f (0)+f(0)x+12S(x)x2]=0 and thatg(x)= f (x)−[f (x)+0+0]=0. So, by the mean value theorem, there exists a numbercstrictly between 0 andxsuch thatg(c)=0. Differentiating(∗∗)with respect tot, withxfixed, we get
g(t )= −f(t )+f(t )−f(t )(x−t )+S(x)(x−t )
Thus,g(c) = −f(c)(x−c)+S(x)(x−c). Becauseg(c) = 0 and c = x, it follows that S(x)=f(c). Hence, we have proved (7.6.4).
The proof for the case whenn >1 is based on the same idea, generalizing(∗)and(∗∗)in the obvious way.
P R O B L E M S F O R S E C T I O N 8 . 4
1. Find the maximum and minimum and draw the graph of
f (x)=4x2−40x+80, x∈[0,8]
⊂SM⊃2. Find the maximum and minimum of each function over the indicated interval:
(a) f (x)= −2x−1, [0,3] (b) f (x)=x3−3x+8, [−1,2]
(c) f (x)=x2+1
x , [12,2] (d) f (x)=x5−5x3, [−1,√ 5 ] (e) f (x)=x3−4500x2+6ã106x, [0,3000]
S E C T I O N 8 . 4 / T H E E X T R E M E V A L U E T H E O R E M 275 3. Suppose the functiongis defined for allx∈[−1,2] byg(x)=15(ex2+e2−x2). Calculateg(x)
and find the extreme points ofg.
4. A sports club plans to charter a plane, and charge its members 10% commission on the price they pay to buy seats. That price is arranged by the charter company. The standard fare for each passenger is $800. For each additional person above 60, all travellers (including the first 60) get a discount of $10. The plane can take at most 80 passengers.
(a) How much commission is earned when there are 61, 70, 80, and 60+xpassengers?
(b) Find the number of passengers that maximizes the total commission earned by the sports club.
5. Let the functionf be defined forx∈[1, e3] by
f (x)=(lnx)3−2(lnx)2+lnx (a) Computef (e1/3),f (e2), andf (e3). Find the zeros off (x).
(b) Find the extreme points off.
(c) Show thatf defined over [e, e3] has an inverse functiongand determineg(2).
HARDER PROBLEMS
⊂SM⊃6. For the following functions determine all numbersx∗in the specified intervals such thatf(x∗)= [f (b)−f (a)]/(b−a):
(a) f (x)=x2 in [1,2] (b) f (x)=
1−x2 in [0,1]
(c) f (x)=2/x in [2,6] (d) f (x)=
9+x2 in [0,4]
7. You are supposed to sail from pointAin a lake to pointB. What does the mean value theorem have to say about your trip?
8. Is the functionf defined for allx∈[−1,1] by
f (x)= x forx∈(−1,1)
0 forx= −1 and forx=1 continuous? Doesf attain a maximum or minimum?
9. Letf be defined for allxin(0,∞)by
f (x)= x+1 forx∈(0,1]
1 forx∈(1,∞)
Prove thatf attains maximum and minimum values. Verify that neverthelessnoneof the con- ditions in the extreme value theorem is satisfied.