Section 4.1 Maximal and Minimal Points of Functions Theory: Optimization problems are of the following form: find the point a in the domain of a function y = f(x), for which the function value f(x) is maximal or minimal. Of course most problems in “real life” depend on more than one variable. Here we will only consider functions in one variable x. Functions depending on more than one variables will be dealt with in further courses. Definition 1 Let y = f(x) be a function defined on a domain D (subset of the real line). A point c in D is called the absolute maximum on D, if f(x) ≤ f(c) for all x ∈ D. In that case f(c) is called the maximum value of f(x) on D. A point c ∈ D is called the absolute minimum on D, if f(x) ≥ f(c), for all x ∈ D. In that case f(c) is called the minimal value of f(x) on D. Absolute maxima and minima are called absolute extreme points of f.
Note: not every function on a given domain might have an absolute max- imum or minimum. Here are some examples:
1 Example. The Function f(x) = x does neither have a maximum nor a minumum on (0, ∞). It has a minimum on, say (0, 2] (namely at c = 2), and it has a minimum and a maximum on [1, 2] or on any other closed interval [a, b], where 0 < a < b. y
x
1 2
Example. Consider the function defined on [0, 2] by 2x if 0 ≤ x < 1, f(x) = 0 if x = 1, 2x − 4 if 1 < x ≤ 2. Then f(x) is bounded, but does neither have a maximum nor a minimum on [0, 1]. y
◦
• x
◦
; 3
Example. Consider the function defined on f(x) = x2 restricted to the open interval (−2, 2). Then f(x) has no maximal value on the open interval (−2, 2). y
◦ ◦
x 4
Theorem 1 Let y = f(x) be a continuous function on a closed bounded interval [a, b]. Then f has a maximum and a minimum. Thus there exists a number c ∈ [a, b] and a number d ∈ [a, b] so that for all x ∈ [a, b] f(c) ≤ f(x) ≤ f(d).
y
• max
• min [ ] x a b
Theorem 1 seems to be obvious, nevertheless a formal proof of it goes beyond the scope of this class (will be covered in Math409). But note that the conclusion of Theorem 1 would be wrong if we restricted ourselves only to continuous functions on rational numbers: Example. Consider the function y = |2 − x2| defined on the closed interval [0, 4]. This is a continuous function on a closed interval.√ So by the Extreme Value Theorem it has a minimum, namely at c = 2. Now if we think f being a function only defined on the rational numbers of the interval, say, [0, 4], then there is no rational number c for which f(c) ≤ f(x) for all rational numbers x in [0, 4]. 5
We now turn to the problem to find the extreme points of a continuous function y = f(x) defined on a closed interval [a, b].
Fermat’s Theorem:
Theorem 2: Let y = f(x) be a continuous function on a closed interval [a, b] and let c be an extreme point, and assume that it is in the open interval (a, b). Then: Either f(x) is not differentiable at c, Or it is differentiable at c, in which case f 0(c) = 0
Proof of Theorem 2. Assume that f(x) has an absolute maximum at some point c, a < c < b, and assume that f 0(c) exists. We need to show that f 0(c) = 0. Recall that f(x) − f(c) f(x) − f(c) f(x) − f(c) f 0(c) = lim = lim = lim . x→c x − c x→c− x − c x→c+ x − c Then for all a < x < c, we have f(x) − f(c) ≤ 0 (since c is absolute max), and thus (since x − c < 0) we have f(x) − f(c) ≥ 0, for all a < x < c. x − c It follows that f(x) − f(c) lim ≥ 0. x→c− x − c On the other hand for all c < x < b, have also have f(x) − f(c) ≤ 0 (since c is absolute max) and thus (since x − c > 0) we have f(x) − f(c) ≤ 0, for all c < x < b, x − c from which we deduce that f(x) − f(c) lim ≤ 0. x→c+ x − c So since f(x) − f(c) f 0(c) = lim ≥ 0, x→c− x − c but also f(x) − f(c) f 0(c) = lim ≤ 0, x→c+ x − c f 0(c) must have the value 0. To show the that f 0(c) = 0, if c is an absolute minimum and if f 0(c) exists, can be done similarly (try it!). 6
Fermat’s Theorem helps us to find the extreme points as follows: We are given a continuous function y = f(x) on an closed interval [a, b]. 1) Find all points c in the open interval (a, b) for which either f 0(c) does not exist, or for which f 0(c) exists and f 0(c) = 0. Let’s say we found finitely many points c1, c2, . . . , cn. We call these points critical points. 2) So by Theorem 2 the maximum and the minimum must either be one of the end points a or b, or it must be one of the critical points. Therefore we need to compare the values f(a), f(b), f(c1), f(c2), . . . , f(cn), to find out which is the largest, or smallest, respec- tively.
y
[ | | | | | ] x a c1 c2 c3 c4 c5 b
Critical points: c1, c2, c3, c4, c5 Extreme values on [a, b]: c2 (absolute minimum), b (absolute maximum). 7
Definition 2 Let y = f(x) be a function on a domain D. We call a number c in D a local maximum of f, if there is an interval [a, b], with a < c < b, so that c is an absolute maximum if f is restricted to [a, b], and we call it local minimum of f, if there is an interval [a, b], with a < c < b, so that c is an absolute minimum if f is restricted to [a, b], The local maxima and minima are called local extrema.
Remark. Local extrema must be critical points by Theorem 1. But not every critical point is a local extreme point. In above picture c1, c2, c3 and c4 are local extrema, but c5 is not an extreme point. Also f(x) = x3 has a critical point at c = 0 (since f 0(0) = 3 · 02 = 0), but it is not a local extreme point. 8 Problems:
Problem 1. Find the absolute extreme points of a) f(x) = x3 − 3x2 + 3x, on [−1, 3].
b) f(x) = |4 − x2| on [−3, 3].
2 c) f(x) = x 5 (5 − x) on [−1, 2]. 9
Problem 2. Find the absolute extrema of √ √ √ f(x) = 3x2 + 2 cos(x2), on I = [ π/2, π].