Math 409-502Harold P. [email protected] interval is called compact if the interval is both closed and bounded.Examples• The interval [−1, 1] is compact.• The interval [0, 1) is not compact.(It is bounded but not closed.)• The interval [0, ∞) is not compact.(It is closed but not bounded.)The key property is that a sequence of points in a compact interval must have a cluster pointthat is still in the interval.Math 409-502 October 22, 2004 — slide #2Continuous functions on compact intervalsTheorem. If f is a continuous function on a compact interval,then the range of f is again a compact interval.The conclusion of the theorem has three parts.1. The range is an interval.2. The range is bounded.3. The range contains the endpoints of the interval.• The first part is the Intermediate Value Theorem.• The second part says that the function has a finite supremum and a finite infimum.• The third part says that the function attains a maximum value and a minimum value.Math 409-502 October 22, 2004 — slide #3Why did the chicken cross the road?Theorem (Intermediate Value Theorem). If f is a continuous function on the (compact) interval[a, b], then every number between f (a) and f (b) is in the range of f .ExampleShow that the function x5− 4x2+ exhas a zero between x = 0 and x = 1.Solution. At x = 0 the function has the value 1, and at x = 1 the function has the value−3 + e < 0. By the theorem, the function takes the intermediate value 0 somewhere in theinterval (0, 1).Using the bisection method, we could locate the zero of the function more precisely.Math 409-502 October 22, 2004 — slide #4Proof of the intermediate value theoremThe book gives a proof (page 173) using the Nested Interval Theorem. Here is a different proof.Without loss of generality, it may be assumed that f (a) < f (b).We need to show that if k is a number such that f (a) < k < f (b), then the number k is in therange of f .Let S denote the set of points c with the property that f (x) < k for all x in the interval [a, c].The set S is non-empty because f (a) < k and f is continuous. Let d denote the supremum of S.Claim: f (d) = k.Because f is continuous, f (d) = limn→ ∞f (d −1n), so f (d) ≤ k(by the limit location theorem). It cannot be that f (d) < k (strict inequality), because then therewould be points to the right of d in the set S (again by the continuity of f ). So the claim holds.Math 409-502 October 22, 2004 — slide #5Homework• Read sections 12.1 and 12.2, pages 172–177.• Do Exercises 12.1/1 and 12.1/5 on page 180.Math 409-502 October 22, 2004 — slide