Homework 8 Solutions

Homework 8 Solutions Problem 1. Find the maximum value of f (x) = x3 − 9x in the interval [−3, 3]. Note: No derivatives! Solution. This was done in class. Problem 2. Find an integer n such that the polynomial equation x3 − x + 3 = 0 has a solution between n and n + 1. Solution. We have (−2)3 − (−2) + 3 = −3 and (−1)3 − (−1) + 3 = 1, so there is a solution to the equation x3 − x + 3 between n = −2 and n + 1 = −1. Problem 3. Prove that there is some number x such that sin x = x − 1. Solution. Let f (x) = x − 1 − sin x. Then f (0) = −1 < 0 and f (π/2) = π/2 > 0, so there is x in (0, π/2) such that f (x) = 0, or sin x = x − 1. Problem 4. (i) Suppose that f is continuous on the interval [0, 1] and that 0 ≤ f (x) ≤ 1 for all x in [0, 1]. Prove that f (x) = x for some number x in [0, 1]. (ii) Let f be continuous and bounded above and below on R. Prove that there is some number x such that f (x) = x. Solution. If f (0) = 0 or if f (1) = 1, then we are done. If not, then f (0) > 0 and f (1) < 1. Let g(x) = x − f (x). Then g is continuous on [0, 1], g(0) = −f (0) < 0 and g(1) = 1 − f (1) > 0. By the intermediate Value Theorem there is x in [0, 1] such that g(x) = 0, or f (x) = x. (ii) If f is bounded below and above on R, then there are numbers a and b such that a < f (x) < b for all x. The continuous function g(x) = x − f (x) satisfies g(a) < 0 < g(b). Apply the Intermediate Value Theorem to g on [a, b]. Problem 5. A function f defined on an interval I has the Intermediate Value Property on I if for any two numbers a < b in I and every y strictly between f (a) and f (b), there is c in (a, b) such that f (c) = y. (i) Prove that the function f given by f (x) = sin 1/x if x 6= 0 and f (0) = 0 has the Intermediate Value Property on the interval [0, B], for any B > 0. (ii) Prove that if f is non decreasing on the interval I and has the Intermediate Value Property on I, then f is continuous on I. (Recall that f is said to be non decreasing on I if f (x) ≤ f (y) whenever x < y in I.) Solution. (i) If 0 < a < b are two numbers in [0, B], then we apply the Intermediate Value Theorem to f (x) = sin 1/x on the interval [a, b] because f is continuous on [a, b]. If 0 = a < b and x is strictly between 0 = f (0) and f (b), let n be a natural number such that 2/b < (2n + 1)π so that the interval J = [2/(2n + 3)π, 2/(2n + 1)π] is contained in [0, b]. The function f (x) = sin 1/x is continuous on J and takes on the values 1 and −1 at the endpoints of J. Since −1 ≤ f (x) ≤ 1, the Intermediate Value Theorem applied to f on J implies that given any y such that −1 < y < 1, there is c in J such that f (c) = y. In particular, if y is strictly between f (a) and f (b), then −1 < y < 1 also, and c in J satisfies 0 = a < c < b, as desired. (ii) Suppose that there is a in I where f fails to be continuous. Then there is a sequence (xn ) in I such that xn → a but f (xn ) does not converge to f (a). We may assume, by taking a subsequence if necessary, that xn increases (or decreases) to a. Then f (xn ) is non decreasing and bounded above by f (a), thus it converges to a number p with p < f (a). Let q be a number such that p < q < f (a). For each xn we have f (xn ) ≤ p < q, so the intermediate value property of f on the interval [xn , a] implies the existence of yn in (xn , a) such that f (yn ) = q. The sequence (yn ) converges to a and f (yn ) = q for all n. Since xn also converges to a, given n there is m such that yn < xm , but f (yn ) = q > p ≥ f (xm ), contradicting that f is non decreasing.