Final Exam Solution - Analysis 1 | MATH 3001, Exams of Advanced Calculus

Download Final Exam Solution - Analysis 1 | MATH 3001 and more Exams Advanced Calculus in PDF only on Docsity! Math 3001 Final Exam Solutions 1. [20 points] Each of the following true/false questions is worth five points. You will only be graded on whether you write TRUE or FALSE; no other information will affect your grade. There are no tricks, but you must read the questions carefully. (a) If f : [a, b] → R is continuous on [a, b] and f(a) < d < f(b), then there is a c ∈ (a, b) such that f(c) = d. TRUE. This is just the Intermediate Value Theorem. (b) If a partition Q is a refinement of a partition P , then U(f,Q) ≤ U(f, P ). TRUE. Easy to see from a picture, or the general principle that the more points you have, the closer your upper sum should be to the actual integral (and thus the smaller it should be). (c) If lim x→∞ f(x) =∞, then lim x→∞ 1 f(x) = 0. TRUE. This was a homework problem. (d) If f is integrable on [a, b] and g is integrable on [c, d] with f([a, b]) ⊂ [c, d], then g ◦ f is integrable on [a, b]. FALSE. This was discussed in class; the theorem is only true if g is assumed to be continuous. If f is the modified Dirichlet function and g is the function g(0) = 0 and g(x) = 1 for x > 0, then f and g are both integrable but g◦f is the Dirichlet function which is not. 2. [30 points] (a) (12 points) State Rolle’s Theorem precisely, and prove it using anything from class aside from Rolle’s Theorem or the Mean Value Theorem. Solution: If f : [a, b] → R is continuous on [a, b] and differentiable on (a, b), and if f(a) = f(b), then there is a point c ∈ (a, b) such that f ′(c) = 0. To prove this, we observe that by the Extreme Value Theorem, f must attain its maximum and minimum on [a, b] (since it’s continuous on a closed and bounded interval). If max f(x) > f(a), then the maximum is attained at some c ∈ (a, b), and as showed in class, we must have f ′(c) = 0 at any interior maximum. Otherwise, if min f(x) < f(a) then there is a minimum attained at some c ∈ (a, b) and again f ′(c) = 0. Finally if max f(x) = min f(x) = f(a), then f must be constant and f ′(c) = 0 for all c ∈ (a, b). (b) (12 points) State precisely what the Fundamental Theorem of Calculus says about the derivative of F (x) = ∫ x 0 f(t) dt, and prove it using anything we proved in class aside from the Fundamental Theorem of Calculus. Solution: Suppose f is integrable on [a, b] and that f is continuous at some c ∈ (a, b). Then F ′(c) = f(c). 1 To prove this, let ε > 0 be any number, and choose δ > 0 such that |x−c| < δ implies |f(x)− f(c)| < ε. Then for any x ∈ (c− δ, c+ δ), we know∫ x c ( f(c)− ε ) dt ≤ ∫ x c f(t) dt ≤ ∫ x c ( f(c) + ε ) dt( f(c)− ε ) (x− c) ≤ F (x)− F (c) ≤ ( f(c) + ε ) (x− c) f(c)− ε ≤ F (x)− F (c) x− c ≤ f(c) + ε∣∣∣∣F (x)− F (c)x− c − f(c) ∣∣∣∣ ≤ ε. For any ε > 0 we can find a δ > 0 such that 0 < |x − c| < δ implies the above inequality, which is just the definition of derivative. So F ′(c) = f(c). (c) (6 points) If f is continuous on [0, 1] and ∫ 1 0 f(x) dx = 0, use the two Theorems above to show that there is a point c ∈ (0, 1) such that f(c) = 0. Solution: We know F (x) = ∫ x 0 f(t) dt is uniformly continuous on [0, 1] and that F ′(c) = f(c) for every c ∈ (0, 1), since f is continuous everywhere. Furthermore we have F (0) = 0 and F (1) = 0 by assumption, so that by Rolle’s Theorem there is a c ∈ (0, 1) such that F ′(c) = 0. In other words, f(c) = 0. 3. [30 points] Recall that a function f is Lipschitz on an interval [a, b] if there is a constant L ≥ 0 such that |f(x)− f(y)| ≤ L|x− y| for all x and y in [a, b]. (a) (5 points) Prove that if f is Lipschitz on [a, b], then f is uniformly continuous on [a, b]. Solution: Let ε > 0 be arbitrary. Choose δ = ε L ; then |x − y| < δ implies |f(x) − f(y)| ≤ L|x− y| < Lδ = ε. (b) (5 points) Give an example of a function that is uniformly continuous on some [a, b] but not Lipschitz on [a, b]. Solution: Let f(x) = √ x on [0, 1]. Then f is continuous on a compact set, so it must also be uniformly continuous. However the quantity f(x)− f(0) x− 0 = √ x x = 1√ x is not bounded by any number L. (c) (12 points) If f and g are both Lipschitz on [a, b] (for possibly different constants) and g is nowhere zero, prove that f/g is Lipschitz. Solution: Since g is Lipschitz, it is continuous on [a, b], and hence it must either always be positive or always negative (because if it could be both, then the Interme- diate Value Theorem implies it crosses the axis). Suppose without loss of generality that g is always positive. By the Extreme Value Theorem, g assumes its minimum at some point of [a, b], and that minimum is also positive. So there is a number m > 0 such that g(x) ≥ m. 2