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ASSIGNMENT 14 for SECTION 001 There are three parts to this assignment. Part A is to be completed online before 7:00 a.m. on Friday, March 4. Part B and Part C, which require full solutions, are to be handed in at the beginning of class on the same date. Part A [10 marks] This part of the assignment focuses on fundamental skills and computations. It can be found online, labelled A14, at webwork.elearning.ubc.ca — sign in using the MATH110 001 2010W button. Part B [5 marks] This part of the assignment is drawn directly from the course texts. It focuses on mathematical exposition; you will be graded primarily on the clarity and elegance of your solutions. From the Calculus: Early Transcendentals text, complete questions 30 and 34 from section 4.6. Part C [15 marks] This part of the assignment consists of more challenging questions. You are expected to provide full solutions with complete arguments and justifications. 1. Let f (x) = ex . (a) The linear approximation L(x) to f (x) at 0 has two criteria: (i) L(0) = f (0), and (ii) L0 (0) = f 0 (0). Find the quadratic approximation Q(x) = a0 + a1 x + a2 x2 to f (x) at 0, using the following criteria: (i) Q(0) = f (0), (ii) Q0 (0) = f 0 (0), and (iii) Q00 (0) = f 00 (0). (b) Consider the “infinite polynomial” M (x) = a0 + a1 x + a2 x2 + a3 x3 + a4 x4 + · · · . Find the constants a0 , a1 , a2 , . . . such that M (0) = f (0) and every derivative of M at 0 is equal to the corresponding derivative of f at 0. 2. An ellipse centred on the origin is described by the equation x2 y2 + = 1. a2 b2 Prove that if the normal line at every point on the ellipse passes through the origin, then the ellipse is a circle. 3. In this question you will explore why the hypotheses of the Mean Value Theorem are necessary. (a) Give an example of a function f which is continuous on a closed interval [a, b], and such that there is no c ∈ (a, b) satisfying f (b) − f (a) . f 0 (c) = b−a (b) Give an example of a function g which is defined on a closed interval [a, b], differentiable on an open interval (a, b), and such that there is no c ∈ (a, b) satisfying g 0 (c) = g(b) − g(a) . b−a