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