Mathematics 242 – Principles of Analysis Midterm Exam 3 December 2, 2005

Directions

Do all work in the blue exam booklet. There are 100 possible regular points and 10 possible Extra Credit points (see question IV below).

I. All parts of this question refer to the function f(x) = x^{2} − 6x + 3. A) (15) Show using the ε, δ definition that

lim_{x→1} f(x) = −2.

B) (15) Show that f(x) is integrable on [3, 4] using any applicable method. (Don’t work

too hard!) II. A) (10) State the Intermediate Value Theorem. B) (10) Let f(x) = a_{1} + a_{2} cos(x), where a_{1}, a_{2} are constants with 0 < a_{1} < a_{2}. Show

that there exists some c ∈ [0, π] such that f(c) = 0.

III. (20) Show that if f is continuous on [a, b] and s = sup f([a, b]), then there exists c ∈ [a, b] where f(c) = s. (Note: This is the portion of the Extreme Value Theorem showing that f attains a maximum.)

IV. True-False. Say whether each of the following statements is true or false, and give a detailed reason for your assertion. Do any three parts. If you submit solutions for all four, then I will consider the other one for Extra Credit.

All parts refer to the function f defined by:

f(x) = \{ \begin{matrix}x^{2} & if x ∈ Q \\ 0 & if x ∉ Q \end{matrix}

A) (10) lim_{x→3} f(x) exists. B) (10) f is differentiable at x = 0. C) (10) f integrable on [1, 2]. D) (10) The conclusion of the Extreme Value Theorem is true for f on the interval

[−\sqrt{2}, \sqrt{2}].

(Note: The conclusion of theorem phrased as “if p, then q” is the statement q; the hypotheses are the statement(s) p. The conclusion of a theorem can be true in a particular example even if the hypotheses are not satisfied.)