Let f be defined on an interval I. Suppose that there exists m > 0 and a> 0 such that: If(x) – f(y)| < m|x – y|ª for allx, y € I (Such a function is said to satisfy a Lipschitz condition of order a on I. d) Prove that if g is differentiable on an interval I, and if g' is bounded on I, then g satisfies a Lipschitz condition of order 1 on I.

2. Let f: I R be a continuous function on a bounded, closed interval I= [a, b). Show that f has a minimum on I.

A function f : R → R is continuous at the point a ∈ R if (and only if)it satisfies the following condition:∀ > 0, ∃δ > 0, |x − a| < δ −→ |f(x) − f(a)| < .(The universe for all variables is R.)Prove that for all a, m, b ∈ R, the function f(x) = mx + b is continuous at a.Remark: A function f : R → R is continuous at the point a ∈ R if (and only if)limx→a f(x) = f(a).

1. Define a function f : Z x Z → Z by f(m, n) = 2n – 4m. (a) Prove that ƒ is not one-to-one. (b) Prove that the range of f is equal to the even numbers. In other words, show rng f = {2k|k E Z}, by using the usual methods for proving an equality of sets.

Let f : (0, 0) → R be the function with f(x) = |1 – In x|. Sketch the graph of y = f(x). Give a brief justification (one sentence each) why f is not injective and why f is not surjective. Come up with a subset A of R of your choice such that the function g : A →→ R with g(x) = |1 – In x| is injective. No justification needed. Come up with a subset B of R of your choice such that the function h : (0, ∞) → B with h(x) = |1 – In x| is surjective. No justification needed.

If f is a continuous function on R with f(1) > 0 and f(4) < 0, then there exists a number c between 1 and 4 such that f(c) = 0. Select one: OTrue O False Fi