We use the notation and assumptions of Theorem 1.42. If e > 0, then /(b) - /(a) <{K + e){b-a).
Proof of Theorem 1.42 from Lemma 1.45. Since f(b) — f(a) < (K +
e)(b - a) for all e > 0, it follows that f(b) - f(a) <K(b-a).

Continuing inductively, we can find a sequence of pairs of points an and bn such that
n an)i
o>n-i < a,n < bn < bn-\,
and bn-an = (bn-i - an_i)/2,
for all n > 1.

Let U be the open interval (a, (3) or the real line Ж. If f : U —> M is differentiable with f'(t) = 0 for all t 6 U, then f is constant.
Proof. Let b, a 6 U with b > a. By applying Theorem 1.42 to / with
К = 0 we see that f(b) — /(a) < 0. By applying Theorem 1.42 to —/ with
К = 0, we see that /(a) — f(b) < 0. Thus /(a) = f(b). But a and b were
arbitrary, so / is constant.
In section 1.1 we noted the importance of the following simple corollary.

Let U be the open interval (a,(3) or the real line Ж. Suppose that a, b 6 U and b > a. If g : U —>M. is differentiable with g'(t) > 0 for all t 6 U then
g(b)-g(a)>0.