5.6 *Exponentials, Powers, and Logarithms 283Theorem 5.6.11 (Exponential Functions, I) For a > 1, the exponential
function f(x) = ax is a continuous, strictly increasing function on JR, with
range (0, +oo).
Proof. Let a > 1 and define f ( x) = ax. All that remains to prove is that f
is continuous everywhere on JR and has range (0, +oo). Continuity follows from
Theorem 5.6.10. Since f is continuous everywhere on JR, and JR is an interval,
the range off must be an interva l, by Theorem 5.3.8. By Corollary 5.6.7, this
range must be (0,+oo). •
We get a similar, but different, result when 0 < a < 1.Theorem 5.6.12 (Exponential Functions, II) For 0 < a < 1, the expo-
nential function f(x) = ax is a continuous, strictly decreasing function on JR
with range (0, +oo).
Proof. Exercise 7.Lemma 5.6.13 Suppose a> 0 and x E JR. Then ax = 1 {::}a= 1 or x = 0.
Proof. Part 1 (-¢=:):By Definition 5.6.5, ix= 1, and by Exercise 5.5. 15 ,
aO = 1.
Part 2 (::::? ): Suppose ax = 1 and a i:-1. The exponential function f (t) =at
is 1-1, by Theorems 5.6.11 and 5.6.12. Thus,
ax = 1 ::::? ax = a^0 ::::? X = 0.
Therefore, if ax = 1, then either a= 1 or x = 0. •
CONSTANT POWER FUNCTIONS
Theorem 5.6.14 (Positive Power Functions) Let t > 0 be fixed. The
power function f(x) = xt, defined with the help of 5.6.5, is positive, strictly in-
creasing, and continuous everywhere on (0 , +oo), lim xt = +oo, and lim xt = 0.
x-+oo x-+O+
Proof. Let t > 0 be fixed throughout the proof that follows.
(a) By Definition 5.6.5, the power function f(x) = xt is defined everywhere
on (0, +oo), and positive there.
(b) To show that f is strictly increasing on (0, +oo), suppose x < y in
(0, +oo). Then, :3 rational numbers q 1 , Q2 such that
x < Q1 < Q2 < y.
Let { tn} be a sequence of positive rational numbers converging tot. By Theorem
5.6.10, xtn ---+ xt and ytn ---+ yt. Since tn is rational, the function gn(x) = xt,. is
strictly increasing on (0, +oo), so
xtn < qin < q~" < ytn.