17.3 Accept on faith that the following familiar functions
are continuous on their domains:

sin(x), cos(x), e^{x}, 2^{x}, log_{e}(x) for x > 0, x^{p} for x > 0 [p any real number].
Use these

facts and theorems in this section to prove that the following functions are
also continuous.

Solution : (a) Notice that cos^{4}(x) = (b ◦ a)(x), where a(x) = cos(x) and b(x) = x^{4}.
Since

a and b are continuous, so is cos^{4}(x) by Theorem 17.5. Since 1 is also
continuous, Theorem

17.4 says c(x) = 1 + cos^{4}(x) is continuous. Finally, log_{e}(1 + cos^{4}(x))
= (d ◦ c)(x) where

d(x) = log _{e}(x). Since d and c are continuous, Theorem 17.5 tells us that log_{e}(1 + cos^{4}(x))

is continuous.

(b) Well, a(x) = x^{2} and b(x) = sin(x) are continuous, so (a
◦ b)(x) = sin^{2}(x) is
continuous

by Theorem 17.5. Similarly, c(x) = x^{6} and d(x) = cos(x) are continuous, so (c
◦ d)(x) =

cos^{6}(x) is continuous by Theorem 17.5. Then Theorem 17.4 tells us that e(x) =
sin^{2}(x) +

cos^{6}(x) is continuous. Finally, is continuous, so Theorem 17.5 says (f
◦ e)(x) =

is continuous.

(c) Well, a(x) = x^{2} and b(x) = 2^{x} are continuous, so by Theorem 17.5, (b
◦ a)(x) =
is

continuous.

17.5 (a) Prove that if m ∈ N, then the function
is continuous on R.

(b) Prove that every polynomial function

is continuous on R.

Proof: [of (a)] By induction on m. For m = 1, we need to show that p(x) = x is
continuous

at any a ∈ R. For any ε > 0, taking δ = ε we get
Hence,
p is

continuous on R.

Now assume that
is continuous and show that
is continuous. But
=

. We showed above that x is continuous, and the induction hypothesis is that
is

continuous. By Theorem 17.4(ii),
is continuous.

By induction, then,
is continuous for every m
∈ N.

(b) By (a),
is continuous for any m
∈ N. Hence, by Theorem 17.3, a is
continuous

By 17.4(i) and induction, any polynomial is continuous .

Formally, we prove by induction that
is continuous for

any k. The case k = 1 was done in the first paragraph. Now assume that

is continuous and show that
is continuous.

But

By (a),
is continuous, by the induction hypothesis,

is continuous. Hence, by Theorem 17.4(i),

is continuous.

Therefore, any polynomial function
is continuous

on R.

17.9 (a) Prove that the following function is continuous at
by verifying the

property of Theorem 17.2.

Solution : Scratch work: We need
So we need

Now if
is small, say
then
-1 < x < 3, so 1 < x + 2 < 5.
Hence,

if
then
< 5 and so. So we want

or 1, whichever is smaller.

Proof: Given ε > 0 let δ =
Then

Hence, f(x) = x^{2} is continuous at
= 2.

17.12 (a) Let f be a continuous real -valued function with domain (a, b). Show

that if f(r) = 0 for each rational number in (a, b), then f(x) = 0 for all

x ∈ (a, b).

(b) Let f and g be continuous real - valued functions on (a, b) such that

f(r) = g(r) for each rational number r in (a, b). Prove that f(x) = g(x)

for all x ∈ (a, b).

Solution : (a) Since f is continuous, for any convergent sequence () in (a, b),
converging

to in (a, b), we have lim f() = f(). Now let
be any number in (a, b).
There is a

sequence of rational numbers () in (a, b) converging to
. [By the denseness
of Q, see

ยง11 Example 3.] But then lim f() = lim 0 = 0 so f() = 0. That is, f(x) = 0
for any

x ∈ (a, b).

(b) Consider h(x) = f(x) - g(x) on (a, b). Then h(r) = 0 for all rational numbers
r ∈ (a, b).

By part (a), h(x) = 0 for all x ∈ (a, b), so f(x) = g(x) for all x ∈ (a, b).

18.5 (a) Let f and g be continuous function on [a, b] such that f(a)
≥ g(a) and

f(b) ≤ g(b). Prove that f() = g() for at least one
in [a, b].

(b) Show that Example 1 can be viewed as a special case of part (a).

Solution: (a) Since f and g are continuous on [a, b], h(x) = f(x)
-g(x) is also
continuous

on [a, b]. Further, since f(a) ≥ g(a) we have h(a) ≥ 0, and since f(b) ≤ g(b), we have

h(b) ≤ 0. So by Theorem 18.2, there is at least one point
∈ [a, b] such that
h() = 0.

At that point, f() = g().

(b) Example 1 says that any continuous function f : [0, 1]
→ [0,1] has a xed
point. We

obtain this statement from (a) by letting g(x) = x.

18.7 Prove that for some x in (0, 1).

Proof: Let f(x) = Notice that f is continuous on R, since x, 2^{x} and 1
are (using

Theorem 17.4(i, ii)). Notice further that f(0) = -1 and f(1) = 1. Since -1 < 0
< 1,

Theorem 18.2 says there is at least one x ∈ (0, 1) where f(x) = 0. For that x,.

18.10 Suppose that f is continuous on [0, 2] and that f(0) = f(2). Prove that

there exist x, y in [0, 2] such that and f(x) = f(y).

Proof: Consider g(x) = f(x + 1) - f(x). Since x + 1 is continuous on [0, 2],
Theorem

17.5 says f(x + 1) is continuous on [0, 1]. Then Theorem 17.4(1) says g(x) is
continuous

on [0, 1].

Notice that g(0)+g(1) = f(1) - f(0)+f(2) - f(1) = f(2) - f(0) = 0, so g(1) =
- g(0).

Either g(1) = g(0) = 0, in which case x = 0 and y = 1 satisfy the statement we
are to

prove, or g(1) and g(0) are on opposite sides of 0. In that case, Theorem 18.2
implies that

there is at least on point such that
At that point, f()
= f(+1),

so letting x = and y =
+ 1 we get the conclusion.