17.3 Accept on faith that the following familiar functions
are continuous on their domains:
sin(x), cos(x), ex, 2x, loge(x) for x > 0, xp for x > 0 [p any real number].
facts and theorems in this section to prove that the following functions are
Solution : (a) Notice that cos4(x) = (b ◦ a)(x), where a(x) = cos(x) and b(x) = x4.
a and b are continuous, so is cos4(x) by Theorem 17.5. Since 1 is also
17.4 says c(x) = 1 + cos4(x) is continuous. Finally, loge(1 + cos4(x))
= (d ◦ c)(x) where
d(x) = loge(x). Since d and c are continuous, Theorem 17.5 tells us that loge(1 + cos4(x))
(b) Well, a(x) = x2 and b(x) = sin(x) are continuous, so (a
◦ b)(x) = sin2(x) is
by Theorem 17.5. Similarly, c(x) = x6 and d(x) = cos(x) are continuous, so (c
◦ d)(x) =
cos6(x) is continuous by Theorem 17.5. Then Theorem 17.4 tells us that e(x) =
cos6(x) is continuous. Finally, is continuous, so Theorem 17.5 says (f
◦ e)(x) =
(c) Well, a(x) = x2 and b(x) = 2x are continuous, so by Theorem 17.5, (b
◦ a)(x) =
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
at any a ∈ R. For any ε > 0, taking δ = ε we get
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
continuous. By Theorem 17.4(ii),
By induction, then,
is continuous for every m
(b) By (a),
is continuous for any m
∈ N. Hence, by Theorem 17.3, a is
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, by the induction hypothesis,
is continuous. Hence, by Theorem 17.4(i),
Solution : (a) Since f is continuous, for any convergent sequence () in (a, b),
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
§11 Example 3.] But then lim f() = lim 0 = 0 so f() = 0. That is, f(x) = 0
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
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
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, 2x and 1
Theorem 17.4(i, ii)). Notice further that f(0) = -1 and f(1) = 1. Since -1 < 0
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],
17.5 says f(x + 1) is continuous on [0, 1]. Then Theorem 17.4(1) says g(x) is
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) =
Either g(1) = g(0) = 0, in which case x = 0 and y = 1 satisfy the statement we
prove, or g(1) and g(0) are on opposite sides of 0. In that case, Theorem 18.2
there is at least on point such that
At that point, f()
so letting x = and y =
+ 1 we get the conclusion.
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