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Math Homework #7 Solutions

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

Solution : (a) Notice that cos4(x) = (b ◦ a)(x), where a(x) = cos(x) and b(x) = x4. Since
a and b are continuous, so is cos4(x) by Theorem 17.5. Since 1 is also continuous, Theorem
17.4 says c(x) = 1 + cos4(x) is continuous. Finally, loge(1 + cos4(x)) = (d ◦ c)(x) where
d(x) = log e(x). Since d and c are continuous, Theorem 17.5 tells us that loge(1 + cos4(x))
is continuous.

(b) Well, a(x) = x2 and b(x) = sin(x) are continuous, so (a ◦ b)(x) = sin2(x) is continuous
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) = sin2(x) +
cos6(x) is continuous. Finally, is continuous, so Theorem 17.5 says (f ◦ e)(x) =
is continuous.

(c) Well, a(x) = x2 and b(x) = 2x are continuous, so by Theorem 17.5, (b ◦ a)(x) = is

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.
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) = x2 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, 2x 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.

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