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Assignment #10
Morgan Schreffler
1 of 2
Page 107
Exercise 1 Define f : (− π2 , π2 ) → R by f (0) = 0, and f (x) =
x−sin(x)
1−cos(x)
if x 6= 0.
(a)Show that f is continuous at x = 0.
(b) Show that f is strictly increasing, and that the image of the interval is all of R.
Proof. (a) To prove that f is continuous at 0, it will suffice to use L’Hôpital’s Rule to
show that limx→0− f (x) = limx→0+ f (x) = f (0) = 0. Let g(x) = x − sin(x), and h(x) =
1 − cos(x). Then g and h are both twice differentiable for all x ∈ R, where g 0 (x) = 1 − cos(x),
g 00 (x) = sin(x), h0 (x) = sin(x), and h00 (x) = cos(x). Now, notice that for each x ∈ (− π2 , 0),
h(x) 6= 0 6= h0 (x) 6= 0 6= h00 (x), lim− g(x) = 0 = lim− h(x), and lim− g 0 (x) = 0 = lim− h0 (x),
x→0
x→0
x→0
x→0
so by two iterations of L’Hôpital’s Rule ,
lim−
x→0
Now,
sin(x)
cos(x)
x − sin(x)
1 − cos(x)
sin(x)
= lim−
= lim−
.
x→0 cos(x)
1 − cos(x) x→0
sin(x)
is continuous at x = 0, and
sin(0)
cos(0)
lim−
x→0
= 0, so
x − sin(x)
= 0.
1 − cos(x)
A similar argument on (0, π2 ) gives us
lim+
x→0
(b) Note that since f (−x) =
(−x)−sin(−x)
1−cos(−x)
x − sin(x)
= 0.
1 − cos(x)
=
−(x−sin(x))
1−cos(x)
= −f (x), f is an odd function. Now,
let A = (−a, a) ⊆ R, and let ω : A → R be an odd function. Then for α, β ∈ A, if α < β ≤ 0
implies ω(α) < ω(β), then ω(−β) = −ω(β) < −ω(α) = ω(−α). In short, if ω is strictly
increasing on (−a, 0], then it is increasing on [0, a) and thus (−a, a). Hence, it suffices to show
that f is strictly increasing on the interval (− π2 , 0]. Consider the numerator g(x) = x−sin(x)
and the denominator h(x) = 1 − cos(x). Note that on (− π2 , 0), g 0 (x) = 1 − cos(x) > 0, so by
Page 104 Exercise 5, g is strictly increasing. Further, h0 (x) = sin(x) < 0, so h is decreasing.
As a result, f =
g
h
is increasing on (− π2 , 0). Finally, note that f (0) = 0, and if c < 0,
f (c) < 0 = f (0), so f is strictly increasing on (− π2 , 0]. To prove that the image of (−2π, 2π)
is R, note that limx→2π− f (x) = +∞ and limx→−2π+ f (x) = −∞, and since f is increasing
on this interval, f ((−π, π)) = (−∞, ∞) = R.
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Assignment #10
Morgan Schreffler
2 of 2
Page 112
Exercise 1 (a)Suppose f is a continuous, real-valued function on the interval [a, b]. Prove
Rb
that there exists c ∈ [a, b] such that a f (x)dx = f (c)(b − a).
Rx
Proof. Let F (x) = a f (t)dt. Now, F is continuous and differentiable on [a, b], so by the
Mean Value Theorem, there is a c ∈ (a, b) such that
F (b)−F (a)
b−a
= F 0 (c) or, equivalently,
F (b) − F (a) = F 0 (c)(b − a). But, by the Fundamental Theorem of Calculus, F (b) − F (a) =
Rb
f (x)dx, and by differentiation of the integral, F 0 (c) = f (c), so we get
a
Z
b
f (x)dx = f (c)(b − a),
a
thus completing the proof.
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