SOLUTION OF HW8
MINGFENG ZHAO
December 04, 2011
1. [Page 186, Problem 2] Use Rolle’s theorem to prove that, regardless of the value of b, there is at
most one point x in the interval [−1, 1] for which x3 − 3x + b = 0.
Proof. Let f (x) = x3 − 3x + b for all x ∈ [−1, 1]. If x3 − 3x + b = 0 has more then one solution in
[−1, 1], so we can assume there exists some −1 ≤ x1 < x2 ≤ 1 such that f (x1 ) = f (x2 ) = 0. Then by
the Rolle’s Theorem, there exists some c ∈ (x1 , x2 ) ⊂ (−1, 1) such that f 0 (c) = 0. But we know that
f 0 (c) = 3c2 − 3 = 3(c2 − 1) < 0, because of −1 < c < 1, so we get a contradiction. Therefore, we can
conclude that there is at most one point x in the interval [−1, 1] for which x3 − 3x + b = 0.
2. [Page 187, Problem 8] Use the mean value theorem to deduce | sin x − sin y| ≤ |x − y|.
Proof. Let f (x) = sin x for all x ∈ R. For any x, y ∈ R and fix. If x = y, then | sin x − sin y| =
| sin x − sin x| = 0 ≤ |x − x| = |x − y|. If x < y, by the Lagrange’s Mean Value Theorem, we know
that there exists some z ∈ (x, y) such that f (x) − f (y) = sin x − sin y = f 0 (z)(x − y). Notice that
f 0 (z) = cos z, then we have
| sin x − sin y| = | cos z||x − y| ≤ |x − y|.
If x > y, by the Lagrange’s Mean Value Theorem, we know that there exists some z ∈ (y, x) such
that f (y) − f (x) = sin y − sin x = f 0 (z)(y − x). Notice that f 0 (z) = cos z, then we have
| sin x − sin y| = | sin y − sin x| = | cos z||x − y| ≤ |x − y|.
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MINGFENG ZHAO
In a summary, we can conclude that for all x, y ∈ R, we get
| sin x − sin y| ≤ |x − y|.
3. [Page 191, Problem 10] Let f (x) =
x2 − 4
.
x2 − 9
a. Find all points x such that f 0 (x) = 0.
b. Examine the sign of f 0 and determine those intervals in which f is monotonic.
c. Examine the sign of f 00 , and determine those intervals in which f 0 is monotonic.
d. Make a sketch of the graph of f .
Proof. a. Since f (x) =
x2 − 9 + 5
5
x2 − 4
=
=1+ 2
, then we have
x2 − 9
x2 − 9
x −9
f 0 (x)
= −
= −
(x2
5
· (2x)
− 9)2
10x
.
(x2 − 9)2
Let f 0 (x) = 0, then x = 0. So only x = 0 is the solution of the f 0 (x) = 0.
b. Since f 0 (x) = −
10x
, then f 0 (x) < 0 for all 0 < x < 3 and all x > 3; f 0 (x) > 0 for all x < −3
(x2 − 9)2
and all −3 < x < 0. So f is monotonic decreasing on (0, 3), and f is monotonic decreasing on (3, ∞);
f is monotonic increasing on (−∞, −3), and f is monotonic increasing on (−3, 0).
c. Since f 0 (x) = −
10x
, then we have
(x2 − 9)2
f 00 (x)
=
−
10(x2 − 9)2 − 10x · 2(x2 − 9) · (2x)
(x2 − 9)4
= −
10(x2 − 9)[(x2 − 9) − 4x2 ]
(x2 − 9)4
= −
10(x2 − 9)(−3x2 − 9)
(x2 − 9)2
=
10(x2 − 9)(3x2 + 9)
.
(x2 − 9)2
SOLUTION OF HW8
3
So we know that f 00 (x) > 0 if |x| > 3, and f 00 (x) < 0 if |x| < 3. Then f 0 is increasing on (−∞, −3),
f 0 is increasing on (3, ∞), f 0 is decreasing on (−3, 3).
d. The graph of f looks like:
4. [Page 195, Problem 16] Find the trapezoid of largest area that can be inscribed in a semicircle, the
lower base being on the diameter.
Proof. Assume the radius of the semicircle is R, and the heighth of the trapezoid is h, then the area
A(h) os the trapezoid should be:
A(h)
=
p
1
h[2R + 2 R2 − h2 ]
2
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MINGFENG ZHAO
= h[R +
p
R2 − h2 ],
for all 0 < h < R.
Then
A0 (h)
= R+
p
1
· (−2h)
R2 − h2 + h · √
2
2 R − h2
= R+
p
R2 − h2 − · √
=
=
h2
R2 − h2
√
R R2 − h2 + R2 − h2 − h2
√
R 2 − h2
√
R R2 − h2 + R2 − 2h2
√
, for all 0 < h < R.
R2 − h2
Assume A0 (h) = 0, then
R
p
R2 − h2 + R2 − 2h2 = 0.
That is,
R
If h <
R
√
2
p
R2 − h2 = 2h2 − R2 .
< R, then 2h2 − R2 < 0, so A0 (h) > 0. If h ≥
R
√
,
2
then 2h2 − R2 ≥ 0, so we have
R2 (R2 − h2 ) = (2h2 − R2 )2 .
That is,
R4 − R2 h2 = 4h4 − 4R2 h2 + R4 .
That is,
4h4 − 3R2 h2 = 0.
√
3R
2 .
So we get h =
A0 (h) > 0 if h <
√
3R
2 ,
Since when h <
R
√
,
2
and A0 (h) > 0 if h >
we know that A0 (h) > 0, so we can conclude that
√
3R
2 .
√
So when h =
3R
2 ,
we know that A(h) should be
the maximum point, that is,
√
A
3R
2
√
!
=

v
u
u
3R 
· R + tR2 −
2
√
3R
2
!2

√
 3 3R2
=
.

4
SOLUTION OF HW8
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Department of Mathematics, University of Connecticut, 196 Auditorium Road, Unit 3009, Storrs, CT
06269-3009
E-mail address: mingfeng.zhao@uconn.edu