(
Iiapler I
Iiijiits
(rcos
=
0,
[xI
(x
3)(x
31
-
3
—
— 9
— 27
—--
-
4)
0)
rsin 0)
ift3
ift=3
desired c from [0, /2] to [0, 71j4]. There is nothing stopping us from selecting the
midpoint of [0, r/4j and evaluating fat that point, thereby narrowing even further
the interval containing c.This process could be continued indefinitely until we find
that c is in a sufficiently small interval. This method of zeroing in on a solution is
called the bisection method, and we will study it further in Section 3.7.
The Intermediate Value Theorem can also lead to some surprising results.
91
Use the Intermediate Value Theorem to show that on a circu
• EXAMPLE
lar wire ring [here are always two points opposite from each other with the same
temperature.
=
=
T(r,0)
—
T(—r, 0)
T(r, 0)
T(—r,0)
—
=
—[T(r, 0)
—
T(—r, 0)]
=
—f(0)
T(r cos 0, r sin 0) — T(r cos(0 + jr), r sin(0 + ar))
SOLUTION Choose coordinates for this problem so that the center of the ring is
the origin, and let r be the radius of the ring. (See Figure 12.) Define T(x, y) to be
the temperature at the point (x, y). Consider a diameter of the circle that makes an
angle 0 with the x-axis, and define f(0) to be the temperature difference between
the points that make angles of 0 and 0 + ; that is,
f(0)
f(r)
f(0)
With this definition
=
=
2. g(x)
4. g(t)
=
=
=
6. hi(t)
f(s)
8. g(t)
10.
x2
It
f(c).
-
9
-
21
t —3
—
—
2!
7x
x—3
=
=
=
=
It
J
t
—
—
3
—
—
—
—
123
It
13
112
1(3
27
3
9
1)2
I—3x + 7
1—2
ift3
ift=3
if t
3
if r > 3
ift
3
if t > 3
if x
3
if s>3
16. From the graph of g (see Figure 13), indicate the values
where g is discontinuous. For each of these values state whether g
is continuous from the right, left, or neither.
15. f(s)
14. f(t)
13. f(t)
12. r(t)
4. The Intermediate Value Theorem says that if a functionf
is continuous on [a, b] and W is a number between f(a) and f(b),
such that
and
then there is a number c between
Thus, either f(0) and f(r) are both zero, or one is positive and the other is nega
tive. If both are zero, then we have found the required two points. Otherwise, we
can apply the Intermediate Value Theorem. Assuming that temperature varies
continuously, we conclude that there exists a c between 0 and ir such that
f (c) 0. Thus, for the two points at the angles c and c + r, the temperatures are
the same.
=
is discontinuous it
3. A function f is said to be continuous on a closed interval
and
bJ if it is continuous at every point of (a, b) and if
2. The function f(x)
1. A functionfis continuous atc if
oncepts Review
gure 12
,
roblem Set 1.6
=
=
It
,
x—3
ti
I
1—3
27
3)I
!‘roblems 1—15, state whether the indicated function is continu—
If it is not continuous, tell why.
lx at3.
1. f(s)
3. h(x)
=
=
I
5. h(t)=——
7. f(t)
9. h(x)
II. r(t)
1.
2
I9ure13
___
.t
I
—6
I
—4
—2
2-
Figurel4
II
2
I
4
I
6
H
—‘r
lr
II)
I!. 1mm the graph of h given in Figure 14, indicate the inter
which h is continuous.
‘ii
Iil.
/(x)
‘0. g(0)
1(x)
=
=
=
=
=
=
=
=
=
x—7
—49
2
r
sin 0
+ 2x
2
x + I
3
—_____________
(x
—
—
—
30)(x
3x+7
x + 3x
—
2
r
7T
3
19. f(x)
23. F(x)
21. H(t)
< 0
—
ifx
I
s
sin 0 + cos 01
I.r
if 0
< 0
ifs>
12—s
ifs
[1]
I
‘ii
=
=
=
=
2x2_18
sinx__—
x +
tan 0
I
c/u+1
[t +
satisfies all the fol
=
=
f that
35. g(t)
ifs > 1
if0x1
(x2
2
s
31. G(x)
29
27. r(O)
!‘robli’ms 24—35, at what points, if any, are the functions
i’nuniious?
22.
—
In I’, i/bins 18—23, the given function is not defined at a certain
iii I/ow should it be deflned in order to make it continuous at
html! (Sc’ Example 1.)
it
‘hi
ft
I’
24. J(x)
25.
a
26. h(0)
‘8
.0. F(x)
32. f(s)
f(t)
33. g(x)=—s
34.
36. Sketch the graph of a function
Imwing conditions.
1) Its domain is [—2, 2].
(h) f(-2) = f(-l) = f(l) = f(2) = 1.
(c)
It is discontinuous at —1 and I.
Id) It is right continuous at —1 and left continuous at 1.
37. Sketch the graph of a function that has domain [0,21 and
is continuous on [0,2) but not on [0,2].
61
and
38. Sketch the graph of a function that has domain [0,61 and
s continuous on [0, 2] and (2, 61 but is not continuous on [0, 6].
39. Sketch the graph of a function that has domain [0,
is continuous on (0,6) but not on [0,61.
40. Let
•
=
( x
1
—
s
ifs is rational
ifs is iational
Section 1.6 Continuity of Functions
f(s)
89
Sketch the graph of this lunction as best you can and decide
where it is continuous.
=
0
x0
x=0
=
. 1
sIn—; c
=
——J;c
= 5
2—V’r
0
10
In ProhlL’!ns 41—48, deter,nine whet/icr the function is cOlitinuous
the given point c. If the fu,mcrion is not continuous, th’ler,nint
whether the discontinuity is removable or nonremovable.
at
5
j-———,
I sin
sinx;c
42. f(s)
=
41. f(s)
44. 5
9
f(x)=;
c=0
=
48. f(x)=
46. F(x)
43. f(x)=’;c=0
5
45. g(s)
s
10,
47. 1
f(x)=sin
;
c=0
x
49. A cell phone company charges $0.12 for connecting a call
plus $0.08 per minute or any part thereof (e.g., a phone call last
ing 2 minutes and 5 seconds costs $0.12 + 3 X $0.08). Sketch a
graph of the cost of making a call as a function of the length of
time i that the call lasts. Discuss the continuity of this function.
mile and $0.20
50. A rental car company charges $20 for one day, allowing
up to 200 miles. lor each additional 100 miles, or any fraction
thereof, the company charges $18. Sketch a graph of the cost for
renting a car for one day as a function of the miles driven. Discuss
the continuity of this function.
51. A cab company charges $2.50 for the first
for each additional mile. Sketch a graph of the cost of a cal, ride
as a function of the number of miles driven. Discuss the con
tinuity of this function.
—
52. Use the Intermediate Value Theorem to prove that
+ 3x
2 = 0 has a real solution between 0 and 1.
—
53. Use the Intermediate Value Theorem to prove that
(cos t)t
3 + 6 sin
5t
3 = 0 has a real solution between Oand 2
7T.
—
—
—
—
54. Use the Intermediate Value Theorem to show that
2 + 14s
7r
8 = 0 has at least one solution in the interval
[0, 5]. Sketch the graph of)’ =
2 + 14s
7s
8 over [0, 51
How many solutions does this equation really have’?
—
—
‘
if and
75 + 14
at c
—
only
if
0 has at
55. Use the Intermediate Value Theorem to show that
cos s = 0 has a solution between 0 and /2. Zoom in on
the graph of y =
cos x to find an interval having length
0.1 that contains this solution.
56. Show that the equation 5
x +
least one real solution.
57. Prove that f is continuous
limf(c + t) = f(c).
I —‘(I
—
58. Prove that if f is continuous at c and f(c) > 0 there is an
interval (c
ö, c + ) such that f(s) > 0 on this interval.
—
59. Prove that if f is continuous on [0, 11 and satisfies
0 Sf f(s)
I there, then f has a fixed point; that is, there is a
number c in [0, 1] such that f(c) = c. Ilimit: Apply the Intermedi
ate Value Theorem to g(x) = s
f(s).
90
Chapter 1 Limits
oj ci
eve rvs here.
60. jind the values
ContinuOUs
ifs < I
jfj s<2
ifs2
and b so that ijie jojjowini’ lunction is
+ 1
3s
f(x)=as+b
=
Then f(—2)
=
= —
and f(2)
=
1.
61. A stretched ejastic string covers the intervaj [0, jj. The
ends are released and the string contracts so that it covers the
interval [a, bj, a
(3, /
I. Prove that this results in at least
one point 01 the string being where it was originally. See Prob—
1cm 50.
62. Let f(s)
1(c)
Does the Intermediate Value Theorem imply the existence of a
0? Explain.
number c between —2 and 2 such that
63. Starting at 4 AM., a jiiker slowjy cjimhed to the top of a
mountain, arriving at noon. The next (lay, he returned along the
same path, starting at 5 i\.M. and getting to the bottom at 11 AM.
Show that it some point along the path his watch showed the
same time on both days.
64. Let D he a bounded. hut otherwise arbitrary, region in the
first quadrant. Given an angle 0. 0
(1
/2, D can be circum
scribed b a rectangle whose base makes angle 0 with the s—axis
,‘
as shown in Figure 15. Prove that at some angle this rectangle is a
square. (This means that any hounded region can he circum—
sen bed b a square.)
Figure 15
=
GMmr
-
L, themi Jim f(s)
L.
alisu’i’l:
r
-
R
to cach of tin’ follomi’ing a,sSi’rtlon.c. Be
it
-,
,,itr
R
GtWni
<R
65. The gravitational t’orce exerted by the earth on an object
having niass in that is a distance Iromn the center ot the earth is
=
=
L.
=
=
f(s)
L. then [(c)
juvtif’ your
I of e
1.7 Chapter Review
10
i4’i!/i true or
Concepts Test
Rc,
m
1
ood
‘/,aIc’cl
Ill’
1. If f(c)
2. II urn
=
0, then for every
i:
> 0 there exists a 6 > 0
3. If 1mm f(s) exists. then f(c) exists.
4. lfliinf(s)
I-leie ( is the gravitational constant, Al is the mass of the earth,
and 1? is the earth’s radius, Is g a continuous function of r?
66. Suppose that f is continuous on [ci, bj and it is never zero
there. Is it possible that fchanges sign on [ci, b]? Explain.
67. Let f(s + y) = f(s) + f(y) for all x andy and suppose
that f is continuous at x = 0.
(a) Prove thatf is continuous everywhere.
(b) Prove that there is a constant m such that f(t) = ml for all
(see Problem 43 of Section 0.5).
68. Prove that if f(s) is a continuous function on an interval
V’(f(x))2.
then so is the function f(s)
69. Show that if g(s) = If(s) is continuous it is not neces—
sirily true that f(s) is continuous.
70. Let f(s)
0 if x is irrational and let f(s) = j/q if s is
the rational number p/q in reduced form (q > 0).
(a) Sketch (as best you can) the graph off on (0,1).
(h) Show that f is continuous at each irrational number in (0, 1),
but is discontinuous at each rational number in (0, 1).
M
I
-c
lim
I)
position
W
f(s)
=
=
I
,i4
f(s)
=
f(b)
1. lLrn .1(s)
‘(‘4
j’’l
—I
f’(ci): lirn
Fl fill posi
=
I
—
x
—
25
2. ever’ inte
iOfl
71. A thin equilateral triangular block of side length 1 unit
has its face in the vertical sy-plane with a vertex V at the origin.
Under the influence of gravity, it will rotate about V until a side
hits the s-axis floor (Figure 16). Lets denote the initial s-coordi
nate of the midpoint Al of the side opposite I, and 1et f(s) de
note the final s-coordinate of this point. Assume that the block
balances when M is directly above V.
(a) Determine the domain and range off:
(h) Where on this domain isfdiscontinuous?
(c) Identify any fixed points off(see Problem 59).
-f
—i
Initial
Figure 16
3.
Answers to Concepts Review:
ger
4. ci: b; [(c)
p(c).
v
5. If f(c) is undefined, then lirn f(s) does not exist.
=
6. The coordinates of the hole in the eraph of
are (a. 10).
7. If p(s) is a polynomial, then urn p(s)
I
i
II. The function f(s)
‘-iv meal miumber.
=
2 sin
2s
—
=
f’
is contin
cos s is continuous at
12. 1ff is continuous at c, then f(c) exists.
=
=
f(c) t’or all c
=
2.3.
s
is continuous at s
c.
=
has two vertical asymptotes.
f
f is continuous at
f(s). then
s). then
[s/2 is continuous at s
c.
tan s has many horizontal asymptotes.
2 is a horizontal asymptote of the graph ot
2.
=
B for
f(s)
jim f(s)
B.
[0,41, then Jim f(s) exists.
13. 1ff’ is continuous on the interval (1, 3), then
us it 2.
14. 1ff’ is continuous on
IS. 1ff’ is a continuous function such that A
ill v. then jim f(s) exists and it satisfies A
sins
-
5
j
=
16. 1ff is continuous on (ci. b) themi jim f(s)
6).
iii (ci,
‘-
17. limit
-t
v
=
18. If the line
f(s), then
=
19. The graph of y
=
—
=
=
=
f( Jim
v
.1
-t
20. The graph of
I
21. jim —s-—
f(s)
22. If lim f(s)
23. If tim
24. The function f(s)
25. If lLrn f(s) = f’(2) > 0, then f(s) < 1 .OOlf’(2) for all s
mm some interval containing 2.
f(s)
f(s)
=
all s’ind
IbI.
lim f(s)
=
=
Al.
0.
exists, then
hm g(s).
Al. then L
=
all s. then lint f(s)
L and Jim
2 + 2s
3v
1 for all s. then Jim f(s)
26. If jim [f(s) + g(s)j exists, then urn f(s) and Jim g(s)
both exist.
27. If 0
=
tor
tor
6, then hmI/’(s)
g(s)
28. If jim f(s)
29. If f(s)
=
30. If f(s) < It)
Hmf(s) < 10.
31. If urn i(s)
2. jim
4. jim
—‘2
6. limn
urn
I
8. im
12.
-,
—
+ 1
—
—
—
+
1
1
I
i
z —6
24
5
—
I
js_-:1I
I{4x]
y-
z
u+1
(1
state i/mat
it doe.r not
32. If I’ is continuous and positive on [a, bj, then I/f must
issurne every value between t/f(a) and 1/f(b).
Sample Test Problems
+ 2
2
u-’i
hi I’,’oblenzs 1—22, fiimd i/ic’ i,icljcatecl li,,mit or
s
,s—2
1. 1mm
.v—’2
—
—
112 — 1
3. jim
u1 a
I
—2/s
5. limi-i
r—’2
—4
tans
7. jim
.v—()
sin
2s
s—4
-
9. 1mm
11.
I
14. jim
sin 5s
15. jim
I)
3s
4
(t
2)
tan 2s
—
17. lim
x—’’°X + 2
t+2
lim
19. jim
21.
‘4
jim f(s)
=
1
s
(s3
,
—
s
16.
jim
._.5
.0*
18. 1mm
-
20. lim
1
—
sin!
X
cos2s
is
COS 1
5
j + sin x
22. jim—
(b)
jim f(s)
lim f(s)
-
(d)
=
7.
Section 1.7 Chapter Review
ifs
ij.v <—I
if —1 < 5 < 1
23. Prove using an e— argument that lim(2s + 1)
24. Let f(s)
Find each value.
-‘
(a) f(i)
(c)
to make it continuous there?
=
X
Jim
=
L
-
91
—2 and if g is contimiLmous
(d)
-
-
j()
Ig(s)_-
lini
v=
(f)
g(’(s))
(b) hm g(s)
3 and lint g(s)
Sg()
4g(s)j
—
f’(.v)
25. Refer to f of Problem 24. (a) What are the values of s at
which fis discontinuous? (h) How should f’he detined at 5 = —I
-
3, find each value.
=
26. (live the i:—
6 definition in each case.
urn g(u) = Al
(b)
=
27. If urn f(s)
5
(a)
at
(a) lirn [2f(x)
jim fs)
(c) g(3)
(e)
f(2)
=
f(4)
=
f(6)
=
2.
=
3.
2.
28. Sketch the graph of a function f’that satisfies all the fol
=
Its domain is [0. (ij.
lowing conditiomis.
(i)
(Ii) f(0)
f(s)
1
ifs0
itt) < s < I
ifs
f is continuous everywhere.
‘is + 6
1—
jim
(c)
=
1 and
jim f(s)
f is continuous except at s
=
(d)
29. Let f(s)
11
Determine a amid 6 so that
=
2 and
,‘
s
r+l
=
3.
32. g(s)
=
=
=
34. (/(s)
36. lI(s)
SiiiL
horizon
30. Use the Intermediate Value Theorem to prove that the
s — 43 — 3s + 1 = 0 his at least one solution
5
between s
ecjuation
=
———
s-—I
tan 2s
s—4
In Problems 31—36, f’iimcl time ec/i(at!ons of’cd/ vertical and
tal asymptotes for the giveil fiazciioim.
31. f(s)
=
=
33. F(s)
35. /m(.v)