Some Consequences Of Continuity
Section 2.5
Difference Equations
to
Differential Equations
Some Consequences
Of Continuity
In this section we consider two properties of functions which are very closely connected
to the notion of continuity. The first of these, the Intermediate Value Theorem, says that
the graph of a continuous function is a connected continuum in the sense of our normal
intuition. That is, the theorem states that as a continuous function changes from one value
to another, it must take on every intermediate value. The second theorem, the Extreme
Value Theorem, says that a continuous function on a closed interval attains a maximum
and a minimum value on that interval. This is related to our intuitive notion that if we
draw a continuous curve with definite beginning and ending points, then the curve has
a point where it is higher than at any other point and a point where it is lower than
at any other point. We shall not attempt formal justifications of these theorems; such
justifications require inquiries into the subtleties of real numbers which are best left to
more advanced courses.
We will begin with a statement of the Intermediate Value Theorem, followed by a
consideration of its application to solving equations.
Intermediate Value Theorem If f is a continuous function on a closed interval [a, b]
and m is any number between f (a) and f (b), then there is a number c in the interval [a, b]
such that f (c) = m.
Example Since f (t) = sin(t) is continuous on [0, π2 ] with f (0) = 0 and f ( π2 ) = 1, the
5
fact that 0 < 2π
< 1 guarantees that there is a number c in [0, π2 ] such that
f (c) =
5
.
2π
Graphically, the situation is as in Figure 2.5.1. Of course, the theorem tells us neither
the value of c nor how we might find it. The Intermediate Value Theorem is an existence
theorem; it guarantees the existence of a certain value, but does not directly provide any
method for calculating the value.
Example Suppose f (t) is the height, in inches, of a certain plant t days after it first
emerges from the soil. From our knowledge of how plants grow, it would be reasonable
to assume that f is a continuous function. Also, we have f (0) = 0. Now if f (10) = 12,
then we know, for example, that there is some time c, 0 < c < 10, such that f (c) = 5.
Of course, this is not surprising, and, in fact, we did not have to bring the subject of
continuous functions into the problem in order to realize that between the time when the
plant was 0 inches tall and the time when it was 12 inches there was a time when it
was 5 inches tall. However, the point of an example like this is to emphasize that the
Intermediate Value Theorem simply states a property that we should expect continuous
1
Copyright c by Dan Sloughter 2000
2
Some Consequences Of Continuity
Section 2.5
1
5
2π
0.8
0.6
0.4
0.2
0.25
0.5
0.75
c 1
1.25
1.5
5
Figure 2.5.1 Intermediate Value Theorem: sin(c) = 2π
functions to have if they are to be used as mathematical models of real world processes
that undergo continuous change.
As a special case, the Intermediate Value Theorem tells us that if f is a continuous
function on a closed interval [a, b] with f (a) and f (b) having opposite signs (that is, one
is negative and the other positive), then there is a point c in the open interval (a, b)
where f (c) = 0. In other words, under these conditions, the Intermediate Value Theorem
guarantees that the equation f (x) = 0. has at least one solution in [a, b]. Although the
theorem does not provide a method for solving the equation, it does provides a basis for
constructing an algorithm for approximating a solution to any desired accuracy.
Bisection Algorithm Suppose f is continuous on [a1 , b1 ] and f (a1 )f (b1 ) < 0 (an easy
way to check that f (a1 ) and f (b1 ) have opposite signs). Then, as above, the equation
f (x) = 0
(2.5.1)
has at least one solution in [a1 , b1 ]. Let
m1 =
a1 + b1
.
2
If f (m1 ) = 0, then we have found a solution to (2.5.1). If f (m1 ) 6= 0, then either
f (a1 )f (m1 ) < 0,
in which case (2.5.1) has a solution in [a1 , m1 ], or
f (m1 )f (b1 ) < 0,
in which case (2.5.1) has a solution in [m1 , b1 ]. In the first case, let a2 = a1 and b2 = m1 ;
in the second case, let a2 = m1 and b2 = b1 . Then
m2 =
a2 + b2
2
(2.5.2)
Section 2.5
Some Consequences Of Continuity
3
will approximate a solution to (2.5.1) with an error less than
b2 − a2
.
2
(2.5.3)
Proceed in a the same manner to define an , bn , and mn for n = 3, 4, 5, . . .. That is, if we
have found an−1 , bn−1 , and mn−1 , and f (mn−1 ) 6= 0, let
an = an−1 and bn = mn−1 if f (an−1 )f (mn−1 ) < 0
(2.5.4)
an = mn−1 and bn = bn−1 if f (mn−1 )f (bn−1 ) < 0.
(2.5.5)
and
Then
an + bn
2
will approximate a solution of (2.5.1) with an error less than
mn =
bn − an
.
2
(2.5.6)
(2.5.7)
Repeat the procedure as many times as necessary to obtain the desired level of accuracy.
Example
Suppose we wish to find a root to the equation
x5 + x = 1.
(2.5.8)
First note that solving (2.5.8) is equivalent to solving
x5 + x − 1 = 0.
(2.5.9)
Letting
f (x) = x5 + x − 1,
we may write (2.5.9) as f (x) = 0. To find an initial interval [a1 , b1 ], we graph f as in
Figure 2.5.2. Noting that f (0) = −1 and f (1) = 1, we may start with a1 = 0 and b1 = 1.
That is, (2.5.8) has a solution in the [0, 1]. Then
m1 =
0+1
= 0.5.
2
Now f (0.5) = −0.468750, so f (0.5)f (1) < 0. Hence a2 = 0.5, b2 = 1, and
m2 =
0.5 + 1.0
= 0.75.
2
Now f (0.75) = −0.012695, so f (0.75)f (1) < 0. Hence a3 = 0.75, b3 = 1, and
m3 =
0.75 + 1.00
= 0.875.
2
4
Some Consequences Of Continuity
Section 2.5
6
4
2
-3
-2
1
-1
2
3
-2
-4
-6
Figure 2.5.2 Graph of f (x) = x5 + x − 1
At this stage we know that 0.875 is an approximation for a solution to (2.5.8) with an
error of no more than
1.00 − 0.75
= 0.125.
2
We may continue in this manner until we attain any desired level of accuracy. The following
table gives the values of an and bn for n = 1, 2, 3, . . . , 10.
an
0.000000000
0.500000000
0.750000000
0.750000000
0.750000000
0.750000000
0.750000000
0.750000000
0.753906250
0.753906250
bn
1.000000000
1.000000000
1.000000000
0.875000000
0.812500000
0.781250000
0.765625000
0.757812500
0.757812500
0.755859380
mn
0.500000000
0.750000000
0.875000000
0.812500000
0.781250000
0.765625000
0.757812500
0.753906250
0.755859380
0.754882815
f (an )
−1.000000000
−0.468750000
−0.012695300
−0.012695300
−0.012695300
−0.012695300
−0.012695300
−0.012695300
−0.002544540
−0.002544540
f (bn )
f (mn )
1.000000000 −0.468750000
1.000000000 −0.012695300
1.000000000
0.387909000
0.387909000
0.166593000
0.166593000
0.072288300
0.072288300
0.028700600
0.028700600
0.007736990
0.007736990 −0.002544540
0.007736990
0.002579770
0.002579770
Rounding to three decimal places, we see that x = 0.755 approximates a solution of (2.5.8)
with an error of no more than, to three decimal places,
0.756 − 0.754
= 0.001.
2
In Section 3.6 we will discuss another method, called Newton’s method, for approximating a solution to an equation of the form f (x) = 0. At that time we will see that
Newton’s method is faster than the bisection algorithm. However, we will also see that
there are conditions under which Newton’s method will fail, whereas the bisection algorithm will always work.
Section 2.5
Some Consequences Of Continuity
5
10
8
6
4
2
-2
-1
1
2
3
4
Figure 2.5.3 Graph of f (x) = x2 on [−1, 3]
We know turn to the Extreme Value Theorem and some of its consequences.
Extreme Value Theorem If f is a continuous function on a closed interval [a, b], then
there exists a point c in [a, b] such that f (c) ≥ f (x) for all values of x in [a, b]. Similarly,
there exists a point d in [a, b] such that f (d) ≤ f (x) for all values of x in [a, b].
In other words, using the notation of the statement of the theorem, f (c) is the maximum value attained by f on [a, b] and f (d) is the minimum value attained by f on [a, b].
As with the Intermediate Value Theorem, this is an existence theorem which does not
indicate any method for finding the points c and d. The importance of the theorem lies in
the fact that it gives conditions under which maximum and minimum values of a function
are guaranteed to exist. Optimization problems, that is, problems concerned with finding
the maximum and minimum values of functions, occur frequently in mathematics and in
the applications of mathematics. As we shall see in Section 3.8, conditions which guarantee
the existence of a solution to an optimization problem, such as those given in the Extreme
Value Theorem, are often an important first step in solving such problems.
Example Consider f (x) = x2 on the interval [−1, 3]. Since f is a continuous function on
this closed interval, the Extreme Value Theorem guarantees the existence of a maximum
value and a minimum value for f . In fact, from our knowledge of the behavior of this
function, in particular that f (0) = 0, f (x) > 0 if x 6= 0, and f (x) > f (y) if |x| > |y|, it is
easy to see that f (x) attains its maximum value when x = 3 and its minimum value when
x = 0 (see Figure 2.5.3). Hence the maximum value of f on [−1, 3] is 9 when x = 3 and
the minimum value is 0 when x = 0.
Example Let A, B, and C be constants with A > 0. Suppose we wish to find the
minimum value of the quadratic polynomial
f (x) = Ax2 + Bx + C
on an interval [a, b]. Completing the square, we may rewrite f as
(2.5.10)
6
Some Consequences Of Continuity
Section 2.5
f (x) = Ax2 + Bx + C
C
B
2
=A x + x+
A
A
!
2
B
B2
C
=A
x+
−
+
2A
4A2
A
2
B
B2
=A x+
+C −
.
2A
4A
2
B 2
is minimized. This latter
Since C − B
4A is a constant, f (x) is minimized when A x + 2A
term is never negative and is minimized when it is 0, that is, when
x+
B
= 0.
2A
Hence the minimum value of f (x) on [a, b] will occur when
x=−
B
,
2A
(2.5.11)
unless this point is not in the interval, in which case the minimum value occurs at one of
the endpoints, x = a or x = b. Note that, geometrically, (2.5.11) is the location of the
vertex of the parabola which is the graph of f . Note that if A < 0, then the maximum
value of f (x) would occur at (2.5.11) if it is in the interval [a, b], and at one of the endpoints
otherwise.
y
x
Figure 2.5.4 A field of length x and width y
Example Suppose we wish to fence in a rectangular field with 500 yards of fencing in
such a way that we maximize the area of the resulting field. If, as in Figure 2.5.4, we let
x denote the length of the field, y its width, and A its area, then
A = xy.
Moreover, since we only have 500 yards of fencing to work with, we know that
2x + 2y = 500.
Section 2.5
Some Consequences Of Continuity
7
16000
14000
12000
10000
8000
6000
4000
2000
50
100
150
200
250
Figure 2.5.5 Graph of A = 250x − x2 on [0, 25]
Hence
y = 250 − x,
(2.5.12)
from which it follows that
A = xy = x(250 − x) = 250x − x2 .
From (2.5.12), and the fact that we must have both x ≥ 0 and y ≥ 0, it follows that
0 ≤ x ≤ 250. Thus our problem becomes one of finding the maximum value of
A = −x2 + 250x
on the closed interval [0, 250]. From our previous example, the maximum value of A will
occur when
250
250
=
= 125.
x=−
(2)(−1)
2
From (2.5.12), we have y = 125 when x = 125. Hence the area of the field is maximized
when its dimensions are 125 yards by 125 yards. For these dimensions, the area of the field
is
A x=125 = (125)(125) = 15, 625 square yards.
See Figure 2.5.5 for the graph of A.
Example
Consider the function f (x) = x2 + 1 on the open interval (0, 1). Then
lim f (x) = lim (x2 + 1) = 1
x→0+
x→0+
and
lim f (x) = lim (x2 + 1) = 2,
x→1−
x→1−
but 0 < f (x) < 2 for all values of x in (0, 1). Hence, as x approaches 0 from the right,
f (x) approaches, but never reaches, 1; similarly, as x approaches 1 from the left, f (x)
8
Some Consequences Of Continuity
Section 2.5
approaches, but never reaches, 2. Thus f is an example of a continuous function on an
open interval which attains neither a maximum nor a minimum value on the interval.
Hence we see why the interval in the statement of the Extreme Value Theorem must be a
closed interval.
Problems
1. Use the bisection algorithm to approximate a solution to each of the following equations
on the given interval. Your answer should have an error of no more than 0.005.
(a) x2 − 2 = 0 on [0, 4]
(c) cos(x) = x on [0, π]
(b) x5 − 6x3 + 2x = 2 on [−1, 1]
√
(d) 2 sin(x) = x + 1 on [0, 2]
2. (a) Plot the graph of g(t) = t2 − cos2 (t) on [−π, π].
(b) How many solutions are there to the equation t2 = cos2 (t)?
(c) Use the bisection algorithm to estimate the solutions to the equation t2 = cos2 (t).
State your answers with an error less than 0.005.
3. Suppose if the market price for a certain product is p dollars, then the demand for
that product will be
50000p + 10000
units.
D(p) =
p2
At the same time, suppose that at a price of p dollars producers will be willing to
supply
1
S(p) = p2 + 2p units.
3
(a) Plot the graphs of D and S on the same graph.
(b) Use the bisection algorithm to estimate the solution to the equation
D(p) = S(p).
This point is called the equilibrium price because it is the price for which the
consumers’ demand for the product is exactly equal to the manufacturers’ supply.
(c) How many units of the product will be manufactured at the equilibrium price?
(d) What would happen if the producers raised the price above the equilibrium price?
What would happen if they lowered the price below the equilibrium price?
(e) What would happen if the producers increased production? What would happen
if they lowered production?
4. A farmer wishes to fence in a rectangular field, using a straight river for one side, with
500 yards of fencing. What should the dimensions of the field be in order to maximize
the area of the field?
5. When a potter sells his pots for p dollars apiece, he can sell D(p) = 750 − 50p of them.
Suppose the pots cost him $5.00 apiece to make. What price should the potter charge
in order to maximize his profit?
Section 2.5
Some Consequences Of Continuity
9
6. Let h(t) = t4 − 1.
(a) Does h have a maximum value on [−1, 2)?
(b) Does h have a minimum value on [−1, 2)?
(c) Are the results of (a) and (b) consistent with the Extreme Value Theorem? Explain.
7. Recall that the Heavidside function is defined by
H(t) =
0, if t < 0,
1, if t ≥ 0.
(a) Note that H(−1) = 0 and H(1) = 1. Is there a point c in (−1, 1) such that
H(c) = 0.5?
(b) Is the result of (a) consistent with the Intermediate Value Theorem?
(c) Does H attain a maximum value on [−1, 1]? Does H attain a minimum value on
[−1, 1]?
(d) Are the results of (c) consistent with the Extreme Value Theorem?
8. Suppose g is defined on [−1, 1] by
g(t) =
|t|, if t 6= 0,
1, if t = 0.
(a) Does g attain a maximum value on [−1, 1]? If so, at what points?
(b) Does g attain a minimum value on [−1, 1]? If so, at what points?
(c) Are the results of (a) and (b) consistent with the Extreme Value Theorem?
9. Suppose f and g are continuous on [0, 1], f (0) < g(0) and f (1) > g(1). Show that
there exists a point c in the open interval (0, 1) such that f (c) = g(c).
10. Suppose f is continuous on [0, 1] and 0 ≤ f (x) ≤ 1 for all x in [0, 1]. Show that there
exists a point c in [0, 1] such that f (c) = c.